Sec 2 G1 Maths Tutor Bukit Timah

Reader note: These are original explanatory exercises for later-stage G1 progression, not official exam papers or immediate Secondary Two syllabus predictions. Programme availability, grades, tutor assignment and subject-level progression require school and centre confirmation. The existing small-group programme reference remains unchanged.

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

Official references and editorial note

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

Official references and editorial note

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

Day one: practise negative squares and bring simple equations into zero-equals form.
Day two: identify a, b and c in five varied equations.
Day three: solve two exact-root cases from the formula and verify both answers.
Day four: examine one repeated root, one negative radicand and one irrational pair.
Day five: solve a new problem and explain the corresponding graph’s intercept behaviour.

Parents can ask: ‘What is b, including its sign?’ ‘Have you put the entire top of the fraction in brackets?’ and ‘Have you checked both candidates?’ These prompts make the mathematical decisions visible. If a child cannot yet solve 3x+5=20 reliably, return to one-variable balance before extending to quadratic techniques.

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

The MOE 2020 Secondary Two G1 outline focuses on linear equations in one variable. The full 2027 K110 graduating syllabus includes quadratic equations by formula. Schools set the current teaching and assessment scope; this article is later-stage preparation.

Does 2027 K110 supply the formula?

Yes. The published 2027 K110 Mathematical Formulae section includes the quadratic-equation formula. Check official requirements for the student’s own eventual examination year.

What if b²−4ac is negative?

There are no real roots. The associated quadratic graph has no real x-intercept.

Why can the formula return two roots?

The ± yields two possible numerator values when the radicand is positive. Each should be checked in the original equation.

Can one root be a fraction or a negative number?

Yes. Roots depend on the signed coefficients. Do not reject a result merely because it is not a positive whole number; substitute it back to check.

Will a 90-minute group guarantee success?

No. The appropriate progression depends on prerequisite knowledge, feedback and independent retention. Ask the tutor how readiness and later transfer are assessed.

Official references and editorial note

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

Day one: practise negative squares and bring simple equations into zero-equals form.
Day two: identify a, b and c in five varied equations.
Day three: solve two exact-root cases from the formula and verify both answers.
Day four: examine one repeated root, one negative radicand and one irrational pair.
Day five: solve a new problem and explain the corresponding graph’s intercept behaviour.

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

Yes. The published 2027 K110 Mathematical Formulae section includes the quadratic-equation formula. Check official requirements for the student’s own eventual examination year.

What if b²−4ac is negative?

There are no real roots. The associated quadratic graph has no real x-intercept.

Why can the formula return two roots?

The ± yields two possible numerator values when the radicand is positive. Each should be checked in the original equation.

Can one root be a fraction or a negative number?

Yes. Roots depend on the signed coefficients. Do not reject a result merely because it is not a positive whole number; substitute it back to check.

Will a 90-minute group guarantee success?

No. The appropriate progression depends on prerequisite knowledge, feedback and independent retention. Ask the tutor how readiness and later transfer are assessed.

Official references and editorial note

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

10 minutes: review signed multiplication, roots and simple equation balance.
20 minutes: introduce the formula using the official sheet and practise coefficient lists.
25 minutes: solve a few exact-root questions, including both ± branches.
15 minutes: distinguish repeated-root and no-real-root cases.
10 minutes: independently solve one changed equation with no visible model.
10 minutes: correct the first sign or bracket error and assign a short later retest.

A five-session parent–student routine

Day one: practise negative squares and bring simple equations into zero-equals form.
Day two: identify a, b and c in five varied equations.
Day three: solve two exact-root cases from the formula and verify both answers.
Day four: examine one repeated root, one negative radicand and one irrational pair.
Day five: solve a new problem and explain the corresponding graph’s intercept behaviour.

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

Yes. The published 2027 K110 Mathematical Formulae section includes the quadratic-equation formula. Check official requirements for the student’s own eventual examination year.

What if b²−4ac is negative?

There are no real roots. The associated quadratic graph has no real x-intercept.

Why can the formula return two roots?

The ± yields two possible numerator values when the radicand is positive. Each should be checked in the original equation.

Can one root be a fraction or a negative number?

Yes. Roots depend on the signed coefficients. Do not reject a result merely because it is not a positive whole number; substitute it back to check.

Will a 90-minute group guarantee success?

No. The appropriate progression depends on prerequisite knowledge, feedback and independent retention. Ask the tutor how readiness and later transfer are assessed.

Official references and editorial note

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

10 minutes: review signed multiplication, roots and simple equation balance.
20 minutes: introduce the formula using the official sheet and practise coefficient lists.
25 minutes: solve a few exact-root questions, including both ± branches.
15 minutes: distinguish repeated-root and no-real-root cases.
10 minutes: independently solve one changed equation with no visible model.
10 minutes: correct the first sign or bracket error and assign a short later retest.

A five-session parent–student routine

Day one: practise negative squares and bring simple equations into zero-equals form.
Day two: identify a, b and c in five varied equations.
Day three: solve two exact-root cases from the formula and verify both answers.
Day four: examine one repeated root, one negative radicand and one irrational pair.
Day five: solve a new problem and explain the corresponding graph’s intercept behaviour.

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

Yes. The published 2027 K110 Mathematical Formulae section includes the quadratic-equation formula. Check official requirements for the student’s own eventual examination year.

What if b²−4ac is negative?

There are no real roots. The associated quadratic graph has no real x-intercept.

Why can the formula return two roots?

The ± yields two possible numerator values when the radicand is positive. Each should be checked in the original equation.

Can one root be a fraction or a negative number?

Yes. Roots depend on the signed coefficients. Do not reject a result merely because it is not a positive whole number; substitute it back to check.

Will a 90-minute group guarantee success?

No. The appropriate progression depends on prerequisite knowledge, feedback and independent retention. Ask the tutor how readiness and later transfer are assessed.

Official references and editorial note

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

Prerequisite mini-check
Read coefficients and signs
Two, one or no real roots
Original worked formula clinics
Five ways a formula goes wrong
Small-group teaching sequence
Short home practice
FAQs
Official source links

The 60-second readiness diagnostic

Begin with three short tasks: solve 3x−5=10; calculate (−3)²; and rewrite x²=5x−6 as an expression equal to zero. Correct responses are x=5, positive nine and x²−5x+6=0. If one-variable balance or a negative square is uncertain, that is the first prerequisite to repair. A pupil cannot depend on a memorised formula to solve a sign problem they do not yet recognise.

Read the equation before typing the formula

First get zero on one side

In x²=5x−6, the coefficients are not read directly from two different sides. Move the terms to obtain x²−5x+6=0. Now a=1, b=−5, c=6. A small visible table of these three signed coefficients helps reduce careless transcription errors, especially when the leading x² coefficient is negative or the ordinary constant is absent.

The negative of b is not necessarily negative

If b=−5, then −b=+5; if b=+5, then −b=−5. The quantity b² is nonnegative in either case, because the squared signed number is (−5)²=25. Those are different uses of a minus sign in the same short calculation. Write parentheses around negative replacements instead of relying on a calculator’s default interpretation.

Everything above the fraction bar shares one denominator

The formula is a single fraction whose numerator is (−b±√(b²−4ac)). The entire numerator is divided by 2a. One wrong set of brackets in the calculator can divide only the radical term, or replace a negative denominator with a positive one. Write the plus and minus branches as separate complete fractions whenever the printed expression looks crowded.

How the quantity under the root changes the answer

The number b²−4ac is the radicand. If it is positive, the two ± branches yield two distinct real roots. If it is zero, the two branches coincide and produce one repeated real root. If it is negative, no real square root exists and there are no real solutions. A corresponding quadratic graph crosses the x-axis twice, touches it once, or does not meet it, respectively. This provides a visual cross-check without requiring pupils to invent non-real outputs.

The root does not become exact merely because the calculator displays many digits. If the formula contains √5, keep it exact until the final accuracy instruction. When three significant figures are requested, round only at the end. The 2027 K110 scheme provides accuracy guidance for graduating candidates; a school’s Secondary Two task may give different instructions.

96 original quadratic formula learning clinics

These examples include exact positive and negative roots, repeated roots, non-perfect-square radicals, no-real-root cases and coefficient-identification checks. They are original teaching material, not past-paper reproductions. Use a few in one lesson, then retest with changed numbers after the model answer is hidden. The purpose is to form a dependable sequence of decisions, not to practise speed with a lengthy equation list.

1. Exact-root clinic 1: x² + 7x + 10=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 7x + 10=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=1, b=7, c=10. The quantity inside the radical is (7)²−4(1)(10)=9. The formula gives x=[-7±√9]/(2)=[-7±3]/(2). Therefore x=-5 or x=-2.

Common trap: A pupil may use positive 7 for b even though the actual coefficient is 7. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: 1(-5)²+(7)(-5)+(10)=0. Substitute x=-2: 1(-2)²+(7)(-2)+(10)=0. This checks the candidates against the original, not merely the formula line.

Independent next step: Rewrite the formula as two complete fractions before using a calculator, ensuring the entire numerator is divided by 2a. Then change the constant c and predict that b²−4ac changes. Ask whether the new equation still has two distinct real roots before calculating.

2. Exact-root clinic 2: −x² + 3x + 10=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x² + 3x + 10=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-1, b=3, c=10. The quantity inside the radical is (3)²−4(-1)(10)=49. The formula gives x=[-3±√49]/(-2)=[-3±7]/(-2). Therefore x=-2 or x=5.

Common trap: A pupil may use positive 3 for b even though the actual coefficient is 3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: -1(-2)²+(3)(-2)+(10)=0. Substitute x=5: -1(5)²+(3)(5)+(10)=0. This checks the candidates against the original, not merely the formula line.

3. Exact-root clinic 3: 2x² − 4x + 2=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x² − 4x + 2=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=2, b=-4, c=2. The quantity inside the radical is (-4)²−4(2)(2)=0. The formula gives x=[4±√0]/(4)=[4±0]/(4). Therefore x=1, a repeated solution.

Common trap: A pupil may use positive 4 for b even though the actual coefficient is -4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: 2(1)²+(-4)(1)+(2)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

4. Exact-root clinic 4: −2x² + 2x + 24=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −2x² + 2x + 24=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-2, b=2, c=24. The quantity inside the radical is (2)²−4(-2)(24)=196. The formula gives x=[-2±√196]/(-4)=[-2±14]/(-4). Therefore x=4 or x=-3.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: -2(4)²+(2)(4)+(24)=0. Substitute x=-3: -2(-3)²+(2)(-3)+(24)=0. This checks the candidates against the original, not merely the formula line.

5. Exact-root clinic 5: 3x² − 48=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 3x² − 48=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=3, b=0, c=-48. The quantity inside the radical is (0)²−4(3)(-48)=576. The formula gives x=[0±√576]/(6)=[0±24]/(6). Therefore x=-4 or x=4.

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: 3(-4)²+(0)(-4)+(-48)=0. Substitute x=4: 3(4)²+(0)(4)+(-48)=0. This checks the candidates against the original, not merely the formula line.

6. Exact-root clinic 6: −3x² − 3x=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −3x² − 3x=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-3, b=-3, c=0. The quantity inside the radical is (-3)²−4(-3)(0)=9. The formula gives x=[3±√9]/(-6)=[3±3]/(-6). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 3 for b even though the actual coefficient is -3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: -3(-1)²+(-3)(-1)+(0)=0. Substitute x=0: -3(0)²+(-3)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

7. Exact-root clinic 7: x² + 2x − 8=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 2x − 8=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=1, b=2, c=-8. The quantity inside the radical is (2)²−4(1)(-8)=36. The formula gives x=[-2±√36]/(2)=[-2±6]/(2). Therefore x=2 or x=-4.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: 1(2)²+(2)(2)+(-8)=0. Substitute x=-4: 1(-4)²+(2)(-4)+(-8)=0. This checks the candidates against the original, not merely the formula line.

8. Exact-root clinic 8: −x² + 8x − 15=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x² + 8x − 15=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-1, b=8, c=-15. The quantity inside the radical is (8)²−4(-1)(-15)=4. The formula gives x=[-8±√4]/(-2)=[-8±2]/(-2). Therefore x=5 or x=3.

Common trap: A pupil may use positive 8 for b even though the actual coefficient is 8. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: -1(5)²+(8)(5)+(-15)=0. Substitute x=3: -1(3)²+(8)(3)+(-15)=0. This checks the candidates against the original, not merely the formula line.

9. Exact-root clinic 9: 2x² + 8x + 6=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x² + 8x + 6=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=2, b=8, c=6. The quantity inside the radical is (8)²−4(2)(6)=16. The formula gives x=[-8±√16]/(4)=[-8±4]/(4). Therefore x=-3 or x=-1.

Common trap: A pupil may use positive 8 for b even though the actual coefficient is 8. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: 2(-3)²+(8)(-3)+(6)=0. Substitute x=-1: 2(-1)²+(8)(-1)+(6)=0. This checks the candidates against the original, not merely the formula line.

10. Exact-root clinic 10: −2x² − 10x=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −2x² − 10x=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-2, b=-10, c=0. The quantity inside the radical is (-10)²−4(-2)(0)=100. The formula gives x=[10±√100]/(-4)=[10±10]/(-4). Therefore x=0 or x=-5.

Common trap: A pupil may use positive 10 for b even though the actual coefficient is -10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=0: -2(0)²+(-10)(0)+(0)=0. Substitute x=-5: -2(-5)²+(-10)(-5)+(0)=0. This checks the candidates against the original, not merely the formula line.

11. Exact-root clinic 11: 3x² − 15x + 18=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 3x² − 15x + 18=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=3, b=-15, c=18. The quantity inside the radical is (-15)²−4(3)(18)=9. The formula gives x=[15±√9]/(6)=[15±3]/(6). Therefore x=3 or x=2.

Common trap: A pupil may use positive 15 for b even though the actual coefficient is -15. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=3: 3(3)²+(-15)(3)+(18)=0. Substitute x=2: 3(2)²+(-15)(2)+(18)=0. This checks the candidates against the original, not merely the formula line.

12. Exact-root clinic 12: −3x² − 21x − 30=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −3x² − 21x − 30=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-3, b=-21, c=-30. The quantity inside the radical is (-21)²−4(-3)(-30)=81. The formula gives x=[21±√81]/(-6)=[21±9]/(-6). Therefore x=-5 or x=-2.

Common trap: A pupil may use positive 21 for b even though the actual coefficient is -21. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: -3(-5)²+(-21)(-5)+(-30)=0. Substitute x=-2: -3(-2)²+(-21)(-2)+(-30)=0. This checks the candidates against the original, not merely the formula line.

13. Exact-root clinic 13: x² − 3x − 10=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 3x − 10=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=1, b=-3, c=-10. The quantity inside the radical is (-3)²−4(1)(-10)=49. The formula gives x=[3±√49]/(2)=[3±7]/(2). Therefore x=-2 or x=5.

Common trap: A pupil may use positive 3 for b even though the actual coefficient is -3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: 1(-2)²+(-3)(-2)+(-10)=0. Substitute x=5: 1(5)²+(-3)(5)+(-10)=0. This checks the candidates against the original, not merely the formula line.

14. Exact-root clinic 14: −x² + 2x − 1=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x² + 2x − 1=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-1, b=2, c=-1. The quantity inside the radical is (2)²−4(-1)(-1)=0. The formula gives x=[-2±√0]/(-2)=[-2±0]/(-2). Therefore x=1, a repeated solution.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: -1(1)²+(2)(1)+(-1)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

15. Exact-root clinic 15: 2x² − 2x − 24=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x² − 2x − 24=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=2, b=-2, c=-24. The quantity inside the radical is (-2)²−4(2)(-24)=196. The formula gives x=[2±√196]/(4)=[2±14]/(4). Therefore x=4 or x=-3.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is -2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: 2(4)²+(-2)(4)+(-24)=0. Substitute x=-3: 2(-3)²+(-2)(-3)+(-24)=0. This checks the candidates against the original, not merely the formula line.

16. Exact-root clinic 16: −2x² + 32=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −2x² + 32=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-2, b=0, c=32. The quantity inside the radical is (0)²−4(-2)(32)=256. The formula gives x=[0±√256]/(-4)=[0±16]/(-4). Therefore x=-4 or x=4.

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: -2(-4)²+(0)(-4)+(32)=0. Substitute x=4: -2(4)²+(0)(4)+(32)=0. This checks the candidates against the original, not merely the formula line.

17. Exact-root clinic 17: 3x² + 3x=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 3x² + 3x=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=3, b=3, c=0. The quantity inside the radical is (3)²−4(3)(0)=9. The formula gives x=[-3±√9]/(6)=[-3±3]/(6). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 3 for b even though the actual coefficient is 3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: 3(-1)²+(3)(-1)+(0)=0. Substitute x=0: 3(0)²+(3)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

18. Exact-root clinic 18: −3x² − 6x + 24=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −3x² − 6x + 24=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-3, b=-6, c=24. The quantity inside the radical is (-6)²−4(-3)(24)=324. The formula gives x=[6±√324]/(-6)=[6±18]/(-6). Therefore x=2 or x=-4.

Common trap: A pupil may use positive 6 for b even though the actual coefficient is -6. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: -3(2)²+(-6)(2)+(24)=0. Substitute x=-4: -3(-4)²+(-6)(-4)+(24)=0. This checks the candidates against the original, not merely the formula line.

19. Exact-root clinic 19: x² − 8x + 15=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 8x + 15=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=1, b=-8, c=15. The quantity inside the radical is (-8)²−4(1)(15)=4. The formula gives x=[8±√4]/(2)=[8±2]/(2). Therefore x=5 or x=3.

Common trap: A pupil may use positive 8 for b even though the actual coefficient is -8. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: 1(5)²+(-8)(5)+(15)=0. Substitute x=3: 1(3)²+(-8)(3)+(15)=0. This checks the candidates against the original, not merely the formula line.

20. Exact-root clinic 20: −x² − 4x − 3=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x² − 4x − 3=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-1, b=-4, c=-3. The quantity inside the radical is (-4)²−4(-1)(-3)=4. The formula gives x=[4±√4]/(-2)=[4±2]/(-2). Therefore x=-3 or x=-1.

Common trap: A pupil may use positive 4 for b even though the actual coefficient is -4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: -1(-3)²+(-4)(-3)+(-3)=0. Substitute x=-1: -1(-1)²+(-4)(-1)+(-3)=0. This checks the candidates against the original, not merely the formula line.

21. Exact-root clinic 21: 2x² + 10x=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x² + 10x=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=2, b=10, c=0. The quantity inside the radical is (10)²−4(2)(0)=100. The formula gives x=[-10±√100]/(4)=[-10±10]/(4). Therefore x=0 or x=-5.

Common trap: A pupil may use positive 10 for b even though the actual coefficient is 10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=0: 2(0)²+(10)(0)+(0)=0. Substitute x=-5: 2(-5)²+(10)(-5)+(0)=0. This checks the candidates against the original, not merely the formula line.

22. Exact-root clinic 22: −2x² + 10x − 12=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −2x² + 10x − 12=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-2, b=10, c=-12. The quantity inside the radical is (10)²−4(-2)(-12)=4. The formula gives x=[-10±√4]/(-4)=[-10±2]/(-4). Therefore x=3 or x=2.

Common trap: A pupil may use positive 10 for b even though the actual coefficient is 10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=3: -2(3)²+(10)(3)+(-12)=0. Substitute x=2: -2(2)²+(10)(2)+(-12)=0. This checks the candidates against the original, not merely the formula line.

23. Exact-root clinic 23: 3x² + 21x + 30=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 3x² + 21x + 30=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=3, b=21, c=30. The quantity inside the radical is (21)²−4(3)(30)=81. The formula gives x=[-21±√81]/(6)=[-21±9]/(6). Therefore x=-5 or x=-2.

Common trap: A pupil may use positive 21 for b even though the actual coefficient is 21. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: 3(-5)²+(21)(-5)+(30)=0. Substitute x=-2: 3(-2)²+(21)(-2)+(30)=0. This checks the candidates against the original, not merely the formula line.

24. Exact-root clinic 24: −3x² + 9x + 30=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −3x² + 9x + 30=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-3, b=9, c=30. The quantity inside the radical is (9)²−4(-3)(30)=441. The formula gives x=[-9±√441]/(-6)=[-9±21]/(-6). Therefore x=-2 or x=5.

Common trap: A pupil may use positive 9 for b even though the actual coefficient is 9. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: -3(-2)²+(9)(-2)+(30)=0. Substitute x=5: -3(5)²+(9)(5)+(30)=0. This checks the candidates against the original, not merely the formula line.

25. Exact-root clinic 25: x² − 2x + 1=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 2x + 1=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=1, b=-2, c=1. The quantity inside the radical is (-2)²−4(1)(1)=0. The formula gives x=[2±√0]/(2)=[2±0]/(2). Therefore x=1, a repeated solution.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is -2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: 1(1)²+(-2)(1)+(1)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

26. Exact-root clinic 26: −x² + x + 12=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x² + x + 12=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-1, b=1, c=12. The quantity inside the radical is (1)²−4(-1)(12)=49. The formula gives x=[-1±√49]/(-2)=[-1±7]/(-2). Therefore x=4 or x=-3.

Common trap: A pupil may use positive 1 for b even though the actual coefficient is 1. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: -1(4)²+(1)(4)+(12)=0. Substitute x=-3: -1(-3)²+(1)(-3)+(12)=0. This checks the candidates against the original, not merely the formula line.

27. Exact-root clinic 27: 2x² − 32=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x² − 32=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=2, b=0, c=-32. The quantity inside the radical is (0)²−4(2)(-32)=256. The formula gives x=[0±√256]/(4)=[0±16]/(4). Therefore x=-4 or x=4.

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: 2(-4)²+(0)(-4)+(-32)=0. Substitute x=4: 2(4)²+(0)(4)+(-32)=0. This checks the candidates against the original, not merely the formula line.

28. Exact-root clinic 28: −2x² − 2x=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −2x² − 2x=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-2, b=-2, c=0. The quantity inside the radical is (-2)²−4(-2)(0)=4. The formula gives x=[2±√4]/(-4)=[2±2]/(-4). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is -2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: -2(-1)²+(-2)(-1)+(0)=0. Substitute x=0: -2(0)²+(-2)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

29. Exact-root clinic 29: 3x² + 6x − 24=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 3x² + 6x − 24=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=3, b=6, c=-24. The quantity inside the radical is (6)²−4(3)(-24)=324. The formula gives x=[-6±√324]/(6)=[-6±18]/(6). Therefore x=2 or x=-4.

Common trap: A pupil may use positive 6 for b even though the actual coefficient is 6. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: 3(2)²+(6)(2)+(-24)=0. Substitute x=-4: 3(-4)²+(6)(-4)+(-24)=0. This checks the candidates against the original, not merely the formula line.

30. Exact-root clinic 30: −3x² + 24x − 45=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −3x² + 24x − 45=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-3, b=24, c=-45. The quantity inside the radical is (24)²−4(-3)(-45)=36. The formula gives x=[-24±√36]/(-6)=[-24±6]/(-6). Therefore x=5 or x=3.

Common trap: A pupil may use positive 24 for b even though the actual coefficient is 24. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: -3(5)²+(24)(5)+(-45)=0. Substitute x=3: -3(3)²+(24)(3)+(-45)=0. This checks the candidates against the original, not merely the formula line.

31. Exact-root clinic 31: x² + 4x + 3=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 4x + 3=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=1, b=4, c=3. The quantity inside the radical is (4)²−4(1)(3)=4. The formula gives x=[-4±√4]/(2)=[-4±2]/(2). Therefore x=-3 or x=-1.

Common trap: A pupil may use positive 4 for b even though the actual coefficient is 4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: 1(-3)²+(4)(-3)+(3)=0. Substitute x=-1: 1(-1)²+(4)(-1)+(3)=0. This checks the candidates against the original, not merely the formula line.

32. Exact-root clinic 32: −x² − 5x=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x² − 5x=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-1, b=-5, c=0. The quantity inside the radical is (-5)²−4(-1)(0)=25. The formula gives x=[5±√25]/(-2)=[5±5]/(-2). Therefore x=0 or x=-5.

Common trap: A pupil may use positive 5 for b even though the actual coefficient is -5. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=0: -1(0)²+(-5)(0)+(0)=0. Substitute x=-5: -1(-5)²+(-5)(-5)+(0)=0. This checks the candidates against the original, not merely the formula line.

33. Exact-root clinic 33: 2x² − 10x + 12=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x² − 10x + 12=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=2, b=-10, c=12. The quantity inside the radical is (-10)²−4(2)(12)=4. The formula gives x=[10±√4]/(4)=[10±2]/(4). Therefore x=3 or x=2.

Common trap: A pupil may use positive 10 for b even though the actual coefficient is -10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=3: 2(3)²+(-10)(3)+(12)=0. Substitute x=2: 2(2)²+(-10)(2)+(12)=0. This checks the candidates against the original, not merely the formula line.

34. Exact-root clinic 34: −2x² − 14x − 20=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −2x² − 14x − 20=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-2, b=-14, c=-20. The quantity inside the radical is (-14)²−4(-2)(-20)=36. The formula gives x=[14±√36]/(-4)=[14±6]/(-4). Therefore x=-5 or x=-2.

Common trap: A pupil may use positive 14 for b even though the actual coefficient is -14. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: -2(-5)²+(-14)(-5)+(-20)=0. Substitute x=-2: -2(-2)²+(-14)(-2)+(-20)=0. This checks the candidates against the original, not merely the formula line.

35. Exact-root clinic 35: 3x² − 9x − 30=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 3x² − 9x − 30=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=3, b=-9, c=-30. The quantity inside the radical is (-9)²−4(3)(-30)=441. The formula gives x=[9±√441]/(6)=[9±21]/(6). Therefore x=-2 or x=5.

Common trap: A pupil may use positive 9 for b even though the actual coefficient is -9. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: 3(-2)²+(-9)(-2)+(-30)=0. Substitute x=5: 3(5)²+(-9)(5)+(-30)=0. This checks the candidates against the original, not merely the formula line.

36. Exact-root clinic 36: −3x² + 6x − 3=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −3x² + 6x − 3=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-3, b=6, c=-3. The quantity inside the radical is (6)²−4(-3)(-3)=0. The formula gives x=[-6±√0]/(-6)=[-6±0]/(-6). Therefore x=1, a repeated solution.

Common trap: A pupil may use positive 6 for b even though the actual coefficient is 6. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: -3(1)²+(6)(1)+(-3)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

37. Exact-root clinic 37: x² − x − 12=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − x − 12=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=1, b=-1, c=-12. The quantity inside the radical is (-1)²−4(1)(-12)=49. The formula gives x=[1±√49]/(2)=[1±7]/(2). Therefore x=4 or x=-3.

Common trap: A pupil may use positive 1 for b even though the actual coefficient is -1. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: 1(4)²+(-1)(4)+(-12)=0. Substitute x=-3: 1(-3)²+(-1)(-3)+(-12)=0. This checks the candidates against the original, not merely the formula line.

38. Exact-root clinic 38: −x² + 16=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x² + 16=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-1, b=0, c=16. The quantity inside the radical is (0)²−4(-1)(16)=64. The formula gives x=[0±√64]/(-2)=[0±8]/(-2). Therefore x=-4 or x=4.

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: -1(-4)²+(0)(-4)+(16)=0. Substitute x=4: -1(4)²+(0)(4)+(16)=0. This checks the candidates against the original, not merely the formula line.

39. Exact-root clinic 39: 2x² + 2x=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x² + 2x=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=2, b=2, c=0. The quantity inside the radical is (2)²−4(2)(0)=4. The formula gives x=[-2±√4]/(4)=[-2±2]/(4). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: 2(-1)²+(2)(-1)+(0)=0. Substitute x=0: 2(0)²+(2)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

40. Exact-root clinic 40: −2x² − 4x + 16=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −2x² − 4x + 16=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-2, b=-4, c=16. The quantity inside the radical is (-4)²−4(-2)(16)=144. The formula gives x=[4±√144]/(-4)=[4±12]/(-4). Therefore x=2 or x=-4.

Common trap: A pupil may use positive 4 for b even though the actual coefficient is -4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: -2(2)²+(-4)(2)+(16)=0. Substitute x=-4: -2(-4)²+(-4)(-4)+(16)=0. This checks the candidates against the original, not merely the formula line.

41. Exact-root clinic 41: 3x² − 24x + 45=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 3x² − 24x + 45=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=3, b=-24, c=45. The quantity inside the radical is (-24)²−4(3)(45)=36. The formula gives x=[24±√36]/(6)=[24±6]/(6). Therefore x=5 or x=3.

Common trap: A pupil may use positive 24 for b even though the actual coefficient is -24. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: 3(5)²+(-24)(5)+(45)=0. Substitute x=3: 3(3)²+(-24)(3)+(45)=0. This checks the candidates against the original, not merely the formula line.

42. Exact-root clinic 42: −3x² − 12x − 9=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −3x² − 12x − 9=0 with the formula x=(−b±√(b²−4ac))/(2a). Identify all coefficients, including the sign of b.

Worked explanation: The signed coefficients are a=-3, b=-12, c=-9. The quantity inside the radical is (-12)²−4(-3)(-9)=36. The formula gives x=[12±√36]/(-6)=[12±6]/(-6). Therefore x=-3 or x=-1.

Common trap: A pupil may use positive 12 for b even though the actual coefficient is -12. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: -3(-3)²+(-12)(-3)+(-9)=0. Substitute x=-1: -3(-1)²+(-12)(-1)+(-9)=0. This checks the candidates against the original, not merely the formula line.

43. Non-perfect-square clinic 1: x² − 7x − 7=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 7x − 7=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-7, c=-7. Compute b²−4ac=(-7)²−4(1)(-7)=77. The roots are x=[7±√77]/(2), giving x≈7.89 and x≈-0.887 to three significant figures.

Common trap: Replacing √77 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √77, substituting both roots in x² − 7x − 7=0 gives zero. Rounded roots such as 7.89 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 77 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

44. Non-perfect-square clinic 2: x² − 7x − 4=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 7x − 4=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-7, c=-4. Compute b²−4ac=(-7)²−4(1)(-4)=65. The roots are x=[7±√65]/(2), giving x≈7.53 and x≈-0.531 to three significant figures.

Common trap: Replacing √65 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √65, substituting both roots in x² − 7x − 4=0 gives zero. Rounded roots such as 7.53 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 65 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

45. Non-perfect-square clinic 3: x² − 7x − 1=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 7x − 1=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-7, c=-1. Compute b²−4ac=(-7)²−4(1)(-1)=53. The roots are x=[7±√53]/(2), giving x≈7.14 and x≈-0.140 to three significant figures.

Common trap: Replacing √53 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √53, substituting both roots in x² − 7x − 1=0 gives zero. Rounded roots such as 7.14 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 53 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

46. Non-perfect-square clinic 4: x² − 7x + 2=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 7x + 2=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-7, c=2. Compute b²−4ac=(-7)²−4(1)(2)=41. The roots are x=[7±√41]/(2), giving x≈6.70 and x≈0.298 to three significant figures.

Common trap: Replacing √41 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √41, substituting both roots in x² − 7x + 2=0 gives zero. Rounded roots such as 6.70 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 41 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

47. Non-perfect-square clinic 5: x² − 7x + 4=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 7x + 4=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-7, c=4. Compute b²−4ac=(-7)²−4(1)(4)=33. The roots are x=[7±√33]/(2), giving x≈6.37 and x≈0.628 to three significant figures.

Common trap: Replacing √33 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √33, substituting both roots in x² − 7x + 4=0 gives zero. Rounded roots such as 6.37 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 33 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

48. Non-perfect-square clinic 6: x² − 7x + 7=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 7x + 7=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-7, c=7. Compute b²−4ac=(-7)²−4(1)(7)=21. The roots are x=[7±√21]/(2), giving x≈5.79 and x≈1.21 to three significant figures.

Common trap: Replacing √21 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √21, substituting both roots in x² − 7x + 7=0 gives zero. Rounded roots such as 5.79 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 21 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

49. Non-perfect-square clinic 7: x² − 7x + 9=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 7x + 9=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-7, c=9. Compute b²−4ac=(-7)²−4(1)(9)=13. The roots are x=[7±√13]/(2), giving x≈5.30 and x≈1.70 to three significant figures.

Common trap: Replacing √13 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √13, substituting both roots in x² − 7x + 9=0 gives zero. Rounded roots such as 5.30 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 13 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

50. Non-perfect-square clinic 8: x² − 5x − 7=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 5x − 7=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-5, c=-7. Compute b²−4ac=(-5)²−4(1)(-7)=53. The roots are x=[5±√53]/(2), giving x≈6.14 and x≈-1.14 to three significant figures.

Common trap: Replacing √53 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √53, substituting both roots in x² − 5x − 7=0 gives zero. Rounded roots such as 6.14 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

51. Non-perfect-square clinic 9: x² − 5x − 4=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 5x − 4=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-5, c=-4. Compute b²−4ac=(-5)²−4(1)(-4)=41. The roots are x=[5±√41]/(2), giving x≈5.70 and x≈-0.702 to three significant figures.

Common trap: Replacing √41 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √41, substituting both roots in x² − 5x − 4=0 gives zero. Rounded roots such as 5.70 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

52. Non-perfect-square clinic 10: x² − 5x − 1=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 5x − 1=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-5, c=-1. Compute b²−4ac=(-5)²−4(1)(-1)=29. The roots are x=[5±√29]/(2), giving x≈5.19 and x≈-0.193 to three significant figures.

Common trap: Replacing √29 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √29, substituting both roots in x² − 5x − 1=0 gives zero. Rounded roots such as 5.19 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 29 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

53. Non-perfect-square clinic 11: x² − 5x + 2=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 5x + 2=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-5, c=2. Compute b²−4ac=(-5)²−4(1)(2)=17. The roots are x=[5±√17]/(2), giving x≈4.56 and x≈0.438 to three significant figures.

Common trap: Replacing √17 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √17, substituting both roots in x² − 5x + 2=0 gives zero. Rounded roots such as 4.56 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 17 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

54. Non-perfect-square clinic 12: x² − 3x − 7=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 3x − 7=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-3, c=-7. Compute b²−4ac=(-3)²−4(1)(-7)=37. The roots are x=[3±√37]/(2), giving x≈4.54 and x≈-1.54 to three significant figures.

Common trap: Replacing √37 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √37, substituting both roots in x² − 3x − 7=0 gives zero. Rounded roots such as 4.54 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 37 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

55. Non-perfect-square clinic 13: x² − 3x − 1=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² − 3x − 1=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=-3, c=-1. Compute b²−4ac=(-3)²−4(1)(-1)=13. The roots are x=[3±√13]/(2), giving x≈3.30 and x≈-0.303 to three significant figures.

Common trap: Replacing √13 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √13, substituting both roots in x² − 3x − 1=0 gives zero. Rounded roots such as 3.30 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

56. Non-perfect-square clinic 14: x² + x − 7=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + x − 7=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=1, c=-7. Compute b²−4ac=(1)²−4(1)(-7)=29. The roots are x=[-1±√29]/(2), giving x≈2.19 and x≈-3.19 to three significant figures.

Common trap: Replacing √29 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √29, substituting both roots in x² + x − 7=0 gives zero. Rounded roots such as 2.19 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

57. Non-perfect-square clinic 15: x² + x − 4=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + x − 4=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=1, c=-4. Compute b²−4ac=(1)²−4(1)(-4)=17. The roots are x=[-1±√17]/(2), giving x≈1.56 and x≈-2.56 to three significant figures.

Common trap: Replacing √17 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √17, substituting both roots in x² + x − 4=0 gives zero. Rounded roots such as 1.56 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

58. Non-perfect-square clinic 16: x² + x − 1=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + x − 1=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=1, c=-1. Compute b²−4ac=(1)²−4(1)(-1)=5. The roots are x=[-1±√5]/(2), giving x≈0.618 and x≈-1.62 to three significant figures.

Common trap: Replacing √5 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √5, substituting both roots in x² + x − 1=0 gives zero. Rounded roots such as 0.618 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 5 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

59. Non-perfect-square clinic 17: x² + 2x − 7=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 2x − 7=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=2, c=-7. Compute b²−4ac=(2)²−4(1)(-7)=32. The roots are x=[-2±√32]/(2), giving x≈1.83 and x≈-3.83 to three significant figures.

Common trap: Replacing √32 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √32, substituting both roots in x² + 2x − 7=0 gives zero. Rounded roots such as 1.83 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 32 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

60. Non-perfect-square clinic 18: x² + 2x − 4=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 2x − 4=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=2, c=-4. Compute b²−4ac=(2)²−4(1)(-4)=20. The roots are x=[-2±√20]/(2), giving x≈1.24 and x≈-3.24 to three significant figures.

Common trap: Replacing √20 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √20, substituting both roots in x² + 2x − 4=0 gives zero. Rounded roots such as 1.24 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 20 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

61. Non-perfect-square clinic 19: x² + 2x − 1=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 2x − 1=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=2, c=-1. Compute b²−4ac=(2)²−4(1)(-1)=8. The roots are x=[-2±√8]/(2), giving x≈0.414 and x≈-2.41 to three significant figures.

Common trap: Replacing √8 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √8, substituting both roots in x² + 2x − 1=0 gives zero. Rounded roots such as 0.414 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 8 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

62. Non-perfect-square clinic 20: x² + 4x − 7=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 4x − 7=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=4, c=-7. Compute b²−4ac=(4)²−4(1)(-7)=44. The roots are x=[-4±√44]/(2), giving x≈1.32 and x≈-5.32 to three significant figures.

Common trap: Replacing √44 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √44, substituting both roots in x² + 4x − 7=0 gives zero. Rounded roots such as 1.32 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 44 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

63. Non-perfect-square clinic 21: x² + 4x − 4=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 4x − 4=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=4, c=-4. Compute b²−4ac=(4)²−4(1)(-4)=32. The roots are x=[-4±√32]/(2), giving x≈0.828 and x≈-4.83 to three significant figures.

Common trap: Replacing √32 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √32, substituting both roots in x² + 4x − 4=0 gives zero. Rounded roots such as 0.828 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

64. Non-perfect-square clinic 22: x² + 4x − 1=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 4x − 1=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=4, c=-1. Compute b²−4ac=(4)²−4(1)(-1)=20. The roots are x=[-4±√20]/(2), giving x≈0.236 and x≈-4.24 to three significant figures.

Common trap: Replacing √20 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √20, substituting both roots in x² + 4x − 1=0 gives zero. Rounded roots such as 0.236 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

65. Non-perfect-square clinic 23: x² + 4x + 2=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 4x + 2=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=4, c=2. Compute b²−4ac=(4)²−4(1)(2)=8. The roots are x=[-4±√8]/(2), giving x≈-0.586 and x≈-3.41 to three significant figures.

Common trap: Replacing √8 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √8, substituting both roots in x² + 4x + 2=0 gives zero. Rounded roots such as -0.586 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

66. Non-perfect-square clinic 24: x² + 6x − 4=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x² + 6x − 4=0. Express both real roots to three significant figures only at the end.

Worked explanation: Take a=1, b=6, c=-4. Compute b²−4ac=(6)²−4(1)(-4)=52. The roots are x=[-6±√52]/(2), giving x≈0.606 and x≈-6.61 to three significant figures.

Common trap: Replacing √52 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

Verify: With unrounded √52, substituting both roots in x² + 6x − 4=0 gives zero. Rounded roots such as 0.606 may give a small nonzero residual: this is consistent with the requested approximation and should not be concealed.

Independent next step: The ± creates two different branches because 52 is positive. Plotting the related quadratic would show two x-axis intercepts. Ask which root is larger and how the negative sign of a, when present, affects division.

67. No-real-root case 1: x² − 3x + 4=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether x² − 3x + 4=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=1, b=-3, c=4. The quantity b²−4ac is (-3)²−4(1)(4)=-7, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -7 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

Independent next step: Compare with a new equation having zero radicand; its two branches coincide at one repeated root. Use a sketch to explain why one touches the axis while the other does not meet it.

68. No-real-root case 2: 2x² − 2x + 3=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether 2x² − 2x + 3=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=2, b=-2, c=3. The quantity b²−4ac is (-2)²−4(2)(3)=-20, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -20 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=2 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

69. No-real-root case 3: 3x² − x + 4=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether 3x² − x + 4=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=3, b=-1, c=4. The quantity b²−4ac is (-1)²−4(3)(4)=-47, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -47 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=3 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

70. No-real-root case 4: x² + 5=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether x² + 5=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=1, b=0, c=5. The quantity b²−4ac is (0)²−4(1)(5)=-20, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -20 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

71. No-real-root case 5: 2x² + x + 6=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether 2x² + x + 6=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=2, b=1, c=6. The quantity b²−4ac is (1)²−4(2)(6)=-47, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -47 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=2 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

72. No-real-root case 6: 3x² + 2x + 7=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether 3x² + 2x + 7=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=3, b=2, c=7. The quantity b²−4ac is (2)²−4(3)(7)=-80, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -80 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=3 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

73. No-real-root case 7: x² + 3x + 10=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether x² + 3x + 10=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=1, b=3, c=10. The quantity b²−4ac is (3)²−4(1)(10)=-31, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -31 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

74. No-real-root case 8: 2x² − 3x + 10=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether 2x² − 3x + 10=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=2, b=-3, c=10. The quantity b²−4ac is (-3)²−4(2)(10)=-71, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -71 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=2 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

75. No-real-root case 9: 3x² − 2x + 10=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether 3x² − 2x + 10=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=3, b=-2, c=10. The quantity b²−4ac is (-2)²−4(3)(10)=-116, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -116 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=3 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

76. No-real-root case 10: x² − x + 11=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Use the formula to decide whether x² − x + 11=0 has real-number solutions. Do not invent a number when the square root is not real.

Worked explanation: Coefficients are a=1, b=-1, c=11. The quantity b²−4ac is (-1)²−4(1)(11)=-43, a negative number. There is no real-number solution because the square root of a negative number is not real.

Common trap: Taking the absolute value of -43 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

77. Interpretation and checking 1: Write standard form before inserting coefficients

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. The equation x²=5x−6 is not yet in the form ax²+bx+c=0.

Worked explanation: Move everything to the left: x²−5x+6=0. Then a=1,b=−5,c=6 and roots are 2 and 3.

Common trap: Reading b as +5 because the 5x appears on the right rather than changing its sign.

Verify: Substitute two and three into x²=5x−6 to verify both.

Independent next step: Repeat the problem with a sign or coefficient changed. Before substituting, rewrite any equation as ax²+bx+c=0, identify the signed values of a, b and c, and check whether the radicand is positive, zero or negative. Then interpret any numerical results in the context of the original question.

78. Interpretation and checking 2: The formula requires a nonzero quadratic coefficient

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. A question presents 0x²+3x−9=0. Is the quadratic formula appropriate?

Worked explanation: No. The coefficient a=0 leaves a linear equation, and 3x−9=0 gives x=3.

Common trap: Using the formula with denominator 2a=0 would create undefined division.

Verify: Identify the highest nonzero power before selecting a method.

79. Interpretation and checking 3: The full numerator is divided by 2a

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. How should x=(−b±√(b²−4ac))/(2a) be keyed or written clearly?

Worked explanation: Write the entire numerator in parentheses before dividing by the full denominator (2a). The plus and minus are separate branches.

Common trap: Writing −b+√D/2a can divide only the radical instead of the whole numerator.

Verify: Test an equation with known roots to confirm the input layout.

80. Interpretation and checking 4: The sign of b affects two parts

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Identify a,b,c in 2x²−5x+2=0.

Worked explanation: a=2,b=−5,c=2. Hence −b=5 while b²=25. The formula returns x=2 or x=1/2.

Common trap: Writing b=+5 changes −b and can reverse the final candidates.

Verify: Check both root candidates in the original equation.

81. Interpretation and checking 5: A repeated root is one distinct real answer

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x²−6x+9=0.

Worked explanation: The radical term has 36−36=0, so x=(6±0)/2=3. It is a repeated real root.

Common trap: A learner may invent two different answers merely because the formula has ±.

Verify: Substitute x=3 and see 9−18+9=0.

82. Interpretation and checking 6: The square-root symbol has a principal value

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. What differs between √9 and solving x²=9?

Worked explanation: √9=3, the nonnegative principal square root. The equation x²=9 has x=3 and x=−3.

Common trap: Calling √9 itself ±3 conflates the expression and the equation.

Verify: Both +3 and −3 square to nine, but √9 denotes +3.

83. Interpretation and checking 7: A negative leading coefficient changes the denominator

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x²+4x−3=0.

Worked explanation: a=−1,b=4,c=−3, so D=16−12=4 and denominator is −2. Roots are x=1 and x=3.

Common trap: Using +2 in the denominator makes both signs incorrect.

Verify: Check −1+4−3=0 and −9+12−3=0.

84. Interpretation and checking 8: A missing constant has value zero

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Find a,b,c for 3x²−2x=0.

Worked explanation: a=3,b=−2,c=0. Formula yields roots x=0 and 2/3.

Common trap: Inventing a constant of one because there are three labels a,b,c.

Verify: Substitute zero and two-thirds into the original.

85. Interpretation and checking 9: Some roots are simple rational fractions

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x²−3x+1=0.

Worked explanation: D=9−8=1. x=(3±1)/4, so roots are 1 and 1/2.

Common trap: Dividing only one numerator term by four gives incorrect results.

Verify: Verify x=1 and x=1/2 exactly.

86. Interpretation and checking 10: An upward-opening quadratic need not cross the axis

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Does x²+4=0 have real solutions?

Worked explanation: No. D=−16 is negative, and the graph y=x²+4 is always at least four.

Common trap: Claiming every quadratic must have two real roots.

Verify: Its lowest graph point is (0,4), above the axis.

87. Interpretation and checking 11: A graph’s crossings are equation roots

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. For y=x²−3x+2, where are the x-intercepts?

Worked explanation: Set y=0 and solve: roots are x=1 and x=2. Thus crossings are (1,0) and (2,0).

Common trap: Mistaking the y-intercept (0,2) for a solution of y=0.

Verify: At both x=1 and x=2, the output is zero.

88. Interpretation and checking 12: Do not round the root expression too early

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Approximate (3+√5)/2 to three significant figures.

Worked explanation: The unrounded value is about 2.6180339; report 2.62 to three significant figures.

Common trap: Rounding √5 to 2 before finishing produces 2.5, a poor approximation.

Verify: Use unrounded intermediate values then round final display.

89. Interpretation and checking 13: A root is a candidate until it is checked

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Check x=2 in x²−5x+6=0.

Worked explanation: Substitute 2²−5(2)+6=4−10+6=0. So two is a solution.

Common trap: Accepting a written x-value without testing the original equality.

Verify: Test x=3 as well; it is a second solution.

90. Interpretation and checking 14: A negative root is not necessarily a mistake

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x²+5x+6=0.

Worked explanation: a=1,b=5,c=6 and D=25−24=1; formula gives (−5±1)/2, so roots −2 and −3.

Common trap: Assuming positive leading a guarantees positive roots.

Verify: Substitute both negative candidates.

91. Interpretation and checking 15: An exact radical and a rounded value are different outputs

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. A formula produces x=(1+√3)/2. Is x=1.37 exact?

Worked explanation: No. The expression with √3 is exact; 1.37 is a rounded approximation to three significant figures.

Common trap: Dropping the approximation sign conceals the loss of precision.

Verify: Calculate with full precision when checking the original equation.

92. Interpretation and checking 16: A possible real-world constraint can reject one root

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. An idealised width equation has roots −2 and 5. Which could represent a positive physical width?

Worked explanation: Only five is positive, assuming no additional constraint. The negative algebraic root is mathematically valid but not physically feasible for width.

Common trap: Deleting the negative root without first reporting it as a mathematical solution may confuse the two stages.

Verify: State algebraic answers first, then apply the context restriction.

93. Interpretation and checking 17: The 2027 K110 formulae list contains the quadratic formula

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Is a guessed mnemonic preferable to checking the official formulae sheet?

Worked explanation: No. The published 2027 K110 Number and Algebra formulae list contains x=(−b±√(b²−4ac))/(2a). The essential skill is accurate substitution and interpretation.

Common trap: Assuming every school’s current formula sheet is identical to an older unofficial screenshot.

Verify: Check official documentation for the student’s own examination year.

94. Interpretation and checking 18: A formula result should be shown clearly

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. A learner writes x=3 or −2 without stating which equation was solved.

Worked explanation: The result may be mathematically correct but should be tied to a stated equation, with enough working to show a,b,c and a valid calculation.

Common trap: Copying two numbers from an answer key is not evidence the method was understood.

Verify: A changed-number problem and a direct substitution check provide evidence.

95. Interpretation and checking 19: Two negative factors under b² remain positive

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. In x²+4x+1=0, what is b²?

Worked explanation: b=4, so b²=16. For b=−4 in x²−4x+1, b² is still 16.

Common trap: Writing −16 after squaring a negative coefficient fails to bracket the full quantity.

Verify: Compare (−4)² with −(4²), which are not equal.

96. Interpretation and checking 20: One solution repeated when D=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 4x²+4x+1=0.

Worked explanation: D=16−16=0; x=−4/8=−1/2. The two branches coincide.

Common trap: Writing +1/2 after losing the negative numerator.

Verify: Check 4×1/4+4×(−1/2)+1=0.

Five different errors that appear as a bad final answer

The equation was never written in standard form

A learner sees x²=5x−6 and inserts b=5, c=−6. That is not the same as rewriting x²−5x+6=0. Require the equality to zero on one side before reading coefficients. This is one-variable equation balance, a prerequisite the student should already understand before attempting quadratic methods.

b was entered without its negative sign

A student reads 2x²−5x+2=0 but records b=5. Mark b=−5 visibly and evaluate −b=5 as a separate line. If x=2 or x=1/2 is the expected root, substitution into the original equation makes a wrong signed formula candidate easy to reject.

The plus/minus was treated as optional

For x²−9=0, both +3 and −3 satisfy the equation. A pupil who enters only the plus branch has not solved for all real candidates. When the radicand is zero, the two branches coincide; ask the learner to explain that difference instead of always writing two distinct numbers.

The denominator was applied to only half the numerator

Compare (−b+√D)/(2a) with −b+√D/(2a). The second expression generally represents something else. Place parentheses around the entire numerator, enter 2a as a whole signed denominator and review the printed calculator expression before accepting the output.

Rounding made the check seem to fail

A root found from √D may be irrational. If the exact result is retained but the submitted display is three significant figures, plugging the rounded display into the original polynomial may give a small nonzero residual. That is expected approximation error. A large residual, on the other hand, suggests a calculation or transcription mistake that needs correcting.

An illustrative three-pupil bridge lesson

The immutable eduKateSG Clementi three-pupil Mathematics programme describes 90-minute tutorials near Sixth Avenue MRT. A G1 bridge lesson can have one pupil repair negative squares, another practise a,b,c recognition and a third solve complete formula problems with independent checks. This is a hypothetical teaching adaptation, not proof of current Secondary 2 G1 extension availability.

10 minutes: review signed multiplication, roots and simple equation balance.
20 minutes: introduce the formula using the official sheet and practise coefficient lists.
25 minutes: solve a few exact-root questions, including both ± branches.
15 minutes: distinguish repeated-root and no-real-root cases.
10 minutes: independently solve one changed equation with no visible model.
10 minutes: correct the first sign or bracket error and assign a short later retest.

A five-session parent–student routine

Day one: practise negative squares and bring simple equations into zero-equals form.
Day two: identify a, b and c in five varied equations.
Day three: solve two exact-root cases from the formula and verify both answers.
Day four: examine one repeated root, one negative radicand and one irrational pair.
Day five: solve a new problem and explain the corresponding graph’s intercept behaviour.

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

Yes. The published 2027 K110 Mathematical Formulae section includes the quadratic-equation formula. Check official requirements for the student’s own eventual examination year.

What if b²−4ac is negative?

There are no real roots. The associated quadratic graph has no real x-intercept.

Why can the formula return two roots?

The ± yields two possible numerator values when the radicand is positive. Each should be checked in the original equation.

Can one root be a fraction or a negative number?

Yes. Roots depend on the signed coefficients. Do not reject a result merely because it is not a positive whole number; substitute it back to check.

Will a 90-minute group guarantee success?

No. The appropriate progression depends on prerequisite knowledge, feedback and independent retention. Ask the tutor how readiness and later transfer are assessed.

Official references and editorial note

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

Prerequisite mini-check
Read coefficients and signs
Two, one or no real roots
Original worked formula clinics
Five ways a formula goes wrong
Small-group teaching sequence
Short home practice
FAQs
Official source links

The 60-second readiness diagnostic

Read the equation before typing the formula

First get zero on one side

The negative of b is not necessarily negative

Everything above the fraction bar shares one denominator

How the quantity under the root changes the answer

96 original quadratic formula learning clinics

1. Exact-root clinic 1: x² + 7x + 10=0

Worked explanation: The signed coefficients are a=1, b=7, c=10. The quantity inside the radical is (7)²−4(1)(10)=9. The formula gives x=[-7±√9]/(2)=[-7±3]/(2). Therefore x=-5 or x=-2.

Common trap: A pupil may use positive 7 for b even though the actual coefficient is 7. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: 1(-5)²+(7)(-5)+(10)=0. Substitute x=-2: 1(-2)²+(7)(-2)+(10)=0. This checks the candidates against the original, not merely the formula line.

2. Exact-root clinic 2: −x² + 3x + 10=0

Common trap: A pupil may use positive 3 for b even though the actual coefficient is 3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: -1(-2)²+(3)(-2)+(10)=0. Substitute x=5: -1(5)²+(3)(5)+(10)=0. This checks the candidates against the original, not merely the formula line.

3. Exact-root clinic 3: 2x² − 4x + 2=0

Common trap: A pupil may use positive 4 for b even though the actual coefficient is -4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: 2(1)²+(-4)(1)+(2)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

4. Exact-root clinic 4: −2x² + 2x + 24=0

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: -2(4)²+(2)(4)+(24)=0. Substitute x=-3: -2(-3)²+(2)(-3)+(24)=0. This checks the candidates against the original, not merely the formula line.

5. Exact-root clinic 5: 3x² − 48=0

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: 3(-4)²+(0)(-4)+(-48)=0. Substitute x=4: 3(4)²+(0)(4)+(-48)=0. This checks the candidates against the original, not merely the formula line.

6. Exact-root clinic 6: −3x² − 3x=0

Worked explanation: The signed coefficients are a=-3, b=-3, c=0. The quantity inside the radical is (-3)²−4(-3)(0)=9. The formula gives x=[3±√9]/(-6)=[3±3]/(-6). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 3 for b even though the actual coefficient is -3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: -3(-1)²+(-3)(-1)+(0)=0. Substitute x=0: -3(0)²+(-3)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

7. Exact-root clinic 7: x² + 2x − 8=0

Worked explanation: The signed coefficients are a=1, b=2, c=-8. The quantity inside the radical is (2)²−4(1)(-8)=36. The formula gives x=[-2±√36]/(2)=[-2±6]/(2). Therefore x=2 or x=-4.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: 1(2)²+(2)(2)+(-8)=0. Substitute x=-4: 1(-4)²+(2)(-4)+(-8)=0. This checks the candidates against the original, not merely the formula line.

8. Exact-root clinic 8: −x² + 8x − 15=0

Common trap: A pupil may use positive 8 for b even though the actual coefficient is 8. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: -1(5)²+(8)(5)+(-15)=0. Substitute x=3: -1(3)²+(8)(3)+(-15)=0. This checks the candidates against the original, not merely the formula line.

9. Exact-root clinic 9: 2x² + 8x + 6=0

Worked explanation: The signed coefficients are a=2, b=8, c=6. The quantity inside the radical is (8)²−4(2)(6)=16. The formula gives x=[-8±√16]/(4)=[-8±4]/(4). Therefore x=-3 or x=-1.

Common trap: A pupil may use positive 8 for b even though the actual coefficient is 8. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: 2(-3)²+(8)(-3)+(6)=0. Substitute x=-1: 2(-1)²+(8)(-1)+(6)=0. This checks the candidates against the original, not merely the formula line.

10. Exact-root clinic 10: −2x² − 10x=0

Common trap: A pupil may use positive 10 for b even though the actual coefficient is -10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=0: -2(0)²+(-10)(0)+(0)=0. Substitute x=-5: -2(-5)²+(-10)(-5)+(0)=0. This checks the candidates against the original, not merely the formula line.

11. Exact-root clinic 11: 3x² − 15x + 18=0

Common trap: A pupil may use positive 15 for b even though the actual coefficient is -15. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=3: 3(3)²+(-15)(3)+(18)=0. Substitute x=2: 3(2)²+(-15)(2)+(18)=0. This checks the candidates against the original, not merely the formula line.

12. Exact-root clinic 12: −3x² − 21x − 30=0

Common trap: A pupil may use positive 21 for b even though the actual coefficient is -21. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: -3(-5)²+(-21)(-5)+(-30)=0. Substitute x=-2: -3(-2)²+(-21)(-2)+(-30)=0. This checks the candidates against the original, not merely the formula line.

13. Exact-root clinic 13: x² − 3x − 10=0

Common trap: A pupil may use positive 3 for b even though the actual coefficient is -3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: 1(-2)²+(-3)(-2)+(-10)=0. Substitute x=5: 1(5)²+(-3)(5)+(-10)=0. This checks the candidates against the original, not merely the formula line.

14. Exact-root clinic 14: −x² + 2x − 1=0

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: -1(1)²+(2)(1)+(-1)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

15. Exact-root clinic 15: 2x² − 2x − 24=0

Common trap: A pupil may use positive 2 for b even though the actual coefficient is -2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: 2(4)²+(-2)(4)+(-24)=0. Substitute x=-3: 2(-3)²+(-2)(-3)+(-24)=0. This checks the candidates against the original, not merely the formula line.

16. Exact-root clinic 16: −2x² + 32=0

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: -2(-4)²+(0)(-4)+(32)=0. Substitute x=4: -2(4)²+(0)(4)+(32)=0. This checks the candidates against the original, not merely the formula line.

17. Exact-root clinic 17: 3x² + 3x=0

Worked explanation: The signed coefficients are a=3, b=3, c=0. The quantity inside the radical is (3)²−4(3)(0)=9. The formula gives x=[-3±√9]/(6)=[-3±3]/(6). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 3 for b even though the actual coefficient is 3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: 3(-1)²+(3)(-1)+(0)=0. Substitute x=0: 3(0)²+(3)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

18. Exact-root clinic 18: −3x² − 6x + 24=0

Common trap: A pupil may use positive 6 for b even though the actual coefficient is -6. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: -3(2)²+(-6)(2)+(24)=0. Substitute x=-4: -3(-4)²+(-6)(-4)+(24)=0. This checks the candidates against the original, not merely the formula line.

19. Exact-root clinic 19: x² − 8x + 15=0

Worked explanation: The signed coefficients are a=1, b=-8, c=15. The quantity inside the radical is (-8)²−4(1)(15)=4. The formula gives x=[8±√4]/(2)=[8±2]/(2). Therefore x=5 or x=3.

Common trap: A pupil may use positive 8 for b even though the actual coefficient is -8. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: 1(5)²+(-8)(5)+(15)=0. Substitute x=3: 1(3)²+(-8)(3)+(15)=0. This checks the candidates against the original, not merely the formula line.

20. Exact-root clinic 20: −x² − 4x − 3=0

Common trap: A pupil may use positive 4 for b even though the actual coefficient is -4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: -1(-3)²+(-4)(-3)+(-3)=0. Substitute x=-1: -1(-1)²+(-4)(-1)+(-3)=0. This checks the candidates against the original, not merely the formula line.

21. Exact-root clinic 21: 2x² + 10x=0

Common trap: A pupil may use positive 10 for b even though the actual coefficient is 10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=0: 2(0)²+(10)(0)+(0)=0. Substitute x=-5: 2(-5)²+(10)(-5)+(0)=0. This checks the candidates against the original, not merely the formula line.

22. Exact-root clinic 22: −2x² + 10x − 12=0

Common trap: A pupil may use positive 10 for b even though the actual coefficient is 10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=3: -2(3)²+(10)(3)+(-12)=0. Substitute x=2: -2(2)²+(10)(2)+(-12)=0. This checks the candidates against the original, not merely the formula line.

23. Exact-root clinic 23: 3x² + 21x + 30=0

Common trap: A pupil may use positive 21 for b even though the actual coefficient is 21. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: 3(-5)²+(21)(-5)+(30)=0. Substitute x=-2: 3(-2)²+(21)(-2)+(30)=0. This checks the candidates against the original, not merely the formula line.

24. Exact-root clinic 24: −3x² + 9x + 30=0

Common trap: A pupil may use positive 9 for b even though the actual coefficient is 9. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: -3(-2)²+(9)(-2)+(30)=0. Substitute x=5: -3(5)²+(9)(5)+(30)=0. This checks the candidates against the original, not merely the formula line.

25. Exact-root clinic 25: x² − 2x + 1=0

Common trap: A pupil may use positive 2 for b even though the actual coefficient is -2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: 1(1)²+(-2)(1)+(1)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

26. Exact-root clinic 26: −x² + x + 12=0

Common trap: A pupil may use positive 1 for b even though the actual coefficient is 1. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: -1(4)²+(1)(4)+(12)=0. Substitute x=-3: -1(-3)²+(1)(-3)+(12)=0. This checks the candidates against the original, not merely the formula line.

27. Exact-root clinic 27: 2x² − 32=0

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: 2(-4)²+(0)(-4)+(-32)=0. Substitute x=4: 2(4)²+(0)(4)+(-32)=0. This checks the candidates against the original, not merely the formula line.

28. Exact-root clinic 28: −2x² − 2x=0

Worked explanation: The signed coefficients are a=-2, b=-2, c=0. The quantity inside the radical is (-2)²−4(-2)(0)=4. The formula gives x=[2±√4]/(-4)=[2±2]/(-4). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is -2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: -2(-1)²+(-2)(-1)+(0)=0. Substitute x=0: -2(0)²+(-2)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

29. Exact-root clinic 29: 3x² + 6x − 24=0

Common trap: A pupil may use positive 6 for b even though the actual coefficient is 6. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: 3(2)²+(6)(2)+(-24)=0. Substitute x=-4: 3(-4)²+(6)(-4)+(-24)=0. This checks the candidates against the original, not merely the formula line.

30. Exact-root clinic 30: −3x² + 24x − 45=0

Common trap: A pupil may use positive 24 for b even though the actual coefficient is 24. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: -3(5)²+(24)(5)+(-45)=0. Substitute x=3: -3(3)²+(24)(3)+(-45)=0. This checks the candidates against the original, not merely the formula line.

31. Exact-root clinic 31: x² + 4x + 3=0

Worked explanation: The signed coefficients are a=1, b=4, c=3. The quantity inside the radical is (4)²−4(1)(3)=4. The formula gives x=[-4±√4]/(2)=[-4±2]/(2). Therefore x=-3 or x=-1.

Common trap: A pupil may use positive 4 for b even though the actual coefficient is 4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: 1(-3)²+(4)(-3)+(3)=0. Substitute x=-1: 1(-1)²+(4)(-1)+(3)=0. This checks the candidates against the original, not merely the formula line.

32. Exact-root clinic 32: −x² − 5x=0

Common trap: A pupil may use positive 5 for b even though the actual coefficient is -5. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=0: -1(0)²+(-5)(0)+(0)=0. Substitute x=-5: -1(-5)²+(-5)(-5)+(0)=0. This checks the candidates against the original, not merely the formula line.

33. Exact-root clinic 33: 2x² − 10x + 12=0

Common trap: A pupil may use positive 10 for b even though the actual coefficient is -10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=3: 2(3)²+(-10)(3)+(12)=0. Substitute x=2: 2(2)²+(-10)(2)+(12)=0. This checks the candidates against the original, not merely the formula line.

34. Exact-root clinic 34: −2x² − 14x − 20=0

Common trap: A pupil may use positive 14 for b even though the actual coefficient is -14. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: -2(-5)²+(-14)(-5)+(-20)=0. Substitute x=-2: -2(-2)²+(-14)(-2)+(-20)=0. This checks the candidates against the original, not merely the formula line.

35. Exact-root clinic 35: 3x² − 9x − 30=0

Common trap: A pupil may use positive 9 for b even though the actual coefficient is -9. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: 3(-2)²+(-9)(-2)+(-30)=0. Substitute x=5: 3(5)²+(-9)(5)+(-30)=0. This checks the candidates against the original, not merely the formula line.

36. Exact-root clinic 36: −3x² + 6x − 3=0

Common trap: A pupil may use positive 6 for b even though the actual coefficient is 6. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: -3(1)²+(6)(1)+(-3)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

37. Exact-root clinic 37: x² − x − 12=0

Common trap: A pupil may use positive 1 for b even though the actual coefficient is -1. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: 1(4)²+(-1)(4)+(-12)=0. Substitute x=-3: 1(-3)²+(-1)(-3)+(-12)=0. This checks the candidates against the original, not merely the formula line.

38. Exact-root clinic 38: −x² + 16=0

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: -1(-4)²+(0)(-4)+(16)=0. Substitute x=4: -1(4)²+(0)(4)+(16)=0. This checks the candidates against the original, not merely the formula line.

39. Exact-root clinic 39: 2x² + 2x=0

Worked explanation: The signed coefficients are a=2, b=2, c=0. The quantity inside the radical is (2)²−4(2)(0)=4. The formula gives x=[-2±√4]/(4)=[-2±2]/(4). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: 2(-1)²+(2)(-1)+(0)=0. Substitute x=0: 2(0)²+(2)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

40. Exact-root clinic 40: −2x² − 4x + 16=0

Common trap: A pupil may use positive 4 for b even though the actual coefficient is -4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: -2(2)²+(-4)(2)+(16)=0. Substitute x=-4: -2(-4)²+(-4)(-4)+(16)=0. This checks the candidates against the original, not merely the formula line.

41. Exact-root clinic 41: 3x² − 24x + 45=0

Common trap: A pupil may use positive 24 for b even though the actual coefficient is -24. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: 3(5)²+(-24)(5)+(45)=0. Substitute x=3: 3(3)²+(-24)(3)+(45)=0. This checks the candidates against the original, not merely the formula line.

42. Exact-root clinic 42: −3x² − 12x − 9=0

Common trap: A pupil may use positive 12 for b even though the actual coefficient is -12. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: -3(-3)²+(-12)(-3)+(-9)=0. Substitute x=-1: -3(-1)²+(-12)(-1)+(-9)=0. This checks the candidates against the original, not merely the formula line.

43. Non-perfect-square clinic 1: x² − 7x − 7=0

Worked explanation: Take a=1, b=-7, c=-7. Compute b²−4ac=(-7)²−4(1)(-7)=77. The roots are x=[7±√77]/(2), giving x≈7.89 and x≈-0.887 to three significant figures.

Common trap: Replacing √77 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

44. Non-perfect-square clinic 2: x² − 7x − 4=0

Worked explanation: Take a=1, b=-7, c=-4. Compute b²−4ac=(-7)²−4(1)(-4)=65. The roots are x=[7±√65]/(2), giving x≈7.53 and x≈-0.531 to three significant figures.

Common trap: Replacing √65 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

45. Non-perfect-square clinic 3: x² − 7x − 1=0

Worked explanation: Take a=1, b=-7, c=-1. Compute b²−4ac=(-7)²−4(1)(-1)=53. The roots are x=[7±√53]/(2), giving x≈7.14 and x≈-0.140 to three significant figures.

Common trap: Replacing √53 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

46. Non-perfect-square clinic 4: x² − 7x + 2=0

Worked explanation: Take a=1, b=-7, c=2. Compute b²−4ac=(-7)²−4(1)(2)=41. The roots are x=[7±√41]/(2), giving x≈6.70 and x≈0.298 to three significant figures.

Common trap: Replacing √41 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

47. Non-perfect-square clinic 5: x² − 7x + 4=0

Worked explanation: Take a=1, b=-7, c=4. Compute b²−4ac=(-7)²−4(1)(4)=33. The roots are x=[7±√33]/(2), giving x≈6.37 and x≈0.628 to three significant figures.

Common trap: Replacing √33 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

48. Non-perfect-square clinic 6: x² − 7x + 7=0

Worked explanation: Take a=1, b=-7, c=7. Compute b²−4ac=(-7)²−4(1)(7)=21. The roots are x=[7±√21]/(2), giving x≈5.79 and x≈1.21 to three significant figures.

Common trap: Replacing √21 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

49. Non-perfect-square clinic 7: x² − 7x + 9=0

Worked explanation: Take a=1, b=-7, c=9. Compute b²−4ac=(-7)²−4(1)(9)=13. The roots are x=[7±√13]/(2), giving x≈5.30 and x≈1.70 to three significant figures.

Common trap: Replacing √13 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

50. Non-perfect-square clinic 8: x² − 5x − 7=0

Worked explanation: Take a=1, b=-5, c=-7. Compute b²−4ac=(-5)²−4(1)(-7)=53. The roots are x=[5±√53]/(2), giving x≈6.14 and x≈-1.14 to three significant figures.

Common trap: Replacing √53 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

51. Non-perfect-square clinic 9: x² − 5x − 4=0

Worked explanation: Take a=1, b=-5, c=-4. Compute b²−4ac=(-5)²−4(1)(-4)=41. The roots are x=[5±√41]/(2), giving x≈5.70 and x≈-0.702 to three significant figures.

Common trap: Replacing √41 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

52. Non-perfect-square clinic 10: x² − 5x − 1=0

Worked explanation: Take a=1, b=-5, c=-1. Compute b²−4ac=(-5)²−4(1)(-1)=29. The roots are x=[5±√29]/(2), giving x≈5.19 and x≈-0.193 to three significant figures.

Common trap: Replacing √29 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

53. Non-perfect-square clinic 11: x² − 5x + 2=0

Worked explanation: Take a=1, b=-5, c=2. Compute b²−4ac=(-5)²−4(1)(2)=17. The roots are x=[5±√17]/(2), giving x≈4.56 and x≈0.438 to three significant figures.

Common trap: Replacing √17 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

54. Non-perfect-square clinic 12: x² − 3x − 7=0

Worked explanation: Take a=1, b=-3, c=-7. Compute b²−4ac=(-3)²−4(1)(-7)=37. The roots are x=[3±√37]/(2), giving x≈4.54 and x≈-1.54 to three significant figures.

Common trap: Replacing √37 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

55. Non-perfect-square clinic 13: x² − 3x − 1=0

Worked explanation: Take a=1, b=-3, c=-1. Compute b²−4ac=(-3)²−4(1)(-1)=13. The roots are x=[3±√13]/(2), giving x≈3.30 and x≈-0.303 to three significant figures.

Common trap: Replacing √13 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

56. Non-perfect-square clinic 14: x² + x − 7=0

Worked explanation: Take a=1, b=1, c=-7. Compute b²−4ac=(1)²−4(1)(-7)=29. The roots are x=[-1±√29]/(2), giving x≈2.19 and x≈-3.19 to three significant figures.

Common trap: Replacing √29 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

57. Non-perfect-square clinic 15: x² + x − 4=0

Worked explanation: Take a=1, b=1, c=-4. Compute b²−4ac=(1)²−4(1)(-4)=17. The roots are x=[-1±√17]/(2), giving x≈1.56 and x≈-2.56 to three significant figures.

Common trap: Replacing √17 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

58. Non-perfect-square clinic 16: x² + x − 1=0

Worked explanation: Take a=1, b=1, c=-1. Compute b²−4ac=(1)²−4(1)(-1)=5. The roots are x=[-1±√5]/(2), giving x≈0.618 and x≈-1.62 to three significant figures.

Common trap: Replacing √5 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

59. Non-perfect-square clinic 17: x² + 2x − 7=0

Worked explanation: Take a=1, b=2, c=-7. Compute b²−4ac=(2)²−4(1)(-7)=32. The roots are x=[-2±√32]/(2), giving x≈1.83 and x≈-3.83 to three significant figures.

Common trap: Replacing √32 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

60. Non-perfect-square clinic 18: x² + 2x − 4=0

Worked explanation: Take a=1, b=2, c=-4. Compute b²−4ac=(2)²−4(1)(-4)=20. The roots are x=[-2±√20]/(2), giving x≈1.24 and x≈-3.24 to three significant figures.

Common trap: Replacing √20 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

61. Non-perfect-square clinic 19: x² + 2x − 1=0

Worked explanation: Take a=1, b=2, c=-1. Compute b²−4ac=(2)²−4(1)(-1)=8. The roots are x=[-2±√8]/(2), giving x≈0.414 and x≈-2.41 to three significant figures.

Common trap: Replacing √8 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

62. Non-perfect-square clinic 20: x² + 4x − 7=0

Worked explanation: Take a=1, b=4, c=-7. Compute b²−4ac=(4)²−4(1)(-7)=44. The roots are x=[-4±√44]/(2), giving x≈1.32 and x≈-5.32 to three significant figures.

Common trap: Replacing √44 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

63. Non-perfect-square clinic 21: x² + 4x − 4=0

Worked explanation: Take a=1, b=4, c=-4. Compute b²−4ac=(4)²−4(1)(-4)=32. The roots are x=[-4±√32]/(2), giving x≈0.828 and x≈-4.83 to three significant figures.

Common trap: Replacing √32 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

64. Non-perfect-square clinic 22: x² + 4x − 1=0

Worked explanation: Take a=1, b=4, c=-1. Compute b²−4ac=(4)²−4(1)(-1)=20. The roots are x=[-4±√20]/(2), giving x≈0.236 and x≈-4.24 to three significant figures.

Common trap: Replacing √20 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

65. Non-perfect-square clinic 23: x² + 4x + 2=0

Worked explanation: Take a=1, b=4, c=2. Compute b²−4ac=(4)²−4(1)(2)=8. The roots are x=[-4±√8]/(2), giving x≈-0.586 and x≈-3.41 to three significant figures.

Common trap: Replacing √8 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

66. Non-perfect-square clinic 24: x² + 6x − 4=0

Worked explanation: Take a=1, b=6, c=-4. Compute b²−4ac=(6)²−4(1)(-4)=52. The roots are x=[-6±√52]/(2), giving x≈0.606 and x≈-6.61 to three significant figures.

Common trap: Replacing √52 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

67. No-real-root case 1: x² − 3x + 4=0

Common trap: Taking the absolute value of -7 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

68. No-real-root case 2: 2x² − 2x + 3=0

Common trap: Taking the absolute value of -20 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=2 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

69. No-real-root case 3: 3x² − x + 4=0

Common trap: Taking the absolute value of -47 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=3 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

70. No-real-root case 4: x² + 5=0

Common trap: Taking the absolute value of -20 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

71. No-real-root case 5: 2x² + x + 6=0

Common trap: Taking the absolute value of -47 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=2 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

72. No-real-root case 6: 3x² + 2x + 7=0

Common trap: Taking the absolute value of -80 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=3 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

73. No-real-root case 7: x² + 3x + 10=0

Common trap: Taking the absolute value of -31 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

74. No-real-root case 8: 2x² − 3x + 10=0

Common trap: Taking the absolute value of -71 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=2 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

75. No-real-root case 9: 3x² − 2x + 10=0

Common trap: Taking the absolute value of -116 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=3 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

76. No-real-root case 10: x² − x + 11=0

Common trap: Taking the absolute value of -43 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

77. Interpretation and checking 1: Write standard form before inserting coefficients

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. The equation x²=5x−6 is not yet in the form ax²+bx+c=0.

Worked explanation: Move everything to the left: x²−5x+6=0. Then a=1,b=−5,c=6 and roots are 2 and 3.

Common trap: Reading b as +5 because the 5x appears on the right rather than changing its sign.

Verify: Substitute two and three into x²=5x−6 to verify both.

78. Interpretation and checking 2: The formula requires a nonzero quadratic coefficient

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. A question presents 0x²+3x−9=0. Is the quadratic formula appropriate?

Worked explanation: No. The coefficient a=0 leaves a linear equation, and 3x−9=0 gives x=3.

Common trap: Using the formula with denominator 2a=0 would create undefined division.

Verify: Identify the highest nonzero power before selecting a method.

79. Interpretation and checking 3: The full numerator is divided by 2a

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. How should x=(−b±√(b²−4ac))/(2a) be keyed or written clearly?

Worked explanation: Write the entire numerator in parentheses before dividing by the full denominator (2a). The plus and minus are separate branches.

Common trap: Writing −b+√D/2a can divide only the radical instead of the whole numerator.

Verify: Test an equation with known roots to confirm the input layout.

80. Interpretation and checking 4: The sign of b affects two parts

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Identify a,b,c in 2x²−5x+2=0.

Worked explanation: a=2,b=−5,c=2. Hence −b=5 while b²=25. The formula returns x=2 or x=1/2.

Common trap: Writing b=+5 changes −b and can reverse the final candidates.

Verify: Check both root candidates in the original equation.

81. Interpretation and checking 5: A repeated root is one distinct real answer

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x²−6x+9=0.

Worked explanation: The radical term has 36−36=0, so x=(6±0)/2=3. It is a repeated real root.

Common trap: A learner may invent two different answers merely because the formula has ±.

Verify: Substitute x=3 and see 9−18+9=0.

82. Interpretation and checking 6: The square-root symbol has a principal value

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. What differs between √9 and solving x²=9?

Worked explanation: √9=3, the nonnegative principal square root. The equation x²=9 has x=3 and x=−3.

Common trap: Calling √9 itself ±3 conflates the expression and the equation.

Verify: Both +3 and −3 square to nine, but √9 denotes +3.

83. Interpretation and checking 7: A negative leading coefficient changes the denominator

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x²+4x−3=0.

Worked explanation: a=−1,b=4,c=−3, so D=16−12=4 and denominator is −2. Roots are x=1 and x=3.

Common trap: Using +2 in the denominator makes both signs incorrect.

Verify: Check −1+4−3=0 and −9+12−3=0.

84. Interpretation and checking 8: A missing constant has value zero

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Find a,b,c for 3x²−2x=0.

Worked explanation: a=3,b=−2,c=0. Formula yields roots x=0 and 2/3.

Common trap: Inventing a constant of one because there are three labels a,b,c.

Verify: Substitute zero and two-thirds into the original.

85. Interpretation and checking 9: Some roots are simple rational fractions

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x²−3x+1=0.

Worked explanation: D=9−8=1. x=(3±1)/4, so roots are 1 and 1/2.

Common trap: Dividing only one numerator term by four gives incorrect results.

Verify: Verify x=1 and x=1/2 exactly.

86. Interpretation and checking 10: An upward-opening quadratic need not cross the axis

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Does x²+4=0 have real solutions?

Worked explanation: No. D=−16 is negative, and the graph y=x²+4 is always at least four.

Common trap: Claiming every quadratic must have two real roots.

Verify: Its lowest graph point is (0,4), above the axis.

87. Interpretation and checking 11: A graph’s crossings are equation roots

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. For y=x²−3x+2, where are the x-intercepts?

Worked explanation: Set y=0 and solve: roots are x=1 and x=2. Thus crossings are (1,0) and (2,0).

Common trap: Mistaking the y-intercept (0,2) for a solution of y=0.

Verify: At both x=1 and x=2, the output is zero.

88. Interpretation and checking 12: Do not round the root expression too early

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Approximate (3+√5)/2 to three significant figures.

Worked explanation: The unrounded value is about 2.6180339; report 2.62 to three significant figures.

Common trap: Rounding √5 to 2 before finishing produces 2.5, a poor approximation.

Verify: Use unrounded intermediate values then round final display.

89. Interpretation and checking 13: A root is a candidate until it is checked

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Check x=2 in x²−5x+6=0.

Worked explanation: Substitute 2²−5(2)+6=4−10+6=0. So two is a solution.

Common trap: Accepting a written x-value without testing the original equality.

Verify: Test x=3 as well; it is a second solution.

90. Interpretation and checking 14: A negative root is not necessarily a mistake

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x²+5x+6=0.

Worked explanation: a=1,b=5,c=6 and D=25−24=1; formula gives (−5±1)/2, so roots −2 and −3.

Common trap: Assuming positive leading a guarantees positive roots.

Verify: Substitute both negative candidates.

91. Interpretation and checking 15: An exact radical and a rounded value are different outputs

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. A formula produces x=(1+√3)/2. Is x=1.37 exact?

Worked explanation: No. The expression with √3 is exact; 1.37 is a rounded approximation to three significant figures.

Common trap: Dropping the approximation sign conceals the loss of precision.

Verify: Calculate with full precision when checking the original equation.

92. Interpretation and checking 16: A possible real-world constraint can reject one root

Worked explanation: Only five is positive, assuming no additional constraint. The negative algebraic root is mathematically valid but not physically feasible for width.

Common trap: Deleting the negative root without first reporting it as a mathematical solution may confuse the two stages.

Verify: State algebraic answers first, then apply the context restriction.

93. Interpretation and checking 17: The 2027 K110 formulae list contains the quadratic formula

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Is a guessed mnemonic preferable to checking the official formulae sheet?

Worked explanation: No. The published 2027 K110 Number and Algebra formulae list contains x=(−b±√(b²−4ac))/(2a). The essential skill is accurate substitution and interpretation.

Common trap: Assuming every school’s current formula sheet is identical to an older unofficial screenshot.

Verify: Check official documentation for the student’s own examination year.

94. Interpretation and checking 18: A formula result should be shown clearly

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. A learner writes x=3 or −2 without stating which equation was solved.

Worked explanation: The result may be mathematically correct but should be tied to a stated equation, with enough working to show a,b,c and a valid calculation.

Common trap: Copying two numbers from an answer key is not evidence the method was understood.

Verify: A changed-number problem and a direct substitution check provide evidence.

95. Interpretation and checking 19: Two negative factors under b² remain positive

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. In x²+4x+1=0, what is b²?

Worked explanation: b=4, so b²=16. For b=−4 in x²−4x+1, b² is still 16.

Common trap: Writing −16 after squaring a negative coefficient fails to bracket the full quantity.

Verify: Compare (−4)² with −(4²), which are not equal.

96. Interpretation and checking 20: One solution repeated when D=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 4x²+4x+1=0.

Worked explanation: D=16−16=0; x=−4/8=−1/2. The two branches coincide.

Common trap: Writing +1/2 after losing the negative numerator.

Verify: Check 4×1/4+4×(−1/2)+1=0.

Five different errors that appear as a bad final answer

The equation was never written in standard form

b was entered without its negative sign

The plus/minus was treated as optional

The denominator was applied to only half the numerator

Rounding made the check seem to fail

An illustrative three-pupil bridge lesson

10 minutes: review signed multiplication, roots and simple equation balance.
20 minutes: introduce the formula using the official sheet and practise coefficient lists.
25 minutes: solve a few exact-root questions, including both ± branches.
15 minutes: distinguish repeated-root and no-real-root cases.
10 minutes: independently solve one changed equation with no visible model.
10 minutes: correct the first sign or bracket error and assign a short later retest.

A five-session parent–student routine

Day one: practise negative squares and bring simple equations into zero-equals form.
Day two: identify a, b and c in five varied equations.
Day three: solve two exact-root cases from the formula and verify both answers.
Day four: examine one repeated root, one negative radicand and one irrational pair.
Day five: solve a new problem and explain the corresponding graph’s intercept behaviour.

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

Yes. The published 2027 K110 Mathematical Formulae section includes the quadratic-equation formula. Check official requirements for the student’s own eventual examination year.

What if b²−4ac is negative?

There are no real roots. The associated quadratic graph has no real x-intercept.

Why can the formula return two roots?

The ± yields two possible numerator values when the radicand is positive. Each should be checked in the original equation.

Can one root be a fraction or a negative number?

Yes. Roots depend on the signed coefficients. Do not reject a result merely because it is not a positive whole number; substitute it back to check.

Will a 90-minute group guarantee success?

No. The appropriate progression depends on prerequisite knowledge, feedback and independent retention. Ask the tutor how readiness and later transfer are assessed.

Official references and editorial note

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

Did you know? A quadratic equation can have two different real solutions, only one repeated real solution, or no real solution at all. The plus-or-minus symbol in the formula is a way to investigate these possibilities, not a guarantee that two different numbers must appear. For a Secondary 2 G1 student, this is an optional later-stage bridge from one-variable equation balance, powers and factorisation.

Quick answer: quadratic formula in one line

For ax²+bx+c=0 with nonzero a, use x=(−b±√(b²−4ac))/(2a). Rearrange the original equation into that exact form. Record the signed coefficients a, b and c. Compute the quantity under the radical without losing parentheses. Evaluate both branches when it has a real square root, then verify candidates in the original equation. Keep sufficient precision until the required final rounding.

Curriculum accuracy: The MOE 2020 G1 Mathematics Secondary Two content lists linear equations in one variable. The 2027 SEC K110 G1 Mathematics graduating syllabus additionally includes quadratic equations solved by the formula, and its Mathematical Formulae section provides that formula. This article is a targeted next-stage resource for ready pupils, not a claim that a 2026 Secondary 2 G1 school paper includes the quadratic formula.

The 60-second readiness diagnostic

Read the equation before typing the formula

First get zero on one side

The negative of b is not necessarily negative

Everything above the fraction bar shares one denominator

How the quantity under the root changes the answer

96 original quadratic formula learning clinics

1. Exact-root clinic 1: x² + 7x + 10=0

Worked explanation: The signed coefficients are a=1, b=7, c=10. The quantity inside the radical is (7)²−4(1)(10)=9. The formula gives x=[-7±√9]/(2)=[-7±3]/(2). Therefore x=-5 or x=-2.

Common trap: A pupil may use positive 7 for b even though the actual coefficient is 7. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: 1(-5)²+(7)(-5)+(10)=0. Substitute x=-2: 1(-2)²+(7)(-2)+(10)=0. This checks the candidates against the original, not merely the formula line.

2. Exact-root clinic 2: −x² + 3x + 10=0

Common trap: A pupil may use positive 3 for b even though the actual coefficient is 3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: -1(-2)²+(3)(-2)+(10)=0. Substitute x=5: -1(5)²+(3)(5)+(10)=0. This checks the candidates against the original, not merely the formula line.

3. Exact-root clinic 3: 2x² − 4x + 2=0

Common trap: A pupil may use positive 4 for b even though the actual coefficient is -4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: 2(1)²+(-4)(1)+(2)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

4. Exact-root clinic 4: −2x² + 2x + 24=0

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: -2(4)²+(2)(4)+(24)=0. Substitute x=-3: -2(-3)²+(2)(-3)+(24)=0. This checks the candidates against the original, not merely the formula line.

5. Exact-root clinic 5: 3x² − 48=0

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: 3(-4)²+(0)(-4)+(-48)=0. Substitute x=4: 3(4)²+(0)(4)+(-48)=0. This checks the candidates against the original, not merely the formula line.

6. Exact-root clinic 6: −3x² − 3x=0

Worked explanation: The signed coefficients are a=-3, b=-3, c=0. The quantity inside the radical is (-3)²−4(-3)(0)=9. The formula gives x=[3±√9]/(-6)=[3±3]/(-6). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 3 for b even though the actual coefficient is -3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: -3(-1)²+(-3)(-1)+(0)=0. Substitute x=0: -3(0)²+(-3)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

7. Exact-root clinic 7: x² + 2x − 8=0

Worked explanation: The signed coefficients are a=1, b=2, c=-8. The quantity inside the radical is (2)²−4(1)(-8)=36. The formula gives x=[-2±√36]/(2)=[-2±6]/(2). Therefore x=2 or x=-4.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: 1(2)²+(2)(2)+(-8)=0. Substitute x=-4: 1(-4)²+(2)(-4)+(-8)=0. This checks the candidates against the original, not merely the formula line.

8. Exact-root clinic 8: −x² + 8x − 15=0

Common trap: A pupil may use positive 8 for b even though the actual coefficient is 8. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: -1(5)²+(8)(5)+(-15)=0. Substitute x=3: -1(3)²+(8)(3)+(-15)=0. This checks the candidates against the original, not merely the formula line.

9. Exact-root clinic 9: 2x² + 8x + 6=0

Worked explanation: The signed coefficients are a=2, b=8, c=6. The quantity inside the radical is (8)²−4(2)(6)=16. The formula gives x=[-8±√16]/(4)=[-8±4]/(4). Therefore x=-3 or x=-1.

Common trap: A pupil may use positive 8 for b even though the actual coefficient is 8. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: 2(-3)²+(8)(-3)+(6)=0. Substitute x=-1: 2(-1)²+(8)(-1)+(6)=0. This checks the candidates against the original, not merely the formula line.

10. Exact-root clinic 10: −2x² − 10x=0

Common trap: A pupil may use positive 10 for b even though the actual coefficient is -10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=0: -2(0)²+(-10)(0)+(0)=0. Substitute x=-5: -2(-5)²+(-10)(-5)+(0)=0. This checks the candidates against the original, not merely the formula line.

11. Exact-root clinic 11: 3x² − 15x + 18=0

Common trap: A pupil may use positive 15 for b even though the actual coefficient is -15. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=3: 3(3)²+(-15)(3)+(18)=0. Substitute x=2: 3(2)²+(-15)(2)+(18)=0. This checks the candidates against the original, not merely the formula line.

12. Exact-root clinic 12: −3x² − 21x − 30=0

Common trap: A pupil may use positive 21 for b even though the actual coefficient is -21. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: -3(-5)²+(-21)(-5)+(-30)=0. Substitute x=-2: -3(-2)²+(-21)(-2)+(-30)=0. This checks the candidates against the original, not merely the formula line.

13. Exact-root clinic 13: x² − 3x − 10=0

Common trap: A pupil may use positive 3 for b even though the actual coefficient is -3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: 1(-2)²+(-3)(-2)+(-10)=0. Substitute x=5: 1(5)²+(-3)(5)+(-10)=0. This checks the candidates against the original, not merely the formula line.

14. Exact-root clinic 14: −x² + 2x − 1=0

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: -1(1)²+(2)(1)+(-1)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

15. Exact-root clinic 15: 2x² − 2x − 24=0

Common trap: A pupil may use positive 2 for b even though the actual coefficient is -2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: 2(4)²+(-2)(4)+(-24)=0. Substitute x=-3: 2(-3)²+(-2)(-3)+(-24)=0. This checks the candidates against the original, not merely the formula line.

16. Exact-root clinic 16: −2x² + 32=0

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: -2(-4)²+(0)(-4)+(32)=0. Substitute x=4: -2(4)²+(0)(4)+(32)=0. This checks the candidates against the original, not merely the formula line.

17. Exact-root clinic 17: 3x² + 3x=0

Worked explanation: The signed coefficients are a=3, b=3, c=0. The quantity inside the radical is (3)²−4(3)(0)=9. The formula gives x=[-3±√9]/(6)=[-3±3]/(6). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 3 for b even though the actual coefficient is 3. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: 3(-1)²+(3)(-1)+(0)=0. Substitute x=0: 3(0)²+(3)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

18. Exact-root clinic 18: −3x² − 6x + 24=0

Common trap: A pupil may use positive 6 for b even though the actual coefficient is -6. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: -3(2)²+(-6)(2)+(24)=0. Substitute x=-4: -3(-4)²+(-6)(-4)+(24)=0. This checks the candidates against the original, not merely the formula line.

19. Exact-root clinic 19: x² − 8x + 15=0

Worked explanation: The signed coefficients are a=1, b=-8, c=15. The quantity inside the radical is (-8)²−4(1)(15)=4. The formula gives x=[8±√4]/(2)=[8±2]/(2). Therefore x=5 or x=3.

Common trap: A pupil may use positive 8 for b even though the actual coefficient is -8. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: 1(5)²+(-8)(5)+(15)=0. Substitute x=3: 1(3)²+(-8)(3)+(15)=0. This checks the candidates against the original, not merely the formula line.

20. Exact-root clinic 20: −x² − 4x − 3=0

Common trap: A pupil may use positive 4 for b even though the actual coefficient is -4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: -1(-3)²+(-4)(-3)+(-3)=0. Substitute x=-1: -1(-1)²+(-4)(-1)+(-3)=0. This checks the candidates against the original, not merely the formula line.

21. Exact-root clinic 21: 2x² + 10x=0

Common trap: A pupil may use positive 10 for b even though the actual coefficient is 10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=0: 2(0)²+(10)(0)+(0)=0. Substitute x=-5: 2(-5)²+(10)(-5)+(0)=0. This checks the candidates against the original, not merely the formula line.

22. Exact-root clinic 22: −2x² + 10x − 12=0

Common trap: A pupil may use positive 10 for b even though the actual coefficient is 10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=3: -2(3)²+(10)(3)+(-12)=0. Substitute x=2: -2(2)²+(10)(2)+(-12)=0. This checks the candidates against the original, not merely the formula line.

23. Exact-root clinic 23: 3x² + 21x + 30=0

Common trap: A pupil may use positive 21 for b even though the actual coefficient is 21. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: 3(-5)²+(21)(-5)+(30)=0. Substitute x=-2: 3(-2)²+(21)(-2)+(30)=0. This checks the candidates against the original, not merely the formula line.

24. Exact-root clinic 24: −3x² + 9x + 30=0

Common trap: A pupil may use positive 9 for b even though the actual coefficient is 9. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: -3(-2)²+(9)(-2)+(30)=0. Substitute x=5: -3(5)²+(9)(5)+(30)=0. This checks the candidates against the original, not merely the formula line.

25. Exact-root clinic 25: x² − 2x + 1=0

Common trap: A pupil may use positive 2 for b even though the actual coefficient is -2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: 1(1)²+(-2)(1)+(1)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

26. Exact-root clinic 26: −x² + x + 12=0

Common trap: A pupil may use positive 1 for b even though the actual coefficient is 1. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: -1(4)²+(1)(4)+(12)=0. Substitute x=-3: -1(-3)²+(1)(-3)+(12)=0. This checks the candidates against the original, not merely the formula line.

27. Exact-root clinic 27: 2x² − 32=0

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: 2(-4)²+(0)(-4)+(-32)=0. Substitute x=4: 2(4)²+(0)(4)+(-32)=0. This checks the candidates against the original, not merely the formula line.

28. Exact-root clinic 28: −2x² − 2x=0

Worked explanation: The signed coefficients are a=-2, b=-2, c=0. The quantity inside the radical is (-2)²−4(-2)(0)=4. The formula gives x=[2±√4]/(-4)=[2±2]/(-4). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is -2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: -2(-1)²+(-2)(-1)+(0)=0. Substitute x=0: -2(0)²+(-2)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

29. Exact-root clinic 29: 3x² + 6x − 24=0

Common trap: A pupil may use positive 6 for b even though the actual coefficient is 6. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: 3(2)²+(6)(2)+(-24)=0. Substitute x=-4: 3(-4)²+(6)(-4)+(-24)=0. This checks the candidates against the original, not merely the formula line.

30. Exact-root clinic 30: −3x² + 24x − 45=0

Common trap: A pupil may use positive 24 for b even though the actual coefficient is 24. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: -3(5)²+(24)(5)+(-45)=0. Substitute x=3: -3(3)²+(24)(3)+(-45)=0. This checks the candidates against the original, not merely the formula line.

31. Exact-root clinic 31: x² + 4x + 3=0

Worked explanation: The signed coefficients are a=1, b=4, c=3. The quantity inside the radical is (4)²−4(1)(3)=4. The formula gives x=[-4±√4]/(2)=[-4±2]/(2). Therefore x=-3 or x=-1.

Common trap: A pupil may use positive 4 for b even though the actual coefficient is 4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: 1(-3)²+(4)(-3)+(3)=0. Substitute x=-1: 1(-1)²+(4)(-1)+(3)=0. This checks the candidates against the original, not merely the formula line.

32. Exact-root clinic 32: −x² − 5x=0

Common trap: A pupil may use positive 5 for b even though the actual coefficient is -5. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=0: -1(0)²+(-5)(0)+(0)=0. Substitute x=-5: -1(-5)²+(-5)(-5)+(0)=0. This checks the candidates against the original, not merely the formula line.

33. Exact-root clinic 33: 2x² − 10x + 12=0

Common trap: A pupil may use positive 10 for b even though the actual coefficient is -10. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=3: 2(3)²+(-10)(3)+(12)=0. Substitute x=2: 2(2)²+(-10)(2)+(12)=0. This checks the candidates against the original, not merely the formula line.

34. Exact-root clinic 34: −2x² − 14x − 20=0

Common trap: A pupil may use positive 14 for b even though the actual coefficient is -14. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-5: -2(-5)²+(-14)(-5)+(-20)=0. Substitute x=-2: -2(-2)²+(-14)(-2)+(-20)=0. This checks the candidates against the original, not merely the formula line.

35. Exact-root clinic 35: 3x² − 9x − 30=0

Common trap: A pupil may use positive 9 for b even though the actual coefficient is -9. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-2: 3(-2)²+(-9)(-2)+(-30)=0. Substitute x=5: 3(5)²+(-9)(5)+(-30)=0. This checks the candidates against the original, not merely the formula line.

36. Exact-root clinic 36: −3x² + 6x − 3=0

Common trap: A pupil may use positive 6 for b even though the actual coefficient is 6. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=1: -3(1)²+(6)(1)+(-3)=0. Because the two roots coincide, the ± branches return the same value. This checks the candidates against the original, not merely the formula line.

37. Exact-root clinic 37: x² − x − 12=0

Common trap: A pupil may use positive 1 for b even though the actual coefficient is -1. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=4: 1(4)²+(-1)(4)+(-12)=0. Substitute x=-3: 1(-3)²+(-1)(-3)+(-12)=0. This checks the candidates against the original, not merely the formula line.

38. Exact-root clinic 38: −x² + 16=0

Common trap: A pupil may use positive 0 for b even though the actual coefficient is 0. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-4: -1(-4)²+(0)(-4)+(16)=0. Substitute x=4: -1(4)²+(0)(4)+(16)=0. This checks the candidates against the original, not merely the formula line.

39. Exact-root clinic 39: 2x² + 2x=0

Worked explanation: The signed coefficients are a=2, b=2, c=0. The quantity inside the radical is (2)²−4(2)(0)=4. The formula gives x=[-2±√4]/(4)=[-2±2]/(4). Therefore x=-1 or x=0.

Common trap: A pupil may use positive 2 for b even though the actual coefficient is 2. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-1: 2(-1)²+(2)(-1)+(0)=0. Substitute x=0: 2(0)²+(2)(0)+(0)=0. This checks the candidates against the original, not merely the formula line.

40. Exact-root clinic 40: −2x² − 4x + 16=0

Common trap: A pupil may use positive 4 for b even though the actual coefficient is -4. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=2: -2(2)²+(-4)(2)+(16)=0. Substitute x=-4: -2(-4)²+(-4)(-4)+(16)=0. This checks the candidates against the original, not merely the formula line.

41. Exact-root clinic 41: 3x² − 24x + 45=0

Common trap: A pupil may use positive 24 for b even though the actual coefficient is -24. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=5: 3(5)²+(-24)(5)+(45)=0. Substitute x=3: 3(3)²+(-24)(3)+(45)=0. This checks the candidates against the original, not merely the formula line.

42. Exact-root clinic 42: −3x² − 12x − 9=0

Common trap: A pupil may use positive 12 for b even though the actual coefficient is -12. If a is negative, they might also use a positive denominator. Both change the equation they are solving.

Verify: Substitute x=-3: -3(-3)²+(-12)(-3)+(-9)=0. Substitute x=-1: -3(-1)²+(-12)(-1)+(-9)=0. This checks the candidates against the original, not merely the formula line.

43. Non-perfect-square clinic 1: x² − 7x − 7=0

Worked explanation: Take a=1, b=-7, c=-7. Compute b²−4ac=(-7)²−4(1)(-7)=77. The roots are x=[7±√77]/(2), giving x≈7.89 and x≈-0.887 to three significant figures.

Common trap: Replacing √77 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

44. Non-perfect-square clinic 2: x² − 7x − 4=0

Worked explanation: Take a=1, b=-7, c=-4. Compute b²−4ac=(-7)²−4(1)(-4)=65. The roots are x=[7±√65]/(2), giving x≈7.53 and x≈-0.531 to three significant figures.

Common trap: Replacing √65 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

45. Non-perfect-square clinic 3: x² − 7x − 1=0

Worked explanation: Take a=1, b=-7, c=-1. Compute b²−4ac=(-7)²−4(1)(-1)=53. The roots are x=[7±√53]/(2), giving x≈7.14 and x≈-0.140 to three significant figures.

Common trap: Replacing √53 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

46. Non-perfect-square clinic 4: x² − 7x + 2=0

Worked explanation: Take a=1, b=-7, c=2. Compute b²−4ac=(-7)²−4(1)(2)=41. The roots are x=[7±√41]/(2), giving x≈6.70 and x≈0.298 to three significant figures.

Common trap: Replacing √41 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

47. Non-perfect-square clinic 5: x² − 7x + 4=0

Worked explanation: Take a=1, b=-7, c=4. Compute b²−4ac=(-7)²−4(1)(4)=33. The roots are x=[7±√33]/(2), giving x≈6.37 and x≈0.628 to three significant figures.

Common trap: Replacing √33 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

48. Non-perfect-square clinic 6: x² − 7x + 7=0

Worked explanation: Take a=1, b=-7, c=7. Compute b²−4ac=(-7)²−4(1)(7)=21. The roots are x=[7±√21]/(2), giving x≈5.79 and x≈1.21 to three significant figures.

Common trap: Replacing √21 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

49. Non-perfect-square clinic 7: x² − 7x + 9=0

Worked explanation: Take a=1, b=-7, c=9. Compute b²−4ac=(-7)²−4(1)(9)=13. The roots are x=[7±√13]/(2), giving x≈5.30 and x≈1.70 to three significant figures.

Common trap: Replacing √13 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

50. Non-perfect-square clinic 8: x² − 5x − 7=0

Worked explanation: Take a=1, b=-5, c=-7. Compute b²−4ac=(-5)²−4(1)(-7)=53. The roots are x=[5±√53]/(2), giving x≈6.14 and x≈-1.14 to three significant figures.

Common trap: Replacing √53 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

51. Non-perfect-square clinic 9: x² − 5x − 4=0

Worked explanation: Take a=1, b=-5, c=-4. Compute b²−4ac=(-5)²−4(1)(-4)=41. The roots are x=[5±√41]/(2), giving x≈5.70 and x≈-0.702 to three significant figures.

Common trap: Replacing √41 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

52. Non-perfect-square clinic 10: x² − 5x − 1=0

Worked explanation: Take a=1, b=-5, c=-1. Compute b²−4ac=(-5)²−4(1)(-1)=29. The roots are x=[5±√29]/(2), giving x≈5.19 and x≈-0.193 to three significant figures.

Common trap: Replacing √29 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

53. Non-perfect-square clinic 11: x² − 5x + 2=0

Worked explanation: Take a=1, b=-5, c=2. Compute b²−4ac=(-5)²−4(1)(2)=17. The roots are x=[5±√17]/(2), giving x≈4.56 and x≈0.438 to three significant figures.

Common trap: Replacing √17 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

54. Non-perfect-square clinic 12: x² − 3x − 7=0

Worked explanation: Take a=1, b=-3, c=-7. Compute b²−4ac=(-3)²−4(1)(-7)=37. The roots are x=[3±√37]/(2), giving x≈4.54 and x≈-1.54 to three significant figures.

Common trap: Replacing √37 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

55. Non-perfect-square clinic 13: x² − 3x − 1=0

Worked explanation: Take a=1, b=-3, c=-1. Compute b²−4ac=(-3)²−4(1)(-1)=13. The roots are x=[3±√13]/(2), giving x≈3.30 and x≈-0.303 to three significant figures.

Common trap: Replacing √13 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

56. Non-perfect-square clinic 14: x² + x − 7=0

Worked explanation: Take a=1, b=1, c=-7. Compute b²−4ac=(1)²−4(1)(-7)=29. The roots are x=[-1±√29]/(2), giving x≈2.19 and x≈-3.19 to three significant figures.

Common trap: Replacing √29 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

57. Non-perfect-square clinic 15: x² + x − 4=0

Worked explanation: Take a=1, b=1, c=-4. Compute b²−4ac=(1)²−4(1)(-4)=17. The roots are x=[-1±√17]/(2), giving x≈1.56 and x≈-2.56 to three significant figures.

Common trap: Replacing √17 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

58. Non-perfect-square clinic 16: x² + x − 1=0

Worked explanation: Take a=1, b=1, c=-1. Compute b²−4ac=(1)²−4(1)(-1)=5. The roots are x=[-1±√5]/(2), giving x≈0.618 and x≈-1.62 to three significant figures.

Common trap: Replacing √5 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

59. Non-perfect-square clinic 17: x² + 2x − 7=0

Worked explanation: Take a=1, b=2, c=-7. Compute b²−4ac=(2)²−4(1)(-7)=32. The roots are x=[-2±√32]/(2), giving x≈1.83 and x≈-3.83 to three significant figures.

Common trap: Replacing √32 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

60. Non-perfect-square clinic 18: x² + 2x − 4=0

Worked explanation: Take a=1, b=2, c=-4. Compute b²−4ac=(2)²−4(1)(-4)=20. The roots are x=[-2±√20]/(2), giving x≈1.24 and x≈-3.24 to three significant figures.

Common trap: Replacing √20 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

61. Non-perfect-square clinic 19: x² + 2x − 1=0

Worked explanation: Take a=1, b=2, c=-1. Compute b²−4ac=(2)²−4(1)(-1)=8. The roots are x=[-2±√8]/(2), giving x≈0.414 and x≈-2.41 to three significant figures.

Common trap: Replacing √8 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

62. Non-perfect-square clinic 20: x² + 4x − 7=0

Worked explanation: Take a=1, b=4, c=-7. Compute b²−4ac=(4)²−4(1)(-7)=44. The roots are x=[-4±√44]/(2), giving x≈1.32 and x≈-5.32 to three significant figures.

Common trap: Replacing √44 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

63. Non-perfect-square clinic 21: x² + 4x − 4=0

Worked explanation: Take a=1, b=4, c=-4. Compute b²−4ac=(4)²−4(1)(-4)=32. The roots are x=[-4±√32]/(2), giving x≈0.828 and x≈-4.83 to three significant figures.

Common trap: Replacing √32 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

64. Non-perfect-square clinic 22: x² + 4x − 1=0

Worked explanation: Take a=1, b=4, c=-1. Compute b²−4ac=(4)²−4(1)(-1)=20. The roots are x=[-4±√20]/(2), giving x≈0.236 and x≈-4.24 to three significant figures.

Common trap: Replacing √20 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

65. Non-perfect-square clinic 23: x² + 4x + 2=0

Worked explanation: Take a=1, b=4, c=2. Compute b²−4ac=(4)²−4(1)(2)=8. The roots are x=[-4±√8]/(2), giving x≈-0.586 and x≈-3.41 to three significant figures.

Common trap: Replacing √8 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

66. Non-perfect-square clinic 24: x² + 6x − 4=0

Worked explanation: Take a=1, b=6, c=-4. Compute b²−4ac=(6)²−4(1)(-4)=52. The roots are x=[-6±√52]/(2), giving x≈0.606 and x≈-6.61 to three significant figures.

Common trap: Replacing √52 by a rounded whole number early changes the numerical roots. Forgetting brackets around 2×(1) can also affect the result when a is negative.

67. No-real-root case 1: x² − 3x + 4=0

Common trap: Taking the absolute value of -7 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

68. No-real-root case 2: 2x² − 2x + 3=0

Common trap: Taking the absolute value of -20 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=2 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

69. No-real-root case 3: 3x² − x + 4=0

Common trap: Taking the absolute value of -47 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=3 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

70. No-real-root case 4: x² + 5=0

Common trap: Taking the absolute value of -20 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

71. No-real-root case 5: 2x² + x + 6=0

Common trap: Taking the absolute value of -47 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=2 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

72. No-real-root case 6: 3x² + 2x + 7=0

Common trap: Taking the absolute value of -80 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=3 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

73. No-real-root case 7: x² + 3x + 10=0

Common trap: Taking the absolute value of -31 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

74. No-real-root case 8: 2x² − 3x + 10=0

Common trap: Taking the absolute value of -71 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=2 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

75. No-real-root case 9: 3x² − 2x + 10=0

Common trap: Taking the absolute value of -116 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=3 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

76. No-real-root case 10: x² − x + 11=0

Common trap: Taking the absolute value of -43 before square-rooting creates solutions to a different equation. The correct response is to recognise the limitation of the real-number system.

Verify: With positive a=1 and a negative radicand, the corresponding upward-opening graph stays above the x-axis. The absence of a crossing is consistent with the formula’s conclusion.

77. Interpretation and checking 1: Write standard form before inserting coefficients

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. The equation x²=5x−6 is not yet in the form ax²+bx+c=0.

Worked explanation: Move everything to the left: x²−5x+6=0. Then a=1,b=−5,c=6 and roots are 2 and 3.

Common trap: Reading b as +5 because the 5x appears on the right rather than changing its sign.

Verify: Substitute two and three into x²=5x−6 to verify both.

78. Interpretation and checking 2: The formula requires a nonzero quadratic coefficient

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. A question presents 0x²+3x−9=0. Is the quadratic formula appropriate?

Worked explanation: No. The coefficient a=0 leaves a linear equation, and 3x−9=0 gives x=3.

Common trap: Using the formula with denominator 2a=0 would create undefined division.

Verify: Identify the highest nonzero power before selecting a method.

79. Interpretation and checking 3: The full numerator is divided by 2a

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. How should x=(−b±√(b²−4ac))/(2a) be keyed or written clearly?

Worked explanation: Write the entire numerator in parentheses before dividing by the full denominator (2a). The plus and minus are separate branches.

Common trap: Writing −b+√D/2a can divide only the radical instead of the whole numerator.

Verify: Test an equation with known roots to confirm the input layout.

80. Interpretation and checking 4: The sign of b affects two parts

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Identify a,b,c in 2x²−5x+2=0.

Worked explanation: a=2,b=−5,c=2. Hence −b=5 while b²=25. The formula returns x=2 or x=1/2.

Common trap: Writing b=+5 changes −b and can reverse the final candidates.

Verify: Check both root candidates in the original equation.

81. Interpretation and checking 5: A repeated root is one distinct real answer

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x²−6x+9=0.

Worked explanation: The radical term has 36−36=0, so x=(6±0)/2=3. It is a repeated real root.

Common trap: A learner may invent two different answers merely because the formula has ±.

Verify: Substitute x=3 and see 9−18+9=0.

82. Interpretation and checking 6: The square-root symbol has a principal value

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. What differs between √9 and solving x²=9?

Worked explanation: √9=3, the nonnegative principal square root. The equation x²=9 has x=3 and x=−3.

Common trap: Calling √9 itself ±3 conflates the expression and the equation.

Verify: Both +3 and −3 square to nine, but √9 denotes +3.

83. Interpretation and checking 7: A negative leading coefficient changes the denominator

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve −x²+4x−3=0.

Worked explanation: a=−1,b=4,c=−3, so D=16−12=4 and denominator is −2. Roots are x=1 and x=3.

Common trap: Using +2 in the denominator makes both signs incorrect.

Verify: Check −1+4−3=0 and −9+12−3=0.

84. Interpretation and checking 8: A missing constant has value zero

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Find a,b,c for 3x²−2x=0.

Worked explanation: a=3,b=−2,c=0. Formula yields roots x=0 and 2/3.

Common trap: Inventing a constant of one because there are three labels a,b,c.

Verify: Substitute zero and two-thirds into the original.

85. Interpretation and checking 9: Some roots are simple rational fractions

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 2x²−3x+1=0.

Worked explanation: D=9−8=1. x=(3±1)/4, so roots are 1 and 1/2.

Common trap: Dividing only one numerator term by four gives incorrect results.

Verify: Verify x=1 and x=1/2 exactly.

86. Interpretation and checking 10: An upward-opening quadratic need not cross the axis

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Does x²+4=0 have real solutions?

Worked explanation: No. D=−16 is negative, and the graph y=x²+4 is always at least four.

Common trap: Claiming every quadratic must have two real roots.

Verify: Its lowest graph point is (0,4), above the axis.

87. Interpretation and checking 11: A graph’s crossings are equation roots

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. For y=x²−3x+2, where are the x-intercepts?

Worked explanation: Set y=0 and solve: roots are x=1 and x=2. Thus crossings are (1,0) and (2,0).

Common trap: Mistaking the y-intercept (0,2) for a solution of y=0.

Verify: At both x=1 and x=2, the output is zero.

88. Interpretation and checking 12: Do not round the root expression too early

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Approximate (3+√5)/2 to three significant figures.

Worked explanation: The unrounded value is about 2.6180339; report 2.62 to three significant figures.

Common trap: Rounding √5 to 2 before finishing produces 2.5, a poor approximation.

Verify: Use unrounded intermediate values then round final display.

89. Interpretation and checking 13: A root is a candidate until it is checked

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Check x=2 in x²−5x+6=0.

Worked explanation: Substitute 2²−5(2)+6=4−10+6=0. So two is a solution.

Common trap: Accepting a written x-value without testing the original equality.

Verify: Test x=3 as well; it is a second solution.

90. Interpretation and checking 14: A negative root is not necessarily a mistake

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve x²+5x+6=0.

Worked explanation: a=1,b=5,c=6 and D=25−24=1; formula gives (−5±1)/2, so roots −2 and −3.

Common trap: Assuming positive leading a guarantees positive roots.

Verify: Substitute both negative candidates.

91. Interpretation and checking 15: An exact radical and a rounded value are different outputs

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. A formula produces x=(1+√3)/2. Is x=1.37 exact?

Worked explanation: No. The expression with √3 is exact; 1.37 is a rounded approximation to three significant figures.

Common trap: Dropping the approximation sign conceals the loss of precision.

Verify: Calculate with full precision when checking the original equation.

92. Interpretation and checking 16: A possible real-world constraint can reject one root

Worked explanation: Only five is positive, assuming no additional constraint. The negative algebraic root is mathematically valid but not physically feasible for width.

Common trap: Deleting the negative root without first reporting it as a mathematical solution may confuse the two stages.

Verify: State algebraic answers first, then apply the context restriction.

93. Interpretation and checking 17: The 2027 K110 formulae list contains the quadratic formula

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Is a guessed mnemonic preferable to checking the official formulae sheet?

Worked explanation: No. The published 2027 K110 Number and Algebra formulae list contains x=(−b±√(b²−4ac))/(2a). The essential skill is accurate substitution and interpretation.

Common trap: Assuming every school’s current formula sheet is identical to an older unofficial screenshot.

Verify: Check official documentation for the student’s own examination year.

94. Interpretation and checking 18: A formula result should be shown clearly

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. A learner writes x=3 or −2 without stating which equation was solved.

Worked explanation: The result may be mathematically correct but should be tied to a stated equation, with enough working to show a,b,c and a valid calculation.

Common trap: Copying two numbers from an answer key is not evidence the method was understood.

Verify: A changed-number problem and a direct substitution check provide evidence.

95. Interpretation and checking 19: Two negative factors under b² remain positive

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. In x²+4x+1=0, what is b²?

Worked explanation: b=4, so b²=16. For b=−4 in x²−4x+1, b² is still 16.

Common trap: Writing −16 after squaring a negative coefficient fails to bracket the full quantity.

Verify: Compare (−4)² with −(4²), which are not equal.

96. Interpretation and checking 20: One solution repeated when D=0

Did you know? The ± symbol represents two branches, but they need not always be distinct real values. Solve 4x²+4x+1=0.

Worked explanation: D=16−16=0; x=−4/8=−1/2. The two branches coincide.

Common trap: Writing +1/2 after losing the negative numerator.

Verify: Check 4×1/4+4×(−1/2)+1=0.

Five different errors that appear as a bad final answer

The equation was never written in standard form

b was entered without its negative sign

The plus/minus was treated as optional

The denominator was applied to only half the numerator

Rounding made the check seem to fail

An illustrative three-pupil bridge lesson

10 minutes: review signed multiplication, roots and simple equation balance.
20 minutes: introduce the formula using the official sheet and practise coefficient lists.
25 minutes: solve a few exact-root questions, including both ± branches.
15 minutes: distinguish repeated-root and no-real-root cases.
10 minutes: independently solve one changed equation with no visible model.
10 minutes: correct the first sign or bracket error and assign a short later retest.

A five-session parent–student routine

Day one: practise negative squares and bring simple equations into zero-equals form.
Day two: identify a, b and c in five varied equations.
Day three: solve two exact-root cases from the formula and verify both answers.
Day four: examine one repeated root, one negative radicand and one irrational pair.
Day five: solve a new problem and explain the corresponding graph’s intercept behaviour.

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

Yes. The published 2027 K110 Mathematical Formulae section includes the quadratic-equation formula. Check official requirements for the student’s own eventual examination year.

What if b²−4ac is negative?

There are no real roots. The associated quadratic graph has no real x-intercept.

Why can the formula return two roots?

The ± yields two possible numerator values when the radicand is positive. Each should be checked in the original equation.

Can one root be a fraction or a negative number?

Yes. Roots depend on the signed coefficients. Do not reject a result merely because it is not a positive whole number; substitute it back to check.

Will a 90-minute group guarantee success?

No. The appropriate progression depends on prerequisite knowledge, feedback and independent retention. Ask the tutor how readiness and later transfer are assessed.

Official references and editorial note

Bring a marked linear-equation problem to eduKate Singapore and ask whether the next useful step is stronger equation balance or an optional quadratic formula preview.

Secondary 2 G1 Mathematics Tutor Bukit Timah | The Quadratic Formula Explained

Official references and editorial note

Official references and editorial note

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

What if b²−4ac is negative?

Why can the formula return two roots?

Can one root be a fraction or a negative number?

Will a 90-minute group guarantee success?

Continue through the relevant Mathematics progression

Official references and editorial note

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

What if b²−4ac is negative?

Why can the formula return two roots?

Can one root be a fraction or a negative number?

Will a 90-minute group guarantee success?

Continue through the relevant Mathematics progression

Official references and editorial note

A five-session parent–student routine

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

What if b²−4ac is negative?

Why can the formula return two roots?

Can one root be a fraction or a negative number?

Will a 90-minute group guarantee success?

Continue through the relevant Mathematics progression

Official references and editorial note

A five-session parent–student routine

Frequently asked questions

Is quadratic formula examinable in current Secondary 2 G1 classes?

Does 2027 K110 supply the formula?

What if b²−4ac is negative?

Why can the formula return two roots?

Can one root be a fraction or a negative number?

Will a 90-minute group guarantee success?

Continue through the relevant Mathematics progression

Official references and editorial note

The 60-second readiness diagnostic

Read the equation before typing the formula

First get zero on one side

The negative of b is not necessarily negative

Everything above the fraction bar shares one denominator

How the quantity under the root changes the answer

96 original quadratic formula learning clinics

1. Exact-root clinic 1: x² + 7x + 10=0

2. Exact-root clinic 2: −x² + 3x + 10=0

3. Exact-root clinic 3: 2x² − 4x + 2=0

4. Exact-root clinic 4: −2x² + 2x + 24=0

5. Exact-root clinic 5: 3x² − 48=0

6. Exact-root clinic 6: −3x² − 3x=0

7. Exact-root clinic 7: x² + 2x − 8=0

8. Exact-root clinic 8: −x² + 8x − 15=0

9. Exact-root clinic 9: 2x² + 8x + 6=0

10. Exact-root clinic 10: −2x² − 10x=0

11. Exact-root clinic 11: 3x² − 15x + 18=0

12. Exact-root clinic 12: −3x² − 21x − 30=0

13. Exact-root clinic 13: x² − 3x − 10=0

14. Exact-root clinic 14: −x² + 2x − 1=0

15. Exact-root clinic 15: 2x² − 2x − 24=0

16. Exact-root clinic 16: −2x² + 32=0

17. Exact-root clinic 17: 3x² + 3x=0

18. Exact-root clinic 18: −3x² − 6x + 24=0

19. Exact-root clinic 19: x² − 8x + 15=0

20. Exact-root clinic 20: −x² − 4x − 3=0

21. Exact-root clinic 21: 2x² + 10x=0

22. Exact-root clinic 22: −2x² + 10x − 12=0

23. Exact-root clinic 23: 3x² + 21x + 30=0

24. Exact-root clinic 24: −3x² + 9x + 30=0

25. Exact-root clinic 25: x² − 2x + 1=0

26. Exact-root clinic 26: −x² + x + 12=0

27. Exact-root clinic 27: 2x² − 32=0

28. Exact-root clinic 28: −2x² − 2x=0

29. Exact-root clinic 29: 3x² + 6x − 24=0

30. Exact-root clinic 30: −3x² + 24x − 45=0

31. Exact-root clinic 31: x² + 4x + 3=0

32. Exact-root clinic 32: −x² − 5x=0