Did you know? Finding the roots of a quadratic does not automatically solve a quadratic inequality. A JC1 H1 Mathematics tutor in Bukit Timah can often spot the underlying confusion in one sentence: the student knows where a graph meets the x-axis, but not where the graph lies above or below it.
Quick answer: to solve a quadratic inequality, find the boundary roots, decide the sign on each interval, and include or exclude the boundaries according to >, ≥, < or ≤. A sketch or a test value can turn a doubtful answer into a reliable one.
For JC1 H1 Mathematics syllabus 8865, inequalities are part of the Pure Mathematics foundation. They also prepare students for questions about conditions, possible values and real-world constraints. Here is a readable method with examples you can check in minutes, not an endless list of memorised rules.
The five-second distinction: roots versus solution sets
Consider f(x) = x² − 5x + 6. Solving the equation f(x) = 0 gives x = 2 or x = 3. That tells us where the curve crosses the horizontal axis.
But if the question asks f(x) > 0, it asks where the graph lies above that axis. Because the parabola opens upwards, the answer is x < 2 or x > 3. The two roots are boundaries, not the entire solution.
If the inequality had instead been f(x) ≤ 0, the answer would be 2 ≤ x ≤ 3. The graph lies on or below the axis between the roots, and the equal sign includes both boundaries.
A one-page method students can use independently
- Bring everything to one side. Write the inequality as f(x) > 0, f(x) ≥ 0, f(x) < 0 or f(x) ≤ 0.
- Find the roots of f(x) = 0 where possible, by factorisation, a quadratic formula, or a permitted graphical approach.
- Find the graph’s direction. For ax² + bx + c, a positive a gives an upward-opening parabola; a negative a gives a downward-opening parabola.
- Decide the sign in each interval. Use the curve sketch or a carefully chosen test value. Do not guess only from the root values.
- Read the boundary symbol. Strict signs (>, <) exclude points where the expression is zero; inclusive signs (≥, ≤) include them.
- Check one value. Substitute an x from your proposed interval into the original inequality to confirm its sign.
This six-step sequence may look slower than a memorised shortcut, but after a few correct examples it becomes fast. More importantly, students can reconstruct it under examination pressure.
Worked example 1: solve x² − 5x + 6 > 0
Step 1: find the boundary roots
Factorise the expression: x² − 5x + 6 = (x − 2)(x − 3). Setting this equal to zero gives the roots x = 2 and x = 3.
Step 2: decide the sign
The coefficient of x² is positive, so the parabola opens upwards. It crosses the x-axis at 2 and 3. The curve is below the axis between those crossings and above the axis to the left of 2 and to the right of 3.
Step 3: state the answer carefully
Since the question asks for a strictly positive value, the roots themselves cannot be included. The answer is x < 2 or x > 3.
Step 4: verify with simple numbers
At x = 0, the expression is 6, which is positive. At x = 2.5, it is 6.25 − 12.5 + 6 = −0.25, negative. At x = 4, it is 16 − 20 + 6 = 2, positive. The tests confirm the three intervals.
A common error is to write 2 < x < 3 because the student remembers the two roots and assumes an inequality must mean ‘between them’. The test at x = 2.5 would immediately disprove that guess.
Worked example 2: solve −x² + 4x − 3 ≥ 0
This time the coefficient of x² is negative, so the parabola opens downwards. Rearrange by multiplying the entire inequality by −1. The inequality sign reverses: x² − 4x + 3 ≤ 0.
Factorise: (x − 1)(x − 3) ≤ 0. The upward-opening version is at or below the axis between its roots. Therefore, 1 ≤ x ≤ 3.
Check x = 2 in the original expression: −4 + 8 − 3 = 1, which is at least zero. Check x = 0: −3, which is not. Both checks agree with the answer.
Why does the sign reverse?
The reason is order on the number line. If 2 > 1, multiplying both sides by −1 gives −2 < −1. Negative multiplication reverses the direction of a comparison. A student who understands this will remember the rule even when a longer question feels unfamiliar.
Worked example 3: can a quadratic be positive for every real x?
Consider x² − 4x + 5. The discriminant is b² − 4ac = (−4)² − 4(1)(5) = −4. It is negative, so there are no real roots. The coefficient of x² is positive, so the graph opens upwards and never meets the x-axis. Therefore x² − 4x + 5 > 0 for every real x.
There is a second, more visual way: x² − 4x + 5 = (x − 2)² + 1. A square cannot be negative, so the expression is always at least 1. Both arguments are useful because they explain the mathematical reason, not merely the calculator output.
In the 2027 H1 8865 syllabus, conditions for a quadratic to have two real roots, equal roots or no real roots, and for a quadratic expression to be always positive or always negative, are part of the equations-and-inequalities content. A tutor should make the connection between the discriminant and the graph explicit.
What the discriminant tells you before solving
- For ax² + bx + c with a ≠ 0, the discriminant is Δ = b² − 4ac.
- If Δ > 0, there are two distinct real roots and the curve crosses the x-axis twice.
- If Δ = 0, there is one repeated real root and the curve touches the x-axis.
- If Δ < 0, there are no real roots; the curve stays entirely above or entirely below the x-axis, depending on the sign of a.
- For a quadratic to be strictly positive for all real x, a > 0 and Δ < 0. For non-negative for all real x, a > 0 and Δ ≤ 0, with a ≠ 0.
These conditions are not a substitute for drawing and thinking. They are a compact summary of what the graph must do.
A real-world model: positive profit is an interval, not a root
Suppose a simplified model gives P(q) = −q² + 12q − 20, where q is a quantity in agreed units. When is P(q) positive? Solve −q² + 12q − 20 > 0. Multiply by −1 and reverse the sign: q² − 12q + 20 < 0. Factorise: (q − 2)(q − 10) < 0. The answer is 2 < q < 10.
The roots q = 2 and q = 10 mark break-even points in this model. Values strictly between them give positive modelled profit. If q must be a whole-number count, the feasible positive-profit quantities are 3, 4, 5, 6, 7, 8 and 9. A mathematical interval only becomes a business answer after its domain and units are interpreted.
Check q = 6: P(6) = −36 + 72 − 20 = 16. Check q = 12: P(12) = −144 + 144 − 20 = −20. The model behaves as the inequality predicts.
This is the kind of interpretation that matters later in JC Mathematics. A correct interval with the wrong units, or one that ignores a whole-number restriction, is an incomplete answer to a contextual question.
How simultaneous equations connect to quadratic reasoning
Imagine a line y = x + 2 and a curve y = x² − 4x + 6. An intersection must satisfy both equations. Equate their right-hand sides: x + 2 = x² − 4x + 6. Rearrange to x² − 5x + 4 = 0 and factorise: (x − 1)(x − 4) = 0.
So x = 1 or x = 4. Substitute back into y = x + 2 to obtain the intersection points (1, 3) and (4, 6). The algebra produces two candidates because a line may meet a parabola twice.
Why include this nearby skill? The 2027 H1 syllabus includes simultaneous equations with one linear and one quadratic equation solved by substitution. Once the student sees equations as graph intersections and inequalities as graph regions, several topics stop looking unrelated.
Graphing calculator: an intelligent check, not a replacement
A graphing calculator can reveal intercepts and regions in which the function is above or below zero. But a poorly chosen viewing window can hide roots or distort the apparent scale. Predict the number and approximate position of intersections before tracing.
The official 2027 H1 Mathematics syllabus expects an approved GC without computer algebra system. It also describes when unsupported answers may be accepted and when graph-based solutions require a sketch. Students should obey the actual question. Exact roots obtained by simple factorisation are often better reported exactly.
If the student says, ‘The graph looks above zero, so the answer is all real x,’ ask to see the window and a test value. This teaches healthy mathematical doubt.
Three errors a good tutor should diagnose immediately
Error 1: stopping at x = 2 and x = 3
The student has answered the equation f(x) = 0 instead of the inequality f(x) > 0. The repair is to sketch the curve and describe where it is above or below the x-axis.
Error 2: forgetting to reverse the inequality
The student changes −x² + 4x − 3 ≥ 0 into x² − 4x + 3 ≥ 0. Multiplication by −1 reverses the comparison, so the correct equivalent statement is x² − 4x + 3 ≤ 0.
Error 3: including strict boundaries
The student reports 2 ≤ x ≤ 3 when asked for x² − 5x + 6 > 0. Even aside from the wrong interval, at a root the expression equals zero, not a strictly positive number. Mark the boundary with an open or filled endpoint on a number line to make the rule visible.
A 20-minute weekly practice cycle
- Four minutes: draw one upward and one downward parabola, labelling where each is positive and negative.
- Six minutes: solve one factorizable strict inequality and one inclusive inequality with a negative leading coefficient.
- Four minutes: check your answer using a number from inside and outside the proposed interval.
- Four minutes: explain one discriminant question or ‘always positive’ condition in words.
- Two minutes: name the first incorrect decision in any wrong solution, then schedule a fresh retry a few days later.
Avoid doing only a large block of nearly identical sums. Alternate roots, signs, sketches and context questions. Mixed practice is where students prove they can choose their own method.
How 3-pax tutorials can help a JC1 student
One learner may factorise fluently but choose the wrong interval. Another may understand the picture but mishandle a negative coefficient. A third may solve both correctly but forget the real-world restriction. In a small group, the tutor can inspect each line of working and set a different next question without losing the discussion that benefits everyone.
The established eduKateSG tuition format is three students per group, normally 90 minutes per week, at 8 Fourth Avenue, Singapore 268674, near Sixth Avenue MRT. Families should confirm current H1 Maths tutor availability, suitable grouping and timetable before enrolling. The immutable tutorial model reference explains the teaching approach; it does not establish a guaranteed JC1 place.
If bracket expansion or factorisation remains unreliable, use the 14-day O-Level-to-H1 algebra bridge first. This inequality lesson works best when that prerequisite is secure.
Frequently asked questions
Are quadratic inequalities tested in H1 Mathematics?
The official 2027 H1 Mathematics 8865 syllabus includes solving quadratic equations and inequalities analytically, and solving inequalities by graphical methods. Always confirm the student’s eventual examination-year document.
Must every inequality be solved using a graph?
No. Factorisation, sign reasoning and algebra are often efficient. A sketch is a useful way to explain and check the sign intervals, while the question’s wording determines how much graphical working is expected.
What if a quadratic has no real roots?
Use the discriminant and direction of opening to decide whether the expression is always positive or always negative. For example, x² − 4x + 5 is positive for every real x because it is (x − 2)² + 1.
What is the best first check after solving?
Substitute a simple number inside the proposed interval into the original inequality. Then test a number outside it. Check boundary inclusion separately. This catches many incorrect ‘between or outside’ guesses.
Can a student without O-Level A-Maths learn this?
Yes, but they may need a measured bridge through factorisation, equations, graphs and signed-number rules before tackling mixed contextual questions. H1 Mathematics is designed to make this pathway possible, not effortless.
A small conclusion with a big payoff
A quadratic inequality is not primarily about memorising whether the answer goes ‘between’ or ‘outside’. It is about reading a mathematical relationship. Roots mark the boundary; a graph explains the sign; the inequality symbol decides whether the boundary belongs; and the real-world context decides which values make sense.
For the broader Bukit Timah H1 Mathematics examination map, see the syllabus-linked overview. For source-of-truth requirements, consult the SEAB 2027 H1 8865 document and later the examination-year syllabus that applies to the learner. Strong explanations make strong checking habits, and those habits travel into every other H1 topic.