When a parameter appears in the coefficient of x², find the value that makes that coefficient zero before using the discriminant. Substitute that value into the original equation. The remaining equation may be linear and have a real solution of its own.
The 2027 SEC G2 Additional Mathematics K232 syllabus covers conditions for the real roots of a quadratic equation. This worked example shows how the answer changes when the question counts all solutions of a parameter-dependent equation.
- Set the coefficient of x² equal to zero to find the special case.
- Solve the special case in the original equation.
- Apply b² − 4ac to the remaining parameter values.
- Distinguish one real solution from a repeated quadratic root.
CHAPTER 1 OF 4 · Back to contents
Check the degree first
For ax² + bx + c = 0, a is the coefficient of x². When a = 0, the equation becomes bx + c = 0. If b is non-zero, there is one solution. The quadratic formula has 2a in the denominator, which explains why it cannot handle that case.
If b also vanishes, inspect what remains. A non-zero constant gives an impossible statement, such as 5 = 0. A zero constant gives 0 = 0, which every real x satisfies.
CHAPTER 2 OF 4 · Back to contents
Worked example | Two ways to get one real solution
Find all real values of k for which (k − 1)x² + 2x + 1 = 0 has exactly one distinct real solution. Start with k = 1. The x² term vanishes and 2x + 1 = 0 gives x = −½.
For k ≠ 1, the equation is quadratic. Its discriminant is 2² − 4(k − 1)(1) = 8 − 4k. Set this equal to zero to obtain k = 2. Substituting k = 2 gives x² + 2x + 1 = (x + 1)² = 0, so the repeated root is x = −1.
The equation has exactly one distinct real solution at k = 1 or k = 2. Only k = 2 gives a quadratic with a repeated root.
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CHAPTER 3 OF 4 · Back to contents
Try changing the question
Now ask for two distinct real solutions. The discriminant must be positive, giving k < 2, and the equation must still be quadratic, giving k ≠ 1. The two conditions together give k < 1 or 1 < k < 2.
To check the boundary, put k = 1 back into the original equation. Although 8 − 4k gives a positive value there, the equation has just one solution because the x² term has gone. Keep the degree condition beside your discriminant working.
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CHAPTER 4 OF 4 · Back to contents
Frequently asked questions
Can I use the discriminant first?
You can work on the quadratic branch first if you write k ≠ 1 explicitly and then solve the excluded case separately.
Does a repeated root count as one distinct solution?
Yes. At k = 2, the two factors give the same value of x. The solution set contains just −1.
How can a parent check the working quickly?
Ask which value makes the x² term disappear and point to where the student handled it. Then ask whether the final line answers the exact wording.
Sources and next reading
Read the official SEAB syllabus
SEC 2027 G3 Mathematics K310 | Same Median, Different Interquartile Range