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S3 Maths: Quadratic Equations – Formula and Discriminant (Paper 4 (error clinic))
S3 Maths: Quadratic Equations – Formula and Discriminant (Paper 4 (error clinic))
Paper 4 (error clinic): ten quadratic-equation questions with answers and one-line notes.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S3 Maths: Quadratic Equations – Formula and Discriminant |
| Level | S3 |
02 Questions (10 items)
- FoundationSolve x^2 – 7x + 12 = 0.
- FoundationSolve x^2 – 6x = 0.
- FoundationSolve x^2 + 4x + 4 = 0.
- IntermediateSolve 2x^2 + x – 6 = 0.
- IntermediateDoes x^2 + 3x + 5 = 0 have real roots?
- IntermediateIf alpha and beta are the roots of x^2 + 8x + 15 = 0, find the sum, the product and both roots.
- ChallengeForm the quadratic equation with roots -2 and 7.
- ChallengeFind k so that x^2 + kx + 16 = 0 has equal roots.
- ChallengeIf alpha and beta are the roots of x^2 – 4x + 1 = 0, find alpha squared plus beta squared.
- Error redoA student solves x^2 – 2x + 5 = 0 and gets a discriminant of -16, then writes that it square-roots to 4, so there are real roots. What went wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | x = 3 or x = 4 | The discriminant is 49 minus 48, which is 1, so the formula gives (7 plus or minus 1) over 2. |
| 2 | x = 0 or x = 6 | Taking the common factor x gives x(x – 6) = 0, and each factor supplies one root. |
| 3 | x = -2, a repeated root | The expression is the perfect square (x + 2)^2 = 0, so there is only one root. |
| 4 | x = 3/2 or x = -2 | The discriminant is 1 plus 48, which is 49, so the formula gives (-1 plus or minus 7) over 4. |
| 5 | No real roots | The discriminant is 9 minus 20, which is -11; being negative, there are no real roots. |
| 6 | sum = -8, product = 15, roots -3 and -5 | The sum is -8 and the product is 15, and the pair -3 and -5 satisfies both conditions. |
| 7 | x^2 – 5x – 14 = 0 | The sum is 5 and the product is -14, so the equation takes minus 5 and plus -14 as its terms. |
| 8 | k = 8 or k = -8 | Equal roots need the discriminant k^2 – 64 to be zero, so k is plus or minus 8. |
| 9 | 14 | Take the square of the sum minus twice the product: 16 minus 2 gives 14. |
| 10 | A negative number has no real square root; this equation has no real roots | A negative discriminant already means no real roots, so there is nothing to square-root and no sign to drop. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Before using the formula, rearrange the equation into ax^2 + bx + c = 0 and write down a, b and c clearly, because signs slip most often here. Work out the discriminant D = b^2 – 4ac first: positive means two distinct real roots, zero means a repeated root and negative means no real roots. Remember the plus-or-minus sign gives two branches, simplify the surd before cancelling with the denominator, and finish by substituting the answers back into the original equation.Tutor Registration →
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