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S3 Maths: Quadratic Equations – Formula and Discriminant (Paper 3 (challenge))
S3 Maths: Quadratic Equations – Formula and Discriminant (Paper 3 (challenge))
Paper 3 (challenge): 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 6x^2 – 5x – 6 = 0.
- FoundationSolve 4x^2 – 4x – 3 = 0.
- FoundationForm the quadratic equation with 1 as the leading coefficient and roots 3 and -5.
- IntermediateFind the range of k so that x^2 – 6x + k = 0 has two distinct real roots.
- IntermediateIf alpha and beta are the roots of 3x^2 + 7x – 2 = 0, find alpha + beta and alpha beta.
- IntermediateIf alpha and beta are the roots of x^2 – 5x + 6 = 0, find alpha squared plus beta squared.
- ChallengeForm the quadratic equation with roots 1/2 and -3, clearing fractions.
- ChallengeA rectangle has area 48 and perimeter 28. Find its side lengths.
- ChallengeThe product of two consecutive positive integers is 156. Find them.
- Error redoSolving x^2 = 4x, a student divides both sides by x and answers only x = 4. Which root is lost?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | x = 3/2 or x = -2/3 | The discriminant is 25 plus 144, which is 169, so its root is 13 and the formula gives (5 plus or minus 13) over 12. |
| 2 | x = 3/2 or x = -1/2 | The discriminant is 16 plus 48, which is 64, so the formula gives (4 plus or minus 8) over 8. |
| 3 | x^2 + 2x – 15 = 0 | The sum of the roots is -2 and the product is -15, so the equation is written with minus the sum and plus the product. |
| 4 | k < 9 | Two distinct real roots need the discriminant 36 – 4k to be positive, which gives k less than 9. |
| 5 | alpha + beta = -7/3 and alpha beta = -2/3 | The sum is minus b over a, and b is 7 here, giving -7/3; the product is c over a, giving -2/3. |
| 6 | 13 | Use the square of the sum minus twice the product: 25 minus 12 gives 13 without finding the roots first. |
| 7 | 2x^2 + 5x – 3 = 0 | The sum is -5/2 and the product is -3/2; start from x^2 + (5/2)x – 3/2 = 0 and multiply the whole equation by 2. |
| 8 | 6 and 8 | Let the sides be x and 14 minus x, giving x^2 – 14x + 48 = 0, whose discriminant is 4. |
| 9 | 12 and 13 | Writing n and n + 1 gives n^2 + n – 156 = 0 with discriminant 625, and the positive root is 12. |
| 10 | The root x = 0 is lost; the correct answer is x = 0 or x = 4 | Before dividing by x you must rule out x equal to zero; factorising as x(x – 4) = 0 keeps both roots. |
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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