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Year 11 Mathematics · Solving quadratic equations

Name: ______________________

Date: ____________

1. Solve (x − 7)(x + 2) = 0.
- ☐ (a) x = −7 or x = 2
- ☐ (b) x = 7 or x = −2
- ☐ (c) x = 7 or x = 2
- ☐ (d) x = −14
2. Solve x^2 − 9 = 0.
- ☐ (a) x = 3 only
- ☐ (b) x = 9 or x = −9
- ☐ (c) x = 3 or x = −3
- ☐ (d) x = 4.5
3. How many real solutions does x^2 + 4x + 5 = 0 have?
- ☐ (a) 0
- ☐ (b) 1
- ☐ (c) 2
- ☐ (d) Infinitely many
4. Which is the correct factorisation of x^2 + x − 12?
- ☐ (a) (x + 6)(x − 2)
- ☐ (b) (x − 4)(x + 3)
- ☐ (c) (x + 12)(x − 1)
- ☐ (d) (x + 4)(x − 3)
5. What is the best first step to solve x^2 = 3x + 10?
- ☐ (a) Take the square root of both sides
- ☐ (b) Rearrange to x^2 − 3x − 10 = 0
- ☐ (c) Divide both sides by x
- ☐ (d) Subtract 10 from the right side only
6. Find the positive solution of x^2 − 3x − 10 = 0.
7. Find the larger solution of x^2 − 11x + 24 = 0.
8. Calculate the discriminant of x^2 − 6x + 9 = 0.

Answer sheet

1. x = 7 or x = −2 — x − 7 = 0 gives x = 7, and x + 2 = 0 gives x = −2.
2. x = 3 or x = −3 — x^2 − 9 = (x − 3)(x + 3) = 0, so x = 3 or x = −3. Both square to 9.
3. 0 — The discriminant is 4^2 − 4 × 1 × 5 = 16 − 20 = −4, which is negative, so there are no real solutions.
4. (x + 4)(x − 3) — 4 × (−3) = −12 and 4 + (−3) = 1, so x^2 + x − 12 = (x + 4)(x − 3).
5. Rearrange to x^2 − 3x − 10 = 0 — To use the null factor law, one side must be zero. Rearranging gives x^2 − 3x − 10 = 0, which factorises to (x − 5)(x + 2) = 0.
6. x = 5 — x^2 − 3x − 10 = (x − 5)(x + 2) = 0, so x = 5 or x = −2. The positive solution is 5.
7. x = 8 — Two numbers that multiply to 24 and add to −11 are −3 and −8, so (x − 3)(x − 8) = 0. The solutions are 3 and 8; the larger is 8.
8. 0 — b^2 − 4ac = (−6)^2 − 4 × 1 × 9 = 36 − 36 = 0, so the equation has exactly one solution (x = 3).