Quadratics

Algebra

These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.

Solve $(x - 3)(x + 2) = 0$.

2 marks
Factorise $x^{2} + 5x + 6$.

2 marks
Solve $x^{2} - 7x + 10 = 0$.

3 marks
Solve $x^{2} - 9 = 0$.

3 marks
Solve $2x^{2} + 5x - 3 = 0$.

4 marks
Solve $x^{2} - 4x - 3 = 0$. Give the exact answers.

4 marks
A rectangle has sides $(x + 3)$ cm and $(x - 1)$ cm, and area $45$ cm$^{2}$. Find the side lengths.

5 marks
For $y = x^{2} - 6x + 5$, find the roots, the turning point and the minimum value. Then solve $x^{2} - 6x + 5 = 4$.

6 marks

Mark scheme

Answer. $x = 3$ or $x = -2$
- One factor is zero.
- $x = 3$ or $x = -2$
Answer. $(x + 2)(x + 3)$
- Numbers adding to $5$ and multiplying to $6$.
- $(x + 2)(x + 3)$
Answer. $x = 5$ or $x = 2$
- $(x - 5)(x - 2) = 0$
- $x = 5$ or $x = 2$
Answer. $x = 3$ or $x = -3$
- $x^{2} = 9$
- $x = \pm 3$
Answer. $x = \dfrac{1}{2}$ or $x = -3$
- $(2x - 1)(x + 3) = 0$
- $x = \dfrac{1}{2}$ or $x = -3$
Answer. $x = 2 \pm \sqrt{7}$
- $x = \dfrac{4 \pm \sqrt{16 + 12}}{2}$
- $x = \dfrac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7}$
Answer. $9$ cm and $5$ cm
- $(x + 3)(x - 1) = 45$
- $x^{2} + 2x - 48 = 0$, so $(x + 8)(x - 6) = 0$.
- $x = 6$ because a side cannot be negative.
- The sides are $9$ cm and $5$ cm.
Answer. Roots $1$ and $5$, turning point $(3, -4)$, minimum $-4$, and $x = 3 \pm 2\sqrt{2}$
- $(x - 1)(x - 5) = 0$, so the roots are $1$ and $5$.
- The turning point is $(3, -4)$, so the minimum value is $-4$.
- $x^{2} - 6x + 1 = 0$
- $x = \dfrac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}$.