Quadratics
Algebra and functions
- AS
- Non-calculator
- 8 questions
- 37 marks
These are original practice questions for the Edexcel A-level Pure content, not a past paper. This site is independent and not affiliated with Pearson.
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Solve $x^2 - 7x + 12 = 0$.
3 marks
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Write $x^2 + 6x - 1$ in the form $(x + a)^2 + b$.
3 marks
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The equation $2x^2 + kx + 8 = 0$ has equal roots. Find the possible values of $k$.
4 marks
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Without solving the equation, determine the nature of the roots of $3x^2 - 2x + 1 = 0$.
4 marks
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Solve $2x^2 - 5x - 3 = 0$.
5 marks
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A rectangle has side lengths $x + 2$ and $x - 1$, and its area is $40$. Find the side lengths.
5 marks
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Find the minimum value of $y = x^2 - 4x + 9$ and the value of $x$ at which it occurs.
6 marks
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Find the set of values of $k$ for which $x^2 + (k - 2)x + k = 0$ has real roots.
7 marks
Mark scheme
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Answer. $x = 3$ or $x = 4$
- $(x - 3)(x - 4) = 0$.
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Answer. $(x + 3)^2 - 10$
- $x^2 + 6x - 1 = (x + 3)^2 - 9 - 1$.
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Answer. $k = 8$ or $k = -8$
- The discriminant is $k^2 - 64$.
- Equal roots require $k^2 - 64 = 0$.
- $k = \pm 8$.
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Answer. There are no real roots.
- The discriminant is $4 - 12 = -8$.
- A negative discriminant means there are no real roots.
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Answer. $x = 3$ or $x = -\dfrac{1}{2}$
- $(2x + 1)(x - 3) = 0$.
- $x = -\dfrac{1}{2}$ or $x = 3$.
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Answer. The sides are $8$ and $5$.
- $(x + 2)(x - 1) = 40$.
- $x^2 + x - 42 = 0$, so $(x + 7)(x - 6) = 0$.
- $x = 6$ because a length cannot be negative.
- The sides are $8$ and $5$.
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Answer. The minimum value is $5$, when $x = 2$.
- $y = (x - 2)^2 - 4 + 9 = (x - 2)^2 + 5$.
- The squared term is at least $0$, so the minimum is $5$ when $x = 2$.
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Answer. $k \le 4 - 2\sqrt{3}$ or $k \ge 4 + 2\sqrt{3}$
- The discriminant is $(k - 2)^2 - 4k = k^2 - 8k + 4$.
- Real roots require $k^2 - 8k + 4 \ge 0$.
- The roots of $k^2 - 8k + 4 = 0$ are $4 \pm 2\sqrt{3}$.
- The quadratic in $k$ is positive outside the roots.