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GCSE · Algebra

5.8D Solving Quadratics: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculators not allowed · about 85 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Solve x² + 6x + 8 = 0

(Total for Question 1 is 2 marks)

Solve x² - 8x + 15 = 0

(Total for Question 2 is 2 marks)

Solve x² - 81 = 0

(Total for Question 3 is 2 marks)

Solve x² + 9x = 0

(Total for Question 4 is 2 marks)

Solve 9x² - 16 = 0

(Total for Question 5 is 3 marks)

Solve 3(x + 1)² = 48

(Total for Question 6 is 3 marks)

Solve 3x² + 11x + 6 = 0

(Total for Question 7 is 3 marks)

Solve 4x² - 4x - 15 = 0

(Total for Question 8 is 3 marks)

Solve x² + 4x - 3 = 0 using the quadratic formula. Give your answers in simplified surd form.

(Total for Question 9 is 3 marks)

A rectangle has length (x + 6) cm and width (x - 1) cm. The area of the rectangle is 44 cm^2.

(a)Show that x² + 5x - 50 = 0(2)

(b)Solve the equation to find the value of x, given that x > 0(2)

(c)Hence find the width of the rectangle(1)

(Total for Question 10 is 5 marks)

Two consecutive positive even integers have a product of 168. Let n be the smaller integer. Form an equation in n and solve it to find the two integers.

(Total for Question 11 is 4 marks)

Solve 5x² - 13x - 6 = 0

(Total for Question 12 is 3 marks)

Solve x² - 10x + 3 = 0 by completing the square, leaving your answer in simplified surd form.

(Total for Question 13 is 3 marks)

g(x) = x² + 10x + 19

(a)Express g(x) in the form (x + a)² + b, where a and b are integers(2)

(b)Hence write down the minimum value of g(x) and the value of x at which it occurs(2)

(Total for Question 14 is 4 marks)

Solve 2x² - 3x - 4 = 0. Give your answers in the form (a ± √b) / c, where a, b and c are integers.

(Total for Question 15 is 3 marks)

A firework is launched from a platform. Its height above the ground, h metres, after t seconds is modelled by h = -5t² + 25t + 2. Find the times at which the firework is at a height of 22 metres.

(Total for Question 16 is 5 marks)

Solve the simultaneous equations:
y = x² - x - 7
y = 3x - 2

(Total for Question 17 is 5 marks)

Prove that x² + 4x + 9 is positive for all real values of x.

(Total for Question 18 is 3 marks)

x² + 2x - 24 is a quadratic expression.

(a)Factorise x² + 2x - 24(2)

(b)Hence write down the solutions of x² + 2x - 24 = 0(1)

(c)Hence, or otherwise, solve (x - 1)² + 2(x - 1) - 24 = 0(3)

(Total for Question 19 is 6 marks)

By using the substitution u = x², or otherwise, solve x⁴ - 13x² + 36 = 0.

(Total for Question 20 is 5 marks)

The equation px² + 12x + p = 0, where p is a non-zero constant, has equal roots. Find the possible values of p.

(Total for Question 21 is 4 marks)

Mark scheme · 5.8D Solving Quadratics: Fluency and Exam Drill

Question 1

M1 correct factorisation (x + 2)(x + 4) = 0 oe
A1 x = -2 and x = -4 cao
Answer: x = -2 or x = -4

Question 2

M1 correct factorisation (x - 3)(x - 5) = 0 oe
A1 x = 3 and x = 5 cao
Answer: x = 3 or x = 5

Question 3

M1 (x - 9)(x + 9) = 0 oe, or x² = 81 seen
A1 x = 9 and x = -9 cao
Answer: x = 9 or x = -9

Question 4

M1 x(x + 9) = 0 oe
A1 x = 0 and x = -9 cao
Answer: x = 0 or x = -9

Question 5

M1 9x² = 16 oe, or correct factorisation (3x - 4)(3x + 4) = 0
A1 x = 4/3 oe
A1 x = -4/3 oe
Answer: x = 4/3 or x = -4/3

Question 6

M1 (x + 1)² = 16 oe
M1 x + 1 = ± 4
A1 x = 3 and x = -5 both cao
Answer: x = 3 or x = -5

Question 7

M1 correct factorisation (3x + 2)(x + 3) = 0 oe
A1 x = -2/3 oe
A1 x = -3 cao
Answer: x = -2/3 or x = -3

Question 8

M1 correct factorisation (2x + 3)(2x - 5) = 0 oe
A1 x = -3/2 oe
A1 x = 5/2 oe
Answer: x = -3/2 or x = 5/2

Question 9

M1 correct substitution into the quadratic formula, x = (-4 ± √16 + 12) / 2 oe
M1 simplifies √28 to 2*√7
A1 x = -2 + √7 and x = -2 - √7 oe cao
Answer: x = -2 + √7 or x = -2 - √7

Question 10

(a) M1 expands (x + 6)(x - 1) to obtain x² + 5x - 6
(a) A1 sets the expansion equal to 44 and rearranges correctly to x² + 5x - 50 = 0 (cso)
(a) Answer: x² + 5x - 50 = 0
(b) M1 correct factorisation (x + 10)(x - 5) = 0, or correct use of the quadratic formula
(b) A1 x = 5, rejecting x = -10 since x > 0
(b) Answer: x = 5
(c) B1 width = 4 cm (ft their x, width = x - 1)
(c) Answer: 4 cm

Question 11

M1 n(n + 2) = 168 oe, using n for the smaller even integer
M1 rearranges to n² + 2n - 168 = 0 and factorises correctly, e.g. (n - 12)(n + 14) = 0, or uses the quadratic formula
A1 n = 12, rejecting n = -14 as it is not a positive integer
B1 states the two integers are 12 and 14
Answer: The two integers are 12 and 14

Question 12

M1 correct factorisation (5x + 2)(x - 3) = 0 oe
A1 x = -2/5 oe
A1 x = 3 cao
Answer: x = -2/5 or x = 3

Question 13

M1 (x - 5)² - 22 = 0 oe correct completed-square form
M1 (x - 5)² = 22 leading to x - 5 = ± √22
A1 x = 5 + √22 and x = 5 - √22 both cao
Answer: x = 5 + √22 or x = 5 - √22

Question 14

(a) M1 (x + 5)² seen (correct value of a)
(a) A1 (x + 5)² - 6 oe cao (correct value of b)
(a) Answer: (x + 5)² - 6
(b) B1 minimum value = -6 (ft from part (a))
(b) B1 x = -5 (ft from part (a))
(b) Answer: Minimum value = -6 at x = -5

Question 15

M1 correct substitution into the quadratic formula, x = (3 ± √9 + 32) / 4 oe
A1 x = (3 + √41) / 4 oe cao
A1 x = (3 - √41) / 4 oe cao
Answer: x = (3 + √41) / 4 or x = (3 - √41) / 4

Question 16

M1 sets up -5t² + 25t + 2 = 22 oe
M1 rearranges to -5t² + 25t - 20 = 0 or t² - 5t + 4 = 0 oe
M1 correct factorisation (t - 1)(t - 4) = 0, or correct use of the quadratic formula
A1 t = 1
A1 t = 4
Answer: t = 1 second and t = 4 seconds

Question 17

M1 equates x² - x - 7 = 3x - 2 oe
dM1 rearranges to a correct three-term quadratic x² - 4x - 5 = 0 oe, dependent on the previous M1
M1 correct factorisation (x - 5)(x + 1) = 0, or correct use of the quadratic formula
A1 x = 5 and x = -1
A1 y = 13 and y = -5 respectively, cao
Answer: (x, y) = (5, 13) or (x, y) = (-1, -5)

Question 18

M1 completes the square correctly, (x + 2)² + 5 oe
M1 states or uses that (x + 2)² ≥ 0 for all real x
A1 correct conclusion: since (x + 2)² ≥ 0, (x + 2)² + 5 ≥ 5, so x² + 4x + 9 > 0 for all real x (cso)
Answer: x² + 4x + 9 = (x + 2)² + 5, and since (x + 2)² ≥ 0 for all x, the expression is always at least 5, so it is always positive

Question 19

(a) M1 two brackets of the form (x + p)(x + q) with pq = -24 attempted
(a) A1 (x + 6)(x - 4) cao
(a) Answer: (x + 6)(x - 4)
(b) B1 x = -6 and x = 4 both stated (ft from part (a))
(b) Answer: x = -6 or x = 4
(c) M1 recognises the substitution u = x - 1 (or an equivalent direct approach) to give u² + 2u - 24 = 0
(c) M1 u = -6 or u = 4, following the factorisation pattern from part (a)
(c) A1 x = -5 and x = 5 both cao
(c) Answer: x = -5 or x = 5

Question 20

M1 substitutes u = x² to form u² - 13u + 36 = 0 oe
M1 correct factorisation (u - 9)(u - 4) = 0
A1 u = 9 and u = 4
A1 x = 3 and x = -3 (from u = 9)
A1 x = 2 and x = -2 (from u = 4)
Answer: x = 3, x = -3, x = 2, x = -2

Question 21

M1 uses the condition for equal roots, b² - 4ac = 0, with a = p, b = 12, c = p
dM1 forms 144 - 4p² = 0 and rearranges to p² = 36 (dependent on the first M1)
A1 p = 6
A1 p = -6
Answer: p = 6 or p = -6

Solving Quadratics: Fluency and Exam Drill - Worksheets, Questions and Revision