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GCSE · Algebra

A3 Quadratic Equations: Factorising, the Formula and Completing the Square

AQA 8365 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Factorise each expression.

(a)x² + 7x + 12(1)

(b)x² - 5x - 24(2)

(Total for Question 1 is 3 marks)

Which of the following is the correct factorisation of x² + 2x - 15?

A) (x+5)(x-3)
B) (x-5)(x+3)
C) (x+15)(x-1)
D) (x-15)(x+1)

(Total for Question 2 is 1 mark)

Solve the equation x² + 3x - 10 = 0 by factorising. Show your working.

(Total for Question 3 is 3 marks)

Factorise fully: 6x² + 7x - 3

(Total for Question 4 is 3 marks)

Factorise each expression using the difference of two squares.

(a)9x² - 16(1)

(b)25x² - 4y²(1)

(Total for Question 5 is 4 marks)

Solve the equation 4x² - 12x + 5 = 0 by factorising.

(Total for Question 6 is 4 marks)

Solve 3x² + 5x - 7 = 0 using the quadratic formula. Give your answers correct to 2 decimal places.

(Total for Question 7 is 4 marks)

Show that the equation x² + 4x + 9 = 0 has no real roots.

(Total for Question 8 is 2 marks)

Find the value(s) of k for which the equation x² + kx + 9 = 0 has equal roots.

(Total for Question 9 is 3 marks)

By completing the square, write x² + 10x + 3 in the form (x+p)² + q, where p and q are constants.

(Total for Question 10 is 2 marks)

Using your answer to question 10, state the coordinates of the minimum point of the curve y = x² + 10x + 3.

(Total for Question 11 is 2 marks)

By completing the square, express 3x² - 18x + 7 in the form a(x+p)² + q, where a, p and q are constants to be found.

(Total for Question 12 is 3 marks)

Solve x² - 8x + 5 = 0 by completing the square. Give your answer in simplified surd form.

(Total for Question 13 is 4 marks)

Solve the inequality x² - 2x - 15 ≤ 0. Give your answer using inequality notation.

(Total for Question 14 is 3 marks)

The curve y = x² - 3x - 4 and the line y = 2x - 8 intersect at two points. Find the coordinates of both points of intersection.

(Total for Question 15 is 5 marks)

Solve the equation 5/x - 2/(x+3) = 1, giving your solutions as exact values.

(Total for Question 16 is 5 marks)

A rectangular vegetable patch has length (x+5) metres and width x metres. The area of the patch is 60 m^2.

(a)Show that x² + 5x - 60 = 0.(2)

(b)Solve the equation to find x, giving your answer correct to 2 decimal places. Reject any solution that is not valid in this context.(4)

(Total for Question 17 is 6 marks)

The equation x² - 2kx + (k² + 4) = 0, where k is a constant, is to be solved for x. Prove that this equation has no real roots for any value of k.

(Total for Question 18 is 4 marks)

By completing the square, prove that x² - 4x + 8 > 3 for all real values of x.

(Total for Question 19 is 4 marks)

Solve the equation x⁴ - 13x² + 36 = 0.

(Total for Question 20 is 5 marks)

Mark scheme · A3 Quadratic Equations: Factorising, the Formula and Completing the Square

Question 1

(a) B1 (x+3)(x+4) cao
(a) Answer: (x+3)(x+4)
(b) M1 identifies pair of numbers with product -24 and sum -5 (-8 and 3), or partial correct bracket
(b) A1 (x-8)(x+3) cao
(b) Answer: (x-8)(x+3)

Question 2

B1 A cao
Answer: A

Question 3

M1 attempt to factorise, e.g. seeking numbers with product -10 and sum 3
A1 (x+5)(x-2) = 0 oe cao
A1 x = -5 and x = 2 (both values) cao
Answer: x = -5 or x = 2

Question 4

M1 correct split of middle term using ac = -18, e.g. 6x² + 9x - 2x - 3
A1 correct partial factorisation, e.g. 3x(2x+3) - 1(2x+3)
A1 (3x-1)(2x+3) cao
Answer: (3x-1)(2x+3)

Question 5

(a) B1 (3x-4)(3x+4) cao
(a) Answer: (3x-4)(3x+4)
(b) B1 (5x-2y)(5x+2y) cao
(b) Answer: (5x-2y)(5x+2y)
(c) M1 takes out common factor 2: 2(x²-25)
(c) A1 2(x-5)(x+5) cao
(c) Answer: 2(x-5)(x+5)

Question 6

M1 attempt to factorise using ac = 20, splitting -12x as -2x-10x
A1 (2x-1)(2x-5) = 0 oe cao
A1 x = 0.5 (or 1/2) cao
A1 x = 2.5 (or 5/2) cao
Answer: x = 1/2 or x = 5/2

Question 7

M1 correct substitution into x = (-b ± √b²-4ac) / (2a) with a=3, b=5, c=-7
A1 discriminant = 109 (or √109 = 10.44...) seen
A1 x = 0.91 awrt
A1 x = -2.57 awrt
Answer: x = 0.91 or x = -2.57 (2dp)

Question 8

M1 discriminant = 4² - 4(1)(9) (= 16-36)
A1 = -20, and since -20 < 0 the equation has no real roots cso
Answer: Discriminant = -20 < 0, so no real roots

Question 9

M1 discriminant = k² - 4(1)(9) (= k²-36)
M1 sets discriminant = 0 and rearranges to k² = 36
A1 k = 6 or k = -6 (both values) oe cao
Answer: k = 6 or k = -6

Question 10

M1 (x+5)² seen
A1 (x+5)² - 22 cao
Answer: (x+5)² - 22

Question 11

B1 x = -5 ft from (x+5)² form
B1 y = -22 ft
Answer: Minimum point (-5, -22)

Question 12

M1 takes out factor 3: 3(x²-6x)+7
M1 3[(x-3)²-9]+7 oe
A1 3(x-3)² - 20 cao (a=3, p=-3, q=-20)
Answer: 3(x-3)² - 20

Question 13

M1 (x-4)² - 16 + 5 (= 0) oe
M1 (x-4)² = 11
A1 x = 4 ± √11 oe (unsimplified acceptable at this stage)
A1 x = 4 + √11 and x = 4 - √11 fully simplified surd form cao
Answer: x = 4 + √11 or x = 4 - √11

Question 14

M1 factorises to (x-5)(x+3) (=0)
A1 critical values x = 5 and x = -3
A1 -3 ≤ x ≤ 5 oe cao (correct inequality direction, since coefficient of x² is positive)
Answer: -3 ≤ x ≤ 5

Question 15

M1 sets x²-3x-4 = 2x-8
A1 rearranges to x² - 5x + 4 = 0 oe
M1 solves by factorising: (x-1)(x-4) = 0
A1 x = 1 and x = 4 (both values)
A1 correct coordinates (1,-6) and (4,0) cao
Answer: (1, -6) and (4, 0)

Question 16

M1 multiplies through by x(x+3): 5(x+3) - 2x = x(x+3)
M1 expands correctly: 5x+15-2x = x²+3x
A1 simplifies to x² = 15 oe (e.g. x²+3x-3x-15=0)
A1 x = √15
A1 x = -√15
Answer: x = √15 or x = -√15

Question 17

(a) M1 area = x(x+5) = 60
(a) A1 x² + 5x - 60 = 0 cso (answer given, correct expansion shown)
(a) Answer: x² + 5x - 60 = 0
(b) M1 correct substitution into quadratic formula with a=1, b=5, c=-60
(b) A1 discriminant = 265 (or √265 = 16.27...) seen
(b) A1 x = 5.64 awrt
(b) B1 rejects x = -10.64 with reason that a length cannot be negative
(b) Answer: x = 5.64 m (2dp)

Question 18

M1 discriminant = (-2k)² - 4(1)(k²+4)
M1 correctly expands to 4k² - 4k² - 16
A1 simplifies to discriminant = -16
A1 since -16 < 0 for all values of k, the equation has no real roots for any k cso
Answer: Discriminant = -16 for all k, so no real roots for any value of k

Question 19

M1 (x-2)² - 4 + 8 seen
A1 correctly simplifies to (x-2)² + 4
M1 states (x-2)² ≥ 0 for all real x, so the minimum value of the expression is 4
A1 since 4 > 3, x²-4x+8 > 3 for all real x cso
Answer: x² - 4x + 8 = (x-2)² + 4 ≥ 4 > 3 for all real x

Question 20

M1 substitutes u = x² to form u² - 13u + 36 = 0
A1 factorises to (u-9)(u-4) = 0, giving u = 9 or u = 4
M1 back-substitutes: x² = 9 and x² = 4
A1 x = 3 and x = -3
A1 x = 2 and x = -2
Answer: x = 3, -3, 2, -2

Quadratic Equations: Factorising, the Formula and Completing the Square - Worksheets, Questions and Revision