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8.4D Completing the Square: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculators not allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Completing the Square: Method Reminder

Revision Library original resource, written for Edexcel 1MA1 topic 8.4.

To complete the square for x^2 + bx + c, halve the coefficient of x to get b/2, then write (x + b/2)^2 - (b/2)^2 + c. For ax^2 + bx + c with a not equal to 1, first factorise a out of the x^2 and x terms before completing the square inside the bracket, then multiply back through. Once in the form a(x+p)^2 + q, you can read off the turning point of the graph directly: minimum (or maximum) at (-p, q). To solve ax^2 + bx + c = 0, isolate the squared bracket and take a square root of both sides, remembering the +/- sign.

Write x² + 16x in the form (x + p)² + q.

(Total for Question 1 is 2 marks)

Write x² + 18x + 4 in the form (x + p)² + q.

(Total for Question 2 is 2 marks)

Write x² - 20x + 3 in the form (x + p)² + q.

(Total for Question 3 is 2 marks)

Write x² + 9x in the form (x + p)² + q, where p and q may be fractions.

(Total for Question 4 is 3 marks)

Write x² - 7x + 1 in the form (x + p)² + q.

(Total for Question 5 is 3 marks)

Write x² - 16x - 5 in the form (x + p)² + q.

(Total for Question 6 is 3 marks)

Write 5x² + 20x + 3 in the form a(x + p)² + q, stating the values of a, p and q.

(Total for Question 7 is 4 marks)

Write 6x² - 24x + 11 in the form a(x + p)² + q, stating the values of a, p and q.

(Total for Question 8 is 4 marks)

The graph of y = (x - 7)² + 2 is a translation of the graph of y = x^2. State the coordinates of its minimum point.

(Total for Question 9 is 2 marks)

By completing the square, find the minimum value of f(x) = x² + 4x + 9, and state the value of x at which this minimum occurs.

(Total for Question 10 is 3 marks)

Solve x² + 14x + 9 = 0 by completing the square. Give your answers in surd form, simplified where possible.

(Total for Question 11 is 4 marks)

Solve x² - 6x - 3 = 0 by completing the square. Give your answers in surd form, simplified where possible.

(Total for Question 12 is 4 marks)

The expression x² + 2x - 8 is to be written in completed-square form.

(a)Write x² + 2x - 8 in the form (x + p)² + q.(2)

(b)Hence solve x² + 2x - 8 = 0, showing your method clearly.(3)

(Total for Question 13 is 5 marks)

Express 3x² - 12x + 7 in the form a(x + p)² + q, stating the values of a, p and q.

(Total for Question 14 is 4 marks)

Show that x² + 4x + 13 is positive for all values of x.

(Total for Question 15 is 3 marks)

The curve with equation y = x² + px + q has a minimum point at (5, -3). Find the values of p and q.

(Total for Question 16 is 4 marks)

A market trader models the weekly profit, L pounds, from selling padlocks at a price of x pounds each, using L = -3x² + 30x - 39.

(a)Show that L can be written as -3(x - 5)² + 36.(2)

(b)Hence find the price, x, that maximises the weekly profit, and state the maximum profit.(2)

(Total for Question 17 is 4 marks)

Given that x² + 24x + k = (x + 12)² for all values of x, find the value of k. Hence state the minimum value of y = x² + 24x + k.

(Total for Question 18 is 3 marks)

The curve y = 4x² - 24x + c has a minimum value of 11. Find the value of c, and state the coordinates of the minimum point.

(Total for Question 19 is 6 marks)

Prove that the equation x² + 2kx + k² + 3 = 0, where k is a constant, has no real solutions for any value of k.

(Total for Question 20 is 4 marks)

The function f(x) = ax² + bx + c can be written as f(x) = 3(x - 2)² + 5.

(a)Find the values of a, b and c.(2)

(b)Prove that the equation f(x) = 4 has no real solutions.(3)

(Total for Question 21 is 5 marks)

Mark scheme · 8.4D Completing the Square: Fluency and Exam Drill

Question 1

M1 (x + 8)² seen, or correct method to halve the coefficient of x
A1 (x + 8)² - 64 cao
Answer: (x + 8)² - 64

Question 2

M1 (x + 9)² seen
A1 (x + 9)² - 77 cao
Answer: (x + 9)² - 77

Question 3

M1 (x - 10)² seen
A1 (x - 10)² - 97 cao
Answer: (x - 10)² - 97

Question 4

M1 attempt to halve 9 to get 9/2 oe
M1 (x + 9/2)² seen
A1 (x + 9/2)² - 81/4 cao (accept -20.25)
Answer: (x + 9/2)² - 81/4

Question 5

M1 (x - 7/2)² seen
M1 attempt to combine -49/4 with +1
A1 (x - 7/2)² - 45/4 cao (accept -11.25)
Answer: (x - 7/2)² - 45/4

Question 6

M1 (x - 8)² seen
M1 attempt to combine -64 with -5
A1 (x - 8)² - 69 cao
Answer: (x - 8)² - 69

Question 7

M1 5(x² + 4x) + 3 seen, factor of 5 taken out correctly
M1 (x + 2)² seen within the bracket
A1 5(x + 2)² - 20 + 3 correctly simplified (dep. on both M marks)
A1 a = 5, p = 2, q = -17 all correct, fully simplified 5(x + 2)² - 17 cao
Answer: 5(x + 2)² - 17 (a = 5, p = 2, q = -17)

Question 8

M1 6(x² - 4x) + 11 seen, factor of 6 taken out correctly
M1 (x - 2)² seen within the bracket
A1 6(x - 2)² - 24 + 11 correctly simplified (dep. on both M marks)
A1 a = 6, p = -2, q = -13 all correct, fully simplified 6(x - 2)² - 13 cao
Answer: 6(x - 2)² - 13 (a = 6, p = -2, q = -13)

Question 9

B1 x-coordinate = 7
B1 y-coordinate = 2 (accept coordinate pair (7, 2) for both marks if fully correct)
Answer: (7, 2)

Question 10

M1 (x + 2)² seen, complete the square attempted
A1 (x + 2)² + 5 correct, hence minimum value = 5
B1 ft x = -2 (follow through from their p value)
Answer: minimum value = 5, at x = -2

Question 11

M1 (x + 7)² - 49 + 9 = 0 oe, correct completed square set to zero
M1 (x + 7)² = 40 (dep. on previous M1)
M1 x + 7 = ± √40, condone unsimplified surd
A1 x = -7 + 2sqrt(10) or x = -7 - 2sqrt(10) oe, both fully simplified, cao
Answer: x = -7 + 2sqrt(10) or x = -7 - 2sqrt(10)

Question 12

M1 (x - 3)² - 9 - 3 = 0 oe, correct completed square set to zero
M1 (x - 3)² = 12 (dep. on previous M1)
M1 x - 3 = ± √12, condone unsimplified surd
A1 x = 3 + 2sqrt(3) or x = 3 - 2sqrt(3) oe, both fully simplified, cao
Answer: x = 3 + 2sqrt(3) or x = 3 - 2sqrt(3)

Question 13

(a) M1 (x + 1)² seen
(a) A1 (x + 1)² - 9 cao
(a) Answer: (x + 1)² - 9
(b) M1 ft (x + 1)² = 9, using the answer to part (a)
(b) M1 x + 1 = ± 3
(b) A1 x = 2 and x = -4, both correct, cao
(b) Answer: x = 2 or x = -4

Question 14

M1 3(x² - 4x) + 7 seen, factor of 3 taken out correctly
M1 (x - 2)² seen within the bracket
A1 3(x - 2)² - 12 + 7 correctly simplified (dep. on both M marks)
A1 a = 3, p = -2, q = -5 all correct, fully simplified 3(x - 2)² - 5 cao
Answer: 3(x - 2)² - 5 (a = 3, p = -2, q = -5)

Question 15

M1 (x + 2)² seen, complete the square attempted
A1 (x + 2)² + 9 correct
B1 dep on A1: valid reasoning that (x + 2)² ≥ 0 for all real x, so (x + 2)² + 9 ≥ 9, and 9 > 0, hence the expression is always positive (must reference (x + 2)² ≥ 0 explicitly)
Answer: x² + 4x + 13 = (x + 2)² + 9 ≥ 9 > 0 for all x, as required

Question 16

M1 recognise y = (x - 5)² - 3, using the given minimum point to write the completed-square form
M1 expand (x - 5)² to get x² - 10x + 25
A1 p = -10 cao
A1 q = 22 cao
Answer: p = -10, q = 22

Question 17

(a) M1 -3(x² - 10x) - 39 seen, factor of -3 taken out correctly
(a) A1 -3(x - 5)² + 36 correctly derived, showing -3[(x-5)² - 25] - 39 = -3(x-5)² + 75 - 39
(a) Answer: L = -3(x - 5)² + 36
(b) B1 ft x = 5 (pounds), using -3(x-5)² ≤ 0 for all real x so L is greatest when (x-5)² = 0
(b) A1 maximum profit = 36 (pounds)
(b) Answer: x = 5 pounds; maximum profit = 36 pounds

Question 18

M1 expand (x + 12)² = x² + 24x + 144, or compare coefficients directly
A1 k = 144 cao
B1 ft minimum value = 0, occurring at x = -12
Answer: k = 144; minimum value of y = 0

Question 19

M1 4(x² - 6x) + c seen, factor of 4 taken out correctly
M1 (x - 3)² seen within the bracket
A1 4(x - 3)² - 36 + c correctly simplified expression for y
M1 dep: set -36 + c = 11, using their simplified constant term
A1 c = 47 cao
B1 ft minimum point = (3, 11)
Answer: c = 47; minimum point (3, 11)

Question 20

M1 recognise x² + 2kx + k² = (x + k)², a perfect square
A1 equation rewritten as (x + k)² + 3 = 0 oe
M1 dep: valid reasoning that (x + k)² ≥ 0 for all real x (and all real k), so (x + k)² + 3 ≥ 3
A1 cso: hence (x + k)² + 3 cannot equal 0, since it is always ≥ 3 > 0, so the equation has no real solutions for any value of k
Answer: (x + k)² + 3 = 0 has no real solutions, since (x + k)² + 3 ≥ 3 > 0 for all real x and k

Question 21

(a) M1 expand 3(x - 2)² = 3x² - 12x + 12, or equivalent correct expansion
(a) A1 a = 3, b = -12, c = 17 all correct, cao
(a) Answer: a = 3, b = -12, c = 17
(b) M1 set up 3(x - 2)² + 5 = 4, leading to 3(x - 2)² = -1 oe
(b) A1 (x - 2)² = -1/3 oe
(b) B1 cso, dep: valid reasoning that (x - 2)² ≥ 0 for all real x, so it cannot equal a negative number, hence f(x) = 4 has no real solutions
(b) Answer: f(x) = 4 has no real solutions, since this would require (x - 2)² = -1/3, which is impossible for real x

Completing the Square: Fluency and Exam Drill - Worksheets, Questions and Revision