Practise Solving quadratic equations. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
Solve \(y^{2} + 6y - 7 = 0\)[3]
Show the answer and mark scheme
Answer: \(y = -7, \ y = 1\)
M1 for \((y \pm 1)(y \pm 7)\)
A1 for \((y - 1)(y + 7)\)
A1 for \(y = -7\) and \(y = 1\)
Worked solution: \(y^{2} + 6y - 7 = (y - 1)(y + 7) = 0\), so \(y = 1\) or \(y = -7\)
Question 2Medium3 marks
Rob solves \(x^2 = 5x\). He divides both sides by \(x\) and writes \(x = 5\).
(a) Explain why Rob's solution is not complete.[1]
(b) Solve \(2x^2 = 7x\).[2]
Show the answer and mark scheme
(a)Answer: Dividing by \(x\) loses the solution \(x = 0\) (you cannot divide by \(x\) if \(x\) might be 0).
C1 for explaining that \(x = 0\) is also a solution, lost when dividing by \(x\)
Worked solution: \(x^2 - 5x = 0\) gives \(x(x - 5) = 0\), so \(x = 0\) or \(x = 5\). Dividing by \(x\) assumes \(x \ne 0\), so the solution 0 was lost.
(b)Answer: \(x = 0\) or \(x = \frac{7}{2}\)
M1 for \(x(2x - 7) = 0\) or \(2x^2 - 7x = 0\)
A1 for \(x = 0\) and \(x = \frac{7}{2}\) oe
Worked solution: \(2x^2 - 7x = 0\), so \(x(2x - 7) = 0\): \(x = 0\) or \(x = 3.5\).
Question 3Hard6 marks
A rectangular lawn measures 12 m by 8 m. A path of width \(x\) metres is laid all the way around the outside of the lawn.
(a) Show that the area of the path, in m2, is \(4x^2 + 40x\).[3]
(b) The area of the path is equal to the area of the lawn. Work out the width of the path.[3]
M1 for the outer dimensions \(12 + 2x\) and \(8 + 2x\)
M1 for \((12 + 2x)(8 + 2x) - 12 \times 8\) with a correct expansion, \(96 + 24x + 16x + 4x^2\)
C1 for \(4x^2 + 40x\) from fully correct working
Worked solution: The outer rectangle is \((12 + 2x)\) by \((8 + 2x)\). Path area \(= (12 + 2x)(8 + 2x) - 96 = 96 + 24x + 16x + 4x^2 - 96 = 4x^2 + 40x\)
(b)Answer: 2 m
M1 for \(4x^2 + 40x = 96\)
M1 for \(x^2 + 10x - 24 = 0\) and a correct method to solve, e.g. \((x + 12)(x - 2) = 0\)
A1 for 2 (with −12 rejected)
Worked solution: \(4x^2 + 40x = 96\), so \(x^2 + 10x - 24 = 0\) and \((x + 12)(x - 2) = 0\). A width cannot be negative, so \(x = 2\): the path is 2 m wide.