Practise 37 exam-style questions on Quadratic equations for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
37 questions
1–5 marks each
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20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Rearrange so that one side is zero, then factorise. Dividing both sides by \(x\) loses the solution \(x = 0\), which is the most common mistake with this type of equation.
Mark scheme
1 mark
M1 for \(x^2 - 7x = 0\) and \(x(x - 7) = 0\)
oe; correct answer with no working scores full marks
With a coefficient of 2 for \(x^2\), find factors of \(2 \times (-15) = -30\) that add to 7 (10 and −3), or use the quadratic formula: \(b^2 - 4ac = 49 + 120 = 169\), a perfect square, so the equation factorises.
Mark scheme
1 mark
M1 for \((2x \pm 3)(x \pm 5)\)
or for a correct substitution into the quadratic formula; correct answer with no working scores full marks
1 mark
M1 for \((2x - 3)(x + 5) = 0\)
or for \(x = \frac{-7 \pm \sqrt{169}}{4}\)
1 mark
A1 for \(x = 1.5\) and \(x = -5\)
exact values; oe, e.g. \(x = \frac{3}{2}\); both values needed
Question 3
3 marks
Solve \(x + \dfrac{6}{x} = 7.5\) Give your solutions correct to 3 significant figures.
Model answer
Multiply every term by \(x\): \(x^2 + 6 = 7.5x\) \(x^2 - 7.5x + 6 = 0\) \(x = \dfrac{7.5 \pm \sqrt{56.25 - 24}}{2} = \dfrac{7.5 \pm \sqrt{32.25}}{2}\) \(x = 6.5894...\) or \(x = 0.9105...\) \(x = 6.59\) or \(x = 0.911\) (3 s.f.)
Explanation
Multiply every term by \(x\) to clear the fraction; this gives a quadratic. Make sure the 7.5 is multiplied by \(x\) too. Check one answer: \(6.5894... + \frac{6}{6.5894...} = 7.5\).
Mark scheme
1 mark
M1 for multiplying by \(x\) and rearranging: \(x^2 - 7.5x + 6 = 0\)
oe, e.g. \(2x^2 - 15x + 12 = 0\); correct answer with no working scores full marks
1 mark
M1 for a correct substitution into the formula
e.g. \(\dfrac{7.5 \pm \sqrt{56.25 - 24}}{2}\)
1 mark
A1 for 6.59 and 0.911
awrt 6.59 and 0.911 (6.5894... and 0.9105...)
Question 4
5 marks
A rectangle has length \((2x - 1)\) cm and width \((x + 3)\) cm. The area of the rectangle is 40 cm².
Work out the perimeter of the rectangle. Give your answer correct to 3 significant figures.
Form an equation from the area, rearrange it to equal 0 and use the formula. Only the positive solution makes sense, because a length cannot be negative. The perimeter simplifies to \(6x + 4\); use the unrounded value of \(x\).
Mark scheme
1 mark
P1 for forming the equation \((2x - 1)(x + 3) = 40\)
oe; correct answer with no working scores full marks
1 mark
P1 for rearranging to \(2x^2 + 5x - 43 = 0\)
oe
1 mark
P1 for a correct substitution into the formula, e.g. \(\dfrac{-5 \pm \sqrt{25 + 344}}{4}\)
oe
1 mark
P1 for \(x = 3.55...\) (rejecting the negative solution)
accept 3.55 or better
1 mark
A1 for 25.3 cm
awrt 25.3 (25.314...)
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