Practise 37 exam-style questions on Quadratic equations for Edexcel GCSE Maths (Non-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
Instant marking
No sign-up needed
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.
Question 1
2 marks
Sam is solving the equation \(x^2 = 5x\) He divides both sides by \(x\) and writes \(x = 5\)
Explain what is wrong with Sam's method and give the correct solutions of \(x^2 = 5x\)
Model answer
Sam has divided by \(x\), but \(x\) could be 0, so he has lost a solution. \(x^2 - 5x = 0\) \(x(x - 5) = 0\) \(x = 0\) or \(x = 5\)
Explanation
Never divide both sides of an equation by an expression that could be zero. Instead, rearrange so one side is 0 and factorise: \(x(x - 5) = 0\) shows both solutions. Check: \(0^2 = 5 \times 0\) is true.
Mark scheme
1 mark
C1 for a correct explanation, e.g. dividing by \(x\) loses the solution \(x = 0\) (you cannot divide by \(x\) if \(x\) could be 0)
oe
1 mark
B1 for \(x = 0\) and \(x = 5\)
both needed
Question 2
2 marks
Use the quadratic formula to explain why the equation \(3x^2 - 5x + 4 = 0\) has no real solutions.
Model answer
In the formula \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), \(a = 3,\ b = -5,\ c = 4\) \(b^2 - 4ac = 25 - 48 = -23\) You cannot find the square root of a negative number, so there are no real solutions.
Explanation
The part of the formula under the square root, \(b^2 - 4ac\), decides how many solutions there are. If it is negative, the square root does not exist (as a real number), so the equation has no real solutions and its graph does not cross the \(x\)-axis.
Mark scheme
1 mark
M1 for working out \(b^2 - 4ac = (-5)^2 - 4 \times 3 \times 4\) \((= 25 - 48 = -23)\)
1 mark
C1 for explaining that \(b^2 - 4ac\) is negative, so its square root (in the formula) has no real value
Expand the bracket and rearrange so that the equation equals zero before using the formula. 41 has no square factors, so \(\sqrt{41}\) cannot be simplified.
Mark scheme
1 mark
M1 for rearranging to \(2x^2 + 3x - 4 = 0\)
correct answer with no working scores full marks
1 mark
M1 for correct substitution into the quadratic formula, \(\dfrac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-4)}}{2 \times 2}\)
or completing the square to \(\left(x + \dfrac{3}{4}\right)^2 = \dfrac{41}{16}\)
1 mark
A1 for \(x = \dfrac{-3 \pm \sqrt{41}}{4}\)
Question 4
4 marks
A ball is thrown upwards. Its height, \(h\) metres, above the ground \(t\) seconds after it is thrown is given by \(h = 20t - 5t^2\)
Work out the length of time for which the ball is more than 15 metres above the ground.
Model answer
\(20t - 5t^2 = 15\) \(5t^2 - 20t + 15 = 0\) \(t^2 - 4t + 3 = 0\) \((t - 1)(t - 3) = 0\) \(t = 1\) or \(t = 3\) The ball is above 15 m between 1 second and 3 seconds. Time \(= 3 - 1 = 2\) seconds
Explanation
The ball passes 15 m twice: once going up and once coming down. Solve for both times by dividing the equation by 5 to make it easier to factorise. The time above 15 m is the difference between the two solutions.
Mark scheme
1 mark
P1 for forming the equation \(20t - 5t^2 = 15\)
oe, or the inequality \(20t - 5t^2 > 15\); correct answer with no working scores full marks
1 mark
P1 for rearranging and simplifying to \(t^2 - 4t + 3 = 0\)
oe, e.g. \(5t^2 - 20t + 15 = 0\)
1 mark
A1 for \(t = 1\) and \(t = 3\)
1 mark
A1 for 2 (seconds)
cao
Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.
Ctrl + Enter to check
AI TutorHints first, answers after you try
Hi! Stuck? Ask me for a hint and I’ll help you work it out. Once you’ve checked your answer I can explain exactly where the marks are.
