Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Algebra
Practise Solving quadratic equations. 1 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.
Solving equations such as \(x^2 - 5x + 6 = 0\). Foundation: factorising \(x^2 + bx + c\) and reading solutions from a graph. Higher: factorising \(ax^2 + bx + c\), the quadratic formula, and equations that need rearranging first (completing the square has its own notes).
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
4
Read solutions from a quadratic graphThe solutions of \(x^2 + bx + c = 0\) are the x-coordinates where \(y = x^2 + bx + c\) crosses the x-axis.
5
Solve simple quadratics by factorisingFactorise, then set each bracket equal to 0, e.g. \((x - 3)(x + 7) = 0\) gives \(x = 3\) or \(x = -7\).
5
Solve quadratics with a missing term\(x^2 = 49\) gives \(x = \pm 7\); \(x^2 - 5x = 0\) gives \(x(x - 5) = 0\), so \(x = 0\) or \(x = 5\).
Notes
Rearrange to equal 0
A quadratic equation has an \(x^2\) term and no higher power of x. Get everything on one side so that it equals 0.
\(x^2 + 3x = 10\) becomes \(x^2 + 3x - 10 = 0\).
A quadratic usually has two solutions (roots). Some have one repeated root, and some have no solutions.
Solving by factorising
Factorise, then use the fact that if two things multiply to give 0, one of them must be 0.
Check each root in the original equation: \(4^2 + 2 \times 4 - 24 = 0\).
For \(x^2 + bx + c = 0\), the two roots add up to \(-b\) and multiply to give c: a quick check.
If your calculator has an equation solver, use it to check, but always write out the method to get the marks.
Higher: if the question asks for answers to 2 d.p. or 3 s.f., it won't factorise, so go straight to the formula.
Where marks are lost
Not rearranging to equal 0 first, e.g. solving \(x(x + 1) = 12\) as \(x = 12\) or \(x + 1 = 12\).
Wrong signs: \((x + 6)(x - 4) = 0\) gives \(x = -6\) and \(x = 4\), not \(6\) and \(-4\).
Dividing both sides by x and losing the solution \(x = 0\).
Forgetting the negative square root: \(x^2 = 25\) gives \(x = 5\) or \(x = -5\).
Higher: in the formula, the whole of \(-b \pm \sqrt{b^2 - 4ac}\) is divided by \(2a\).
Exam technique
'Solve' needs values of x: if you only factorise, you only get the method mark.
Write the formula with the numbers substituted before using your calculator: that line is the method mark.
In context questions, say which solution you reject and why.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
Solve \(y^2 - 4 = 0\)[2]
Show the answer and mark scheme
Answer: \(y = 2, \ y = -2\)
M1 for \((y - 2)(y + 2) = 0\) or \(y^2 = 4\)
A1 for \(y = 2\) and \(y = -2\)
Worked solution: \(y^2 = 4\), so \(y = \pm 2\)
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 \(x = 3.5\).