Solving quadratic equations
Solving by factorising and by the quadratic formula.
Review these first
Learning objectives
What you need to be able to do
Teacher-mapped phrasing — check against the official Cambridge syllabus for exact wording.
- 2.5.1Solve quadratic equations by factorising, completing the square and using the formula.
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Solving quadratic equations
A quadratic equation has the form ax² + bx + c = 0 and usually has two solutions.
Step zero — rearrange to equal zero
Nothing works until everything is on one side. x² = 5x − 6 must become x² − 5x + 6 = 0 first.
Solving by factorising
Factorise, then use the fact that if two things multiply to zero, at least one must be zero. x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3.
The quadratic formula
When factorising fails, use x = (−b ± √(b² − 4ac)) / 2a Write down a, b and c explicitly first, with their signs, before substituting. Most formula errors are sign errors made while substituting, not arithmetic errors afterwards.
The discriminant
The part under the root, b² − 4ac, tells you how many solutions exist:
- Positive → two distinct real solutions.
- Zero → one repeated solution.
- Negative → no real solutions (the curve never crosses the x-axis).
Presenting answers
Give both solutions. If the question asks for a specific number of decimal places or significant figures, round only at the very end — rounding early shifts the answer.
Think of it like this
The "equals zero" step is the whole trick: zero is the only number where a product tells you something definite about its factors. If two numbers multiply to 12 you know nothing; if they multiply to 0, one of them *is* 0.
Worked examples
Method, step by step
Solve x² − 5x + 6 = 0 by factorising.
- 1Find two numbers that multiply to 6 and add to −5: these are −2 and −3.
- 2Factorise: (x − 2)(x − 3) = 0.
- 3For the product to be zero, one bracket must be zero.
- 4So x − 2 = 0 giving x = 2, or x − 3 = 0 giving x = 3.
x = 2 or x = 3
Solve 2x² + 3x − 4 = 0, giving answers to 2 decimal places.
- 1Identify a = 2, b = 3, c = −4.
- 2Substitute into x = (−b ± √(b² − 4ac)) / 2a: x = (−3 ± √(9 − 4×2×(−4))) / 4.
- 3The discriminant is 9 + 32 = 41, and √41 ≈ 6.4031.
- 4x = (−3 + 6.4031)/4 = 0.8508 or x = (−3 − 6.4031)/4 = −2.3508.
x = 0.85 or x = −2.35 (2 d.p.)
Common misconceptions
- Using the formula without rearranging to = 0 first, so a, b and c are wrong from the start.
- Giving only one solution. A quadratic normally has two, and both are usually needed for full marks.
- Mis-signing b when b is negative. For `x² − 3x + 2`, b = −3, so −b = +3.
In the exam
- If a question says "give your answers to 2 decimal places", that is a strong hint the equation does not factorise — go straight to the formula.
- State a, b and c on their own line before substituting. It costs one line and prevents the most common error in the topic.