Quadratic equations come up on every single GCSE Maths Higher paper, and on most Foundation papers too. If you can solve them confidently — and if you can pick the right method for the question — you've already secured around eight to twelve marks before you even open the geometry section. This guide walks through everything I teach my Year 11s in their first month with me, including the three or four mistakes I see most often when marking exam papers.
What a quadratic equation actually is
A quadratic is any equation you can rearrange into the form:
ax² + bx + c = 0
The "quadratic" bit comes from quadratus, Latin for square — the giveaway is that x² term. The coefficients a, b and c are just numbers. The only rule is that a ≠ 0 (otherwise the x² disappears and you've got a linear equation, which is much easier).
So 2x² + 7x − 4 = 0 is quadratic (a = 2, b = 7, c = −4). And x² = 9 is also quadratic — once you rearrange it to x² − 9 = 0 (so a = 1, b = 0, c = −9). On exam papers, the equation often arrives disguised as a worded question, and step one is always to get it into the standard form above.
The three methods you need
There are three ways of solving a quadratic on the GCSE syllabus. I'll be honest: most students try to use the same method on every question, and that's the single biggest reason they lose marks. Different equations want different methods. Here's how to decide.
- Factorising — fastest when it works. Use it first if you can spot it.
- Completing the square — needed when the question asks for it, and useful for finding the turning point of a parabola.
- The quadratic formula — the universal solver. Slower, but always works.
Method 1 — Factorising
Factorising means rewriting ax² + bx + c as two brackets multiplied together. If you can do that, the rest is easy: if two things multiply to give zero, at least one of them must be zero.
Worked example
Solve x² − 5x + 6 = 0
I need two numbers that multiply to give +6 and add to give −5. Pause and run through the factor pairs of 6: (1, 6), (2, 3). Of those, the pair that adds to −5 is −2 and −3.
So the factorised form is:
(x − 2)(x − 3) = 0
Either bracket equals zero, so either x = 2 or x = 3. Two solutions.
Factorising gets harder when a ≠ 1. For something like 2x² + 7x + 3 = 0 you need two numbers that multiply to give a × c = 6 and add to give b = 7. Those are 6 and 1. Split the middle term: 2x² + 6x + x + 3 = 0, then factor in pairs: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0. So x = −½ or x = −3.
Method 2 — Completing the square
This one looks scary the first time. It's not. The trick is to take the x² + bx part and rewrite it as a perfect square (x + b/2)², which expands to x² + bx + (b/2)² — so we've added an extra (b/2)² that we then have to subtract back.
Worked example
Express x² + 6x − 5 in the form (x + p)² + q.
Half of the coefficient of x is +3, so the square part is (x + 3)². But (x + 3)² = x² + 6x + 9, which has an extra +9 we don't want. So we subtract 9 to compensate:
x² + 6x − 5 = (x + 3)² − 9 − 5 = (x + 3)² − 14
So p = 3 and q = −14.
Why bother? Two big reasons. First, once an equation is in the form (x + p)² + q = 0, you can solve it directly: (x + p)² = −q, then take the square root of both sides. Second, the completed-square form tells you the turning point of the parabola straight off — for (x + 3)² − 14, the minimum point is at (−3, −14). That's a free A* mark whenever it comes up.
Method 3 — The quadratic formula
This is the one to learn off by heart. It works for every quadratic, including the awful ones with decimal coefficients or solutions that don't simplify to whole numbers.
x = (−b ± √(b² − 4ac)) ÷ 2a
Worked example
Solve 2x² + 5x − 3 = 0 using the quadratic formula.
Here a = 2, b = 5, c = −3. Substitute carefully — this is where most marks go missing:
x = (−5 ± √(5² − 4 × 2 × (−3))) ÷ (2 × 2)
x = (−5 ± √(25 + 24)) ÷ 4
x = (−5 ± √49) ÷ 4
x = (−5 ± 7) ÷ 4
So either x = (−5 + 7) ÷ 4 = ½, or x = (−5 − 7) ÷ 4 = −3. Two solutions: x = 0.5 or x = −3.
Quick sanity check: substitute x = 0.5 back in. 2(0.25) + 5(0.5) − 3 = 0.5 + 2.5 − 3 = 0. ✓
The discriminant — and what it tells you
The bit under the square root in the quadratic formula, b² − 4ac, is called the discriminant. You don't always have to compute the whole formula — sometimes the question only asks how many solutions exist, and the discriminant alone tells you:
- If
b² − 4ac > 0→ two distinct real solutions (the parabola crosses the x-axis at two points) - If
b² − 4ac = 0→ exactly one repeated solution (the parabola just touches the x-axis) - If
b² − 4ac < 0→ no real solutions (the parabola never crosses the x-axis at all)
Discriminant questions are a regular feature on the AQA and Edexcel Higher papers. They're often phrased as "show that the equation has no real solutions" or "find the values of k for which the equation has equal roots". Both reduce to setting up an inequality or an equation involving b² − 4ac.
The mistakes that cost the most marks
Exam-day checklist
- Get the equation into the form
ax² + bx + c = 0before doing anything else. - Glance at the coefficients. Are a, b, c small whole numbers? Try factorising. Otherwise jump to the formula.
- If the question explicitly says "by completing the square" or "leave your answer in surd form" or "give exact values", use completing the square or the formula — never decimal-rounded answers.
- Always write out the substitution into the quadratic formula step-by-step. Mark schemes credit method even when the final number is wrong.
- For worded problems, finish by going back to the original context and rejecting any invalid solution with a brief reason.
Where this leads next
Once quadratics are solid, three topics open up almost immediately: simultaneous equations (where you'll often substitute into a quadratic), graphing quadratic functions (where the completed-square form gives you the turning point for free), and quadratic inequalities (which always start with solving the equation first). Working through them in that order is the fastest way to lift a Higher grade from a 5 or 6 up to a 7 or 8.
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