How to Solve Quadratic Equations

La méthode, étape par étape

1. 01

   Rearrange into standard form

   x² + 5x = 24 → x² + 5x - 24 = 0

   Every method that follows assumes one side is zero, because the whole approach rests on the fact that a product equals zero only when one of its factors is zero. An equation left as x² + 5x = 24 cannot use that, since two brackets multiplying to 24 tells you nothing about either one.

2. 02

   Look for two numbers that multiply to ac and add to b

   need: product -24, sum +5 → +8 and -3

   This is the search that factorising actually is. Here a = 1, so the pair goes straight into the brackets. Listing the factor pairs of 24 and testing their sums is faster than staring at it: 1 and 24, 2 and 12, 3 and 8, 4 and 6 — and 8 with −3 gives −24 and +5.

3. 03

   Write the brackets and set each to zero

   (x + 8)(x - 3) = 0 → x + 8 = 0 or x - 3 = 0

   If two things multiply to zero, at least one of them is zero — that single fact is what converts a quadratic into two linear equations. It is also why the rearranging in step one was not optional. Expanding the brackets back to x² + 5x − 24 is a five-second check that the factorising is right.

4. 04

   Read off both roots

   x = -8 or x = 3

   A quadratic has two roots unless the brackets are identical, and giving only one is a lost mark even when the one given is correct. Each bracket is solved as an ordinary linear equation, so the sign flips: (x + 8) gives x = −8, not +8.

5. 05

   Use the formula when it will not factorise

   x = (-b ± √(b² - 4ac)) / (2a)

   Most quadratics have no whole-number factor pair, and hunting for one that does not exist is where exam time disappears. If a minute of searching produces nothing, move to the formula — it works on every quadratic, factorisable or not, and the question usually signals it by asking for answers to a number of decimal places.

6. 06

   Check both roots in the original equation

   x = -8: 64 - 40 = 24 ✓ x = 3: 9 + 15 = 24 ✓

   Substituting into the original equation — the one before rearranging — is what makes this a check rather than a repeat of your own working. Both roots must satisfy it. A root that fails usually points to a sign error in the brackets rather than to the method.

01

Choosing between the three methods

There are three routes to the roots, and part of the skill is deciding quickly which to take. Factorising is fastest when it works, the formula always works, and completing the square is what you use when the question asks for the turning point or for an exact surd answer.

The wording of the question is usually the clue. "Give your answers to 2 decimal places" almost always means the formula, because a factorisable quadratic would have given exact answers. "Write in the form (x + a)² + b" means completing the square, and no other method will earn the marks.

• Factorising — fastest, but only for quadratics with a whole-number factor pair
• Quadratic formula — always works; expect it when decimal places are requested
• Completing the square — for turning points, exact surd answers, and proofs

02

The formula, worked in full

Take 3x² + 5x − 1 = 0, which has no sensible factor pair. Here a = 3, b = 5 and c = −1. The discriminant is b² − 4ac = 25 − 4 × 3 × (−1) = 25 + 12 = 37, so the roots are x = (−5 ± √37) / 6.

Since √37 ≈ 6.0828, the two roots are x = (−5 + 6.0828) / 6 ≈ 0.180 and x = (−5 − 6.0828) / 6 ≈ −1.85, both to three significant figures. Substituting 0.180 back gives 3(0.0325) + 0.902 − 1 ≈ 0, which is the check.

The most common error is with c when it is negative: −4 × 3 × (−1) is +12, not −12. Writing the substitution out with the brackets in place, rather than typing straight into a calculator, is what prevents it.

03

What the discriminant tells you before you solve

The part under the square root, b² − 4ac, decides how many real roots the equation has, and questions often ask about it directly rather than asking for the roots themselves.

It is worth calculating first even when the roots are wanted, because a negative discriminant means there is no real answer to find and continuing is wasted effort.

• b² − 4ac > 0 — two distinct real roots; the curve crosses the x-axis twice
• b² − 4ac = 0 — one repeated root; the curve touches the x-axis once
• b² − 4ac < 0 — no real roots; the curve does not meet the x-axis

04

How we teach this

In our GCSE Mathematics programme we set a one-minute limit on hunting for a factor pair. Students lose more marks to searching for factorisations that do not exist than to any error in the formula itself, and a stated time limit converts an open-ended search into a decision.

We also teach the discriminant before the formula rather than after it. A student who calculates b² − 4ac first knows what kind of answer to expect, and that expectation is what catches an arithmetic slip in the line that follows.