IGCSE Maths quadratic equations: factorise, formula, complete the square

What each tier expects

The key ideas

Every method rests on one fact: if two numbers multiply to give zero, at least one of them is zero. So once you have (x - 3)(x + 5) = 0, either x - 3 = 0 or x + 5 = 0, giving x = 3 or x = -5. That only works when one side is zero, which is why the first step is always to rearrange.

Factorising is fastest when it works. For x^2 + bx + c, find two numbers that multiply to c and add to b. When a is not 1, multiply a by c, find two numbers that multiply to ac and add to b, split the middle term and factorise in pairs.

The quadratic formula works for every quadratic. Write down a, b and c with their signs before you substitute, and work out the discriminant b^2 - 4ac first. If it is negative there are no real solutions; if it is a perfect square the quadratic would have factorised.

Completing the square rewrites x^2 + bx + c as (x + b/2)^2 - (b/2)^2 + c. It gives exact answers and also the turning point of the graph y = x^2 + bx + c, which is (-b/2, c - (b/2)^2). Cambridge Extended asks for turning points this way.

Worked example 1: solve 6x^2 - 5x - 6 = 0 by factorising

1. Here a = 6, b = -5, c = -6, so ac = -36. Two numbers that multiply to -36 and add to -5 are -9 and 4.
2. Split the middle term: 6x^2 - 9x + 4x - 6 = 0.
3. Factorise in pairs: 3x(2x - 3) + 2(2x - 3) = 0, so (3x + 2)(2x - 3) = 0.
4. Solve each bracket: x = -2/3 or x = 3/2.
5. Check x = 3/2: 6(2.25) - 5(1.5) - 6 = 13.5 - 7.5 - 6 = 0. Correct.

Worked example 2: solve 2x^2 + 3x - 7 = 0, giving answers to 2 decimal places

1. Identify a = 2, b = 3, c = -7.
2. Discriminant: b^2 - 4ac = 9 - 4(2)(-7) = 9 + 56 = 65. It is positive but not a square number, so use the formula.
3. x = (-3 ± sqrt(65))/4.
4. sqrt(65) = 8.0622..., so x = 5.0622.../4 = 1.2656... or x = -11.0622.../4 = -2.7656...
5. Answer: x = 1.27 or x = -2.77. Check x = 1.2656: 2(1.6017) + 3.7968 - 7 is approximately 0.

Worked example 3: complete the square to solve x^2 + 6x - 4 = 0 exactly

1. Half of 6 is 3, so x^2 + 6x = (x + 3)^2 - 9.
2. The equation becomes (x + 3)^2 - 9 - 4 = 0, so (x + 3)^2 = 13.
3. Square root both sides, keeping both signs: x + 3 = ± sqrt(13).
4. Answer: x = -3 + sqrt(13) or x = -3 - sqrt(13), about 0.606 and -6.61.
5. Bonus: the graph y = x^2 + 6x - 4 has its minimum turning point at (-3, -13).

Worked example 4: a quadratic from a context

1. A rectangle is 3 cm longer than it is wide and its area is 40 cm^2. Let the width be x cm.
2. Form the equation: x(x + 3) = 40, so x^2 + 3x - 40 = 0.
3. Factorise: (x + 8)(x - 5) = 0, so x = -8 or x = 5.
4. A width cannot be negative, so reject x = -8 and say why. The rectangle is 5 cm by 8 cm, and 5 × 8 = 40.

Common mistakes that cost marks

• Dividing both sides by x and losing a solution: x^2 = 5x gives x = 0 or x = 5, not just 5.
• Not rearranging to zero first: x(x + 3) = 40 does not mean x = 40 or x + 3 = 40.
• Sign errors in the formula, especially -b when b is already negative, and writing the fraction bar under sqrt only instead of under the whole numerator.
• Rounding the square root early, which can move the third significant figure.
• Giving only the positive square root when completing the square.
• Leaving a context answer unjudged: a negative length or time must be rejected with a reason.

Exam technique and how a tutor helps

If a question says 'give your answers correct to 2 decimal places', it is telling you the quadratic will not factorise, so go straight to the formula. If it says 'show your working', write the formula with numbers substituted, because that line carries a method mark. On Cambridge Paper 2, which is non-calculator, expect factorising or surd answers such as -3 ± sqrt(13).

The fastest check is to substitute one answer back into the original equation. With a calculator, most models also have an equation solver; use it only to check, since the working earns the marks.

In one-to-one lessons a tutor usually finds that the difficulty is not the formula itself but factorising when a is not 1, or setting up the equation from words. Lessons then mix short drills on that step with past-paper questions where the quadratic is hidden inside an area, a fraction equation or a speed problem.

Self-check: can you do these?

• Solve x^2 + x - 30 = 0. (Answer: x = 5 or x = -6)
• Solve x(3x - 2) = 5. (Answer: x = 5/3 or x = -1)
• Write 2x^2 + 6x - 1 in the form a(x + b)^2 + c. (Answer: 2(x + 1.5)^2 - 5.5)
• Solve x^2 - 4x - 1 = 0 to 2 decimal places. (Answer: x = 4.24 or x = -0.24)
• Explain how the discriminant tells you whether a quadratic has real solutions.