Solving Quadratic Equations by Completing the Square — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers rewriting a quadratic in the form (x + p)² + q and using it to solve equations. By the end of this guide you should be able to complete the square on a quadratic where the x² coefficient is 1, and solve the resulting equation exactly.

You should also be able to handle a quadratic with a coefficient in front of x², read the turning point of a parabola straight from the completed square form, give answers in surd form, and know when completing the square is the right method to choose.

The organising idea is that completing the square gets x into one place. In x² + 6x + 1 the unknown appears twice, and there is no way to undo both occurrences at once — which is precisely why quadratics cannot be solved by the ordinary balance method. Rewriting it as (x + 3)² − 8 leaves a single x, buried inside a bracket that can be unwrapped one layer at a time: add 8, square-root, subtract 3. Everything else in this topic follows from that single purpose, including why the method reveals the turning point, since a squared bracket is smallest when its contents are zero.

Key terms and definitions

Completing the square — rewriting a quadratic as a squared bracket plus or minus a number.

Completed square form — an expression written as (x + p)² + q.

Coefficient — the number in front of a term. In 3x² the coefficient of x² is 3.

Turning point (vertex) — the minimum or maximum point of the parabola.

Surd — an exact root such as √5, left in that form rather than rounded.

Discriminant — the quantity that decides how many real solutions a quadratic has.

Core concepts

The basic method

For x² + bx + c, halve the coefficient of x to get the number inside the bracket, then subtract the square of that number to keep the value unchanged:

x² + bx + c = (x + b/2)² − (b/2)² + c

For x² + 6x + 1: halving 6 gives 3, so the bracket is (x + 3)². But (x + 3)² expands to x² + 6x + 9, which is 9 too big, so subtract 9 and add the original 1.

The result is (x + 3)² − 8.

Checking by expansion is worth doing until the method is secure: (x + 3)² − 8 = x² + 6x + 9 − 8 = x² + 6x + 1. ✓

Why the subtraction is needed

The squared bracket always brings an unwanted constant with it, and subtracting it is what keeps the expression equal to the original.

Halving b and squaring always produces exactly the surplus, which is why the same two steps work every time. Writing the bracket without compensating changes the expression into a different one entirely.

Negative coefficients of x

The method is unchanged; the halved value simply comes out negative.

For x² − 10x + 3: half of −10 is −5, so the bracket is (x − 5)². That expands to x² − 10x + 25, so subtract 25 and add 3.

The result is (x − 5)² − 22.

Note the sign inside the bracket matches the sign of b, while the number subtracted outside is always positive, since it is a square.

Solving the equation

Once the quadratic is in completed square form, unwrap it one layer at a time.

To solve x² + 6x + 1 = 0: rewrite as (x + 3)² − 8 = 0, add 8 to get (x + 3)² = 8, take the square root of both sides to get x + 3 = ±√8, then subtract 3 to get x = −3 ± √8.

The plus-or-minus is essential. A squared bracket equal to 8 could have contained either the positive or the negative root, so there are two solutions. Omitting it loses half the answer and is the commonest error in the topic.

Simplifying the surd finishes the job properly: √8 = 2√2, so the solutions are x = −3 + 2√2 and x = −3 − 2√2.

When the coefficient of x² is not 1

Factor it out of the first two terms before completing the square.

For 2x² + 8x + 5: take 2 out of the first two terms to get 2(x² + 4x) + 5.

Complete the square inside the bracket: x² + 4x becomes (x + 2)² − 4.

So the expression is 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.

The step people miss is multiplying the −4 by the 2 when it comes out of the bracket. Expanding your answer to check catches it immediately.

The turning point

Completed square form gives the turning point without any further work, which is often the real reason a question asks for it.

For y = (x + p)² + q, the turning point is at x = −p, y = q.

A squared bracket is never negative, so the smallest it can be is zero, which happens when x = −p. At that moment y takes its smallest value, q.

So y = (x − 3)² − 5 has its minimum at (3, −5) — note the sign reversal inside the bracket and no reversal outside.

If the x² coefficient is negative, the bracket makes the expression as large as possible instead, so the turning point is a maximum.

The turning point also gives the line of symmetry, at x = −p.

How many solutions

The completed square form shows immediately how many real solutions exist.

If (x + p)² = a positive number, there are two solutions. If it equals zero, there is one repeated solution. If it equals a negative number, there are no real solutions, since a square cannot be negative.

So x² + 6x + 13 = 0 becomes (x + 3)² + 4 = 0, giving (x + 3)² = −4, which is impossible — the curve never reaches the x-axis.

Choosing the method

Three methods solve quadratics, and the question usually signals which is wanted.

Factorising is quickest when it works, and it works when the quadratic has whole-number roots.

Completing the square is required when the question says so, when it asks for the turning point, or when it asks you to show a result in the form (x + p)² + q.

The quadratic formula is the fallback for any quadratic, and it is itself derived by completing the square on the general form — which is why the two give identical answers.

Look for the words "in the form", "exact", "surd form" or "turning point": each points at completing the square.

Worked examples

Example 1: Completing the square and solving

Solve x² + 8x + 5 = 0, giving your answers in surd form.

Half of 8 is 4, so the bracket is (x + 4)², which expands to x² + 8x + 16.

That is 16 too much, so subtract 16 and add the 5: the equation becomes (x + 4)² − 11 = 0.

Add 11: (x + 4)² = 11.

Square-root both sides, keeping the plus-or-minus: x + 4 = ±√11.

Subtract 4: x = −4 ± √11.

The surd does not simplify, so this is the exact answer. A decimal would not be acceptable where surd form was asked for.

Example 2: A coefficient in front of x²

Write 3x² − 12x + 7 in the form a(x + p)² + q.

Factor 3 from the first two terms: 3(x² − 4x) + 7.

Inside the bracket, half of −4 is −2, so x² − 4x = (x − 2)² − 4.

Substituting back: 3[(x − 2)² − 4] + 7.

Multiply the 3 through the outer bracket: 3(x − 2)² − 12 + 7.

So the answer is 3(x − 2)² − 5.

Check by expanding: 3(x² − 4x + 4) − 5 = 3x² − 12x + 12 − 5 = 3x² − 12x + 7. ✓

Example 3: Reading the turning point

The curve y = x² − 6x + 11 is written in completed square form. Find its turning point and state whether it is a minimum or a maximum.

Half of −6 is −3, so the bracket is (x − 3)², expanding to x² − 6x + 9.

That is 9 too much, so subtract 9 and add 11: y = (x − 3)² + 2.

The squared bracket is smallest when x = 3, where it equals zero, and y is then 2.

The turning point is (3, 2), and since the x² coefficient is positive it is a minimum.

Note also that the minimum value of 2 is above the x-axis, so the curve never crosses it and the equation x² − 6x + 11 = 0 has no real solutions.

Common mistakes and how to avoid them

Forgetting the plus-or-minus. Square-rooting gives two solutions, not one.

Not subtracting the square. The bracket brings a surplus constant that must be removed.

Failing to multiply the outside number through. When the x² coefficient is factored out, it multiplies the constant that comes out of the bracket too.

Getting the sign of the turning point wrong. In (x + p)² + q the turning point is at x = −p, with the sign reversed inside the bracket but not outside.

Rounding when an exact answer is wanted. Leave surds in surd form and simplify them.

Assuming there are always two solutions. A squared bracket equal to a negative number gives none.

Choosing the wrong method. "In the form", "exact" and "turning point" all point to completing the square.

Exam technique for "Completing the Square"

Write the halved coefficient down as a separate step, then the square you must subtract. Both carry method marks.

Expand your completed square form to check it returns the original quadratic. It takes one line and catches the surplus-constant error.

Write the ± as soon as you take the square root, rather than adding it at the end.

Simplify surds fully and leave them exact unless a decimal is explicitly requested.

For a turning point, state both coordinates and say whether it is a minimum or a maximum, using the sign of the x² coefficient as the reason.

When the x² coefficient is not 1, factor it from the first two terms only, and multiply it back through carefully at the end.

Quick revision summary

Completing the square gets x into one place, which is what makes a quadratic solvable by unwrapping.

For x² + bx + c: halve b for inside the bracket, then subtract the square of that half. So x² + 6x + 1 = (x + 3)² − 8.

To solve: isolate the bracket, square-root with a ± sign, then subtract the number inside. x² + 6x + 1 = 0 gives x = −3 ± 2√2.

When the x² coefficient is not 1, factor it out of the first two terms, complete the square inside, then multiply the outside number back through.

The turning point of y = (x + p)² + q is at (−p, q) — a minimum if the x² coefficient is positive, a maximum if negative — and the line of symmetry is x = −p.

The form shows the number of solutions: a positive right-hand side gives two, zero gives one, and a negative gives none.

Use this method when the question says "in the form", "exact", "surd form" or "turning point".