GCSE · Maths · AQA · Spec 8300 · Higher

Quadratic equations with rearrangement, completing the square, quadratic formula (Higher)

x² + 4x − 3 = 0 won't factorise — no whole-number pair works. Yet it has two exact solutions, and you're about to find them two ways.

Your first move

What does this equation need first?

Walk each of these down the tree: (a) x² + 2x = 15 (b) x² − 7x + 12 = 0 (c) x² + 4x − 3 = 0. Where does each one end up?

3 possible first moves.

On A quadratic equation to solve. 2 branches to choose from.

Three equations, three different first moves. Nobody starts with the formula — they start by looking.

Watch out: Brackets that equal something other than 0 are not ready to solve, however factorised they look. (x − 1)(x + 4) = 6 still goes down the left-hand branch.

Why zero?

The brackets that equal 12

Here's an equation that is already factorised, but equals 12 instead of 0: (x + 1)(x + 2) = 12.

Which of these is closest to what you'd do?

Set each bracket equal to 12: x + 1 = 12 and x + 2 = 12, so x = 11 or x = 10.One bracket must be 12 and the other must be 1, because 12 × 1 = 12.Multiply out, rearrange so one side is 0, then factorise again.Once it's (x + 5)(x − 2) = 0, solve the first bracket. That gives the answer.

How sure are you?

I'm sureI think soI'm guessing

Set each bracket equal to 12: x + 1 = 12 and x + 2 = 12, so x = 11 or x = 10.

This is a common misconception

Test it. x = 11 gives 12 × 13 = 156, and x = 10 gives 11 × 12 = 132. Neither makes 12. Two numbers can multiply to 12 in endless ways — 3 × 4, 2 × 6, 1.5 × 8 — so nothing tells you what each bracket has to be. You'd just be guessing.

One bracket must be 12 and the other must be 1, because 12 × 1 = 12.

This is a common misconception

That's one way to make 12 out of endless ways, and here it can't even happen: if x + 1 = 12 then x + 2 = 13, and if x + 2 = 1 then x + 1 = 0. Picking a pair is guesswork. Only 0 is different — if a product is 0, one of the factors really has to be 0.

Multiply out, rearrange so one side is 0, then factorise again.

This matches the evidence

Exactly. x² + 3x + 2 = 12 becomes x² + 3x − 10 = 0, which factorises to (x + 5)(x − 2) = 0. Now the product is 0, so one factor must be 0: x = −5 or x = 2. Check: (−4) × (−3) = 12 and 3 × 4 = 12.

Once it's (x + 5)(x − 2) = 0, solve the first bracket. That gives the answer.

This is a common misconception

Half right. Either factor could be the zero one, and you don't know which — so each one gives a solution. x + 5 = 0 gives x = −5 and x − 2 = 0 gives x = 2. Stop after one and you've thrown away half the answer.

Zero is the only product that tells you about its factors: if A × B = 0, then A = 0 or B = 0. So rearrange to 0 first, then solve every factor.

The missing corner

Why completing the square takes something away

x² + 6x is nearly a perfect square. Half of 6 is 3, so build (x + 3)² — fill in every cell, then collect them.

Grid: each cell is its row times its column
×	x	+3
x
+3

Collect the cells: (x + 3)² =

Type x² as x^2 if you cannot type ². Spaces do not matter.

Completing the square, start to finish

Problem

Solve x² + 4x − 3 = 0 — equation (c) from the tree, the one that won't factorise.

Your turn

When x² has a number in front

Write 2x² − 8x + 3 in the form a(x + p)² + q. Then use it to solve 2x² − 8x + 3 = 0.

2x² − 8x + 3This is an expression — there's no equals sign, so there's no other side to do anything to.
missing step

Which line is step 2?

= 2(x² − 4x) + 3= x² − 4x + 1.5= 2(x² − 8x) + 3

Where the quadratic formula comes from

ax² + bx + c = 0

Every quadratic equation looks like this once it equals 0 (with a ≠ 0). Now do exactly what you did to (c) — just with letters instead of numbers.

1 / 6

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

Get it equal to zero. Factorise if it's friendly. If it isn't, complete the square — and the quadratic formula is just that same move, done once for every quadratic.

What you need to know

Before factorising, one side must be 0. If it isn't, rearrange by doing the same thing to both sides.
If two values multiply to 0, one of them is 0. Either factor could be the zero one, so solve every factor.
Completing the square and the quadratic formula are useful when a quadratic can't easily be factorised.
x² + bx + c = (x + b/2)² − (b/2)² + c. The (b/2)² is always subtracted, whether b/2 is positive or negative.
When you square root, write ±: a positive number has a positive and a negative square root.
For ax² + bx + c = 0 (a ≠ 0), x = (−b ± √(b² − 4ac)) / 2a. It comes from completing the square on the general equation.

The big picture

First make one side 0. If the quadratic factorises easily, set each factor to 0 — a product is only 0 when a factor is. If it doesn't, complete the square: x² + bx + c = (x + b/2)² − (b/2)² + c. Get the squared bracket alone and take both square roots, + and −. When a ≠ 1, factor a out of an expression; dividing through is only valid in an equation. Doing all this to ax² + bx + c = 0 gives the quadratic formula, x = (−b ± √(b² − 4ac)) / 2a. Substitute with brackets, and check with technology.

Key points

1Order of work: equals zero? → factorises easily? → if not, complete the square or use the formula.

2A factorised quadratic equal to any number other than 0 can't be solved bracket by bracket — multiply out and rearrange first.

3To complete the square on x² + bx + c, halve b, write (x + b/2)², then subtract (b/2)² and add c.

4When a ≠ 1 and you're rewriting an expression, factor a out of the x² and x terms; ax² + bx + c = a(x + b/(2a))² + c − b²/(4a).

5Dividing every term by a is fine in an equation, because both sides stay equal — but it changes an expression.

6Substitute into the formula with every value in brackets: (−3)² = 9, but −3² = −9. Then use technology to check your solutions.

.we-_R_18slubsnpfiv5ubrb_ [data-we-step]{display:flex !important}.we-_R_18slubsnpfiv5ubrb_ [data-we-solution]{display:block !important}.we-_R_18slubsnpfiv5ubrb_ [data-we-reveal]{display:none !important}

Worked example

Problem

Solve 2x² = 3x + 4. Give your answers to 2 decimal places.

⚠ Watch out

Writing x + 3 = 5 from (x + 3)² = 25 and stopping. 25 has two square roots, 5 and −5, so x + 3 = ±5, giving x = 2 or x = −8. Check both: 5² = 25 and (−5)² = 25.

🧠

Memory hook

Zero first. Friendly? Factorise. Stubborn? Finish the square — and the formula is just the square, finished for everyone.

✓

Check yourself

Solve x² − 10x + 18 = 0 by completing the square, leaving the square root in your answer. Then solve it again with the quadratic formula and check you get the same two solutions.

Flashcards

(12)

What must be true before you solve a quadratic by factorising?

One side must equal 0. If it doesn't, rearrange first.

Why can you set each factor equal to 0?

If two values multiply to 0, one of them is 0 — and it could be either, so find the solution from each factor.

When are completing the square and the quadratic formula useful?

When the quadratic can't easily be factorised.

Completed square form of x² + bx + c

(x + b/2)² − (b/2)² + c

Why is (b/2)² subtracted when you complete the square?

(x + b/2)² expands to x² + bx + (b/2)², one piece more than you started with — so it's taken away again. Squaring makes it positive even when b/2 is negative, so it is always subtracted.

Solving by completing the square: what goes in front of the square root?

± — a positive number has a positive and a negative square root, and each gives a solution.

First step to write ax² + bx + c (a ≠ 1) in completed square form

Factor a out of the x² and x terms: a(x² + (b/a)x) + c.

Dividing every term by a: when is it allowed?

In an equation, because both sides are divided and stay equal. Not when rewriting an expression — that changes its value.

The quadratic formula (for ax² + bx + c = 0, a ≠ 0)

x = (−b ± √(b² − 4ac)) / 2a

Where does the quadratic formula come from?

Completing the square on the general equation ax² + bx + c = 0.

Why put brackets round values you substitute into the formula?

To avoid slips when squaring and subtracting negatives: (−4)² = 16, but −4² = −16.

How can technology help with quadratic equations?

It can find the solutions, which makes it a quick way to check your own working.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

Learning with Lightbulb is opening soon

You can use this lesson now. Join the waitlist and we'll let you know when the full Lightbulb experience is ready.

Keep me posted

What does this equation need first?

The brackets that equal 12

Why completing the square takes something away

When x² has a number in front

More AQA GCSE Maths topics