A18, A11AlgebraFoundation & HigherGrades 4–8

Quadratic Equations

Q: How do I choose a method?

Try factorising first if the numbers are small and the question does not mention decimals or surds. If it asks for decimal places, use the formula. If it asks for surd form or the turning point, completing the square is often neatest.

Q: What does b^2 - 4ac tell me?

It is called the discriminant. Positive: two solutions. Zero: one repeated solution. Negative: no real solutions (the graph never meets the x-axis).

Q: Can a quadratic have only one solution?

Yes, when it is a perfect square such as (x - 4)^2 = 0. The graph just touches the x-axis at x = 4.

A quadratic equation has an $x^{2}$ term and no higher power, such as $x^{2} - 5 x + 6 = 0$ . Unlike a linear equation it usually has two solutions, because a U-shaped graph can cross the $x$ -axis twice. There are three methods, and choosing the right one quickly is a big part of the skill.

55 min Video by Maths Genie AQA GCSE Maths

The big picture

Every method relies on getting the equation into the form $a x^{2} + b x + c = 0$ first. The key fact is that if two things multiply to give zero, at least one of them must be zero — so a factorised quadratic can be split into two simple linear equations.

Factorising is fastest when it works. Completing the square also finds the turning point of the graph. The quadratic formula works for every quadratic, and is the method to use when the question asks for answers to decimal places — a strong hint that it will not factorise.

Factorising QuadraticsCurated by Kepler, not produced by us. Watch it, then read the notes below.

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What you'll learn

Rearrange a quadratic equation into the form $a x^{2} + b x + c = 0$
Solve quadratic equations by factorising
Solve by completing the square (Higher)
Solve using the quadratic formula (Higher)

Solving by factorising

Rearrange so one side is zero. Factorise, then set each bracket equal to zero and solve.

Two special cases: if there is no constant term, take out $x$ as a common factor — never divide by $x$ , or you lose the solution $x = 0$ . If there is no $x$ term, as in $x^{2} = 49$ , square root and remember $\pm$ .

1Rearrange:

x^{2} + 2 x - 15 = 0

2Factorise:

(x + 5) (x - 3) = 0

3So

x + 5 = 0

x - 3 = 0

Answer

x = - 5

x = 3

Your turnSame method, new numbers — try it before moving on.

Solve $x^{2} - 7 x + 10 = 0$ .

Your answer

Separate the solutions with a comma.

Your turnSame method, new numbers — try it before moving on.

Solve $2 x^{2} + 5 x - 3 = 0$ .

Your answer

Separate with a comma; 1/2 or 0.5 are both fine.

Tip — The solutions have the opposite sign to the numbers in the brackets: $(x + 5)$ gives $x = - 5$ .

Completing the square (Higher)

Rewrite $x^{2} + b x + c$ as $(x + \frac{b}{2})^{2} - (\frac{b}{2})^{2} + c$ . Then the equation can be solved by rearranging and square rooting.

If the $x^{2}$ coefficient is not 1, take it out as a factor from the $x^{2}$ and $x$ terms first.

1Complete the square:

(x + 3)^{2} - 9 - 4 = 0

, so

(x + 3)^{2} = 13

2Square root:

x + 3 = \pm 13

3Subtract 3.

Answer

x = - 3 \pm 13

Your turnSame method, new numbers — try it before moving on.

Write $x^{2} - 10 x + 7$ in the form $(x - a)^{2} - b$ .

Your answer

Type it like (x-3)^2-4.

Why the turning point appears

$(x + 3)^{2} - 13$ can never be less than $- 13$ , because a square is never negative. So the graph of $y = x^{2} + 6 x - 4$ has its lowest point at $(- 3, - 13)$ . Completing the square solves the equation and finds the turning point in one go.

The quadratic formula (Higher)

The solutions of $a x^{2} + b x + c = 0$ are given by the quadratic formula, shown below.

Write down $a$ , $b$ and $c$ with their signs before substituting, and put negative values in brackets. Work out $b^{2} - 4 a c$ first — if it is negative, there are no real solutions.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The one method that works for every quadratic — know it by heart.

a = 2

b = - 3

c = - 7

b^{2} - 4 a c = (- 3)^{2} - 4 (2) (- 7) = 9 + 56 = 65

x = \frac{3 \pm 65}{4}

, giving

2.766 \dots

and

- 1.265 \dots

Answer

x = 2.77

x = - 1.27

Your turnSame method, new numbers — try it before moving on.

Solve $x^{2} + 4 x - 3 = 0$ , giving your answers to 2 decimal places.

Your answer

Separate with a comma, to 2 d.p.

Tip — The formula has $2 a$ under the whole numerator — draw a long fraction line so the $- b$ is divided too.

Forming quadratic equations

Area problems and number problems often lead to quadratics. Form the equation, rearrange to $= 0$ , solve, and then reject any solution that does not make sense — a length cannot be negative.

1Width

x

, length

x + 3

x (x + 3) = 40

x^{2} + 3 x - 40 = 0

, so

(x + 8) (x - 5) = 0

x = - 8

is impossible for a length.

AnswerThe width is 5 cm.

Your turnSame method, new numbers — try it before moving on.

The product of two consecutive positive integers is 132. What is the smaller integer?

Your answer

Think like an examiner

"Give your answers to 2 decimal places" means use the formula (or completing the square) — do not waste time trying to factorise.
Show the substitution into the formula in full. A correct substitution earns a method mark even if the calculator work slips.
If you get $b^{2} - 4 a c < 0$ on an exam question, recheck your signs — GCSE questions almost always have real solutions.

Methods

pq = 0 \Rightarrow p = 0 or q = 0

Why factorising works.

x^{2} + b x = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2}

Completing the square.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The quadratic formula — learn it.

Watch out for these

Solving

x^{2} = 5 x

by dividing by

x

to get

x = 5

Factorise

x (x - 5) = 0

x = 0

x = 5

Solving

x^{2} = 36

x = 6

only.

x = \pm 6

(- 3)^{2}

typed as

- 3^{2} = - 9

in the formula.

Use brackets:

(- 3)^{2} = 9

Factorising without rearranging:

x^{2} + x = 6 \Rightarrow x (x + 1) = 6

, so

x = 6

x = - 1

Make one side zero first:

(x + 3) (x - 2) = 0

Stretch yourself

A right-angled triangle has sides $x$ cm, $(x + 7)$ cm and hypotenuse $(x + 8)$ cm. Find the perimeter of the triangle.

Hint — Use Pythagoras to form a quadratic equation, then solve it.

Questions students ask

How do I choose a method?

Try factorising first if the numbers are small and the question does not mention decimals or surds. If it asks for decimal places, use the formula. If it asks for surd form or the turning point, completing the square is often neatest.

What does

b^{2} - 4 a c

tell me?

It is called the discriminant. Positive: two solutions. Zero: one repeated solution. Negative: no real solutions (the graph never meets the

x

-axis).

Can a quadratic have only one solution?

Yes, when it is a perfect square such as

(x - 4)^{2} = 0

. The graph just touches the

x

-axis at

x = 4

Key takeaways

Rearrange to $a x^{2} + b x + c = 0$ before doing anything else.
Factorise and set each factor to zero.
Never divide by $x$ ; take it out as a factor.
Completing the square gives exact (surd) answers and the turning point.
The quadratic formula works every time — learn it and substitute carefully.

Exam-style questions

Practise it the way it is examined

3 original questions · 11 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 43 marks

(a)Factorise $x^{2} - 3 x - 28$ .[2]
(b)Hence, or otherwise, solve $x^{2} - 3 x - 28 = 0$ .[1]

Question 2Grade 63 marks

Solve $3 x^{2} + 5 x - 1 = 0$ . Give your answers to 2 decimal places.[3]

Question 3Grade 85 marks

The diagram shows a path of width $x$ metres around a rectangular lawn 8 m by 5 m. The total area of the lawn and path is 70 m².

The path has the same width

x

on every side.

(a)Show that $4 x^{2} + 26 x - 30 = 0$ .[3]
(b)Find the width of the path.[2]

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Quadratic Equations

Solving by factorising

Completing the square (Higher)

The quadratic formula (Higher)

Forming quadratic equations

Practise it the way it is examined

Skill drill

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Related topics

Quadratic Equations

Solving by factorising

Completing the square (Higher)

The quadratic formula (Higher)

Forming quadratic equations

Practise it the way it is examined

Skill drill

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