1MA1 · A18AlgebraFoundation & HigherGrades 4–8

Solving Quadratic Equations

Q: Can a quadratic have no solutions?

Yes — when b^2 - 4ac < 0, there are no real solutions and the graph does not cross the x-axis.

A quadratic equation contains an $x^{2}$ term, such as $x^{2} - 5 x - 14 = 0$ . Unlike linear equations it can have two solutions, one, or none. The two main methods are factorising — quick when it works — and the quadratic formula, which always works. Recognising which to use, and setting the equation up correctly first, is most of the skill.

65 min Quadratics

The big picture

Every method depends on one fact: if two numbers multiply to give zero, at least one of them must be zero. So once a quadratic is written as $(x - 7) (x + 2) = 0$ , the solutions $x = 7$ and $x = - 2$ follow immediately. That is why the equation must always be rearranged to " $= 0$ " first.

Foundation papers focus on factorising quadratics with $a = 1$ ; Higher papers add $a \neq = 1$ , the quadratic formula, and context problems where one solution must be rejected — a negative length, for instance.

What you'll learn

Rearrange quadratic equations into the form $a x^{2} + b x + c = 0$
Solve quadratics by factorising
Solve quadratics using the quadratic formula (Higher)
Form and solve quadratic equations from contexts, interpreting both solutions

Setting up: make it equal zero

Collect every term on one side so the other side is 0, ideally with a positive $x^{2}$ coefficient.

$x^{2} = 3 x + 10$ becomes $x^{2} - 3 x - 10 = 0$ .

If there is a common factor, divide through by it first: $2 x^{2} - 6 x - 20 = 0$ becomes $x^{2} - 3 x - 10 = 0$ .

Tip — Never divide both sides by $x$ . It throws away the solution $x = 0$ . Factorise instead: $x^{2} = 5 x$ gives $x (x - 5) = 0$ , so $x = 0$ or $x = 5$ .

Solving by factorising

Factorise the quadratic, set each bracket equal to zero, and solve each small equation.

For $a x^{2} + b x + c$ with $a \neq = 1$ , factorise by splitting the middle term, then solve each bracket.

1Factorise:

(x - 7) (x + 2) = 0

2Either

x - 7 = 0

x + 2 = 0

Answer

x = 7

x = - 2

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

Solve $x^{2} + x - 20 = 0$ .

Your answer

Separate with a comma.

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

Solve $2 x^{2} - 7 x + 3 = 0$ .

Your answer

Separate with a comma.

Why this works

The solutions have the opposite signs to the numbers in the brackets: $(x - 7)$ gives $x = + 7$ . Setting each bracket to zero, rather than reading the number off, avoids sign slips.

The quadratic formula

For $a x^{2} + b x + c = 0$ , the solutions are given by the formula below. It works for every quadratic, including those that do not factorise.

Write down $a$ , $b$ and $c$ with their signs before substituting, and put brackets around negative values.

The expression under the square root, $b^{2} - 4 a c$ , is the $discriminant$ : positive gives two solutions, zero gives one, negative gives none.

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

The quadratic formula.

a = 2

b = - 3

c = - 4

x = \frac{3 \pm 9 + 32}{4} = \frac{3 \pm 41}{4}

Answer

x = 2.35

x = - 0.85

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

Solve $3 x^{2} + x - 5 = 0$ , giving your answers to 2 decimal places.

Your answer

Separate with a comma, to 2 d.p.

Tip — "Give your answers to 2 decimal places" is the signal that the quadratic will not factorise — go straight to the formula.

Quadratics in context

Define the unknown, form an equation from the information, rearrange to zero, and solve.

Check both solutions against the context. Lengths, times and counts usually cannot be negative, so one solution may need rejecting — with a reason.

1Let the width be

x

x (x + 3) = 40

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

, so

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

x = 5

x = - 8

; a width cannot be negative.

AnswerThe width is 5 cm.

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

A rectangle is 4 cm longer than it is wide, and its area is 45 cm². Find its width, in cm.

Your answer

Think like an examiner

Show the equation rearranged to $= 0$ before factorising or using the formula.
Show the substitution into the formula, with brackets round negatives.
State both solutions, and explain any rejection in context.

Quadratic equations

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

Why factorising works.

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

The quadratic formula.

b^{2} - 4 a c

Discriminant: number of solutions.

Watch out for these

Factorising

x^{2} - 3 x = 10

without moving the 10.

Rearrange to

x^{2} - 3 x - 10 = 0

first.

Dividing

x^{2} = 5 x

x

to get

x = 5

only.

Factorise to keep

x = 0

as well.

Using

- b = - 3

when

b = - 3

- b = - (- 3) = 3

Dividing only

b^{2} - 4 a c

2 a

The whole numerator

- b \pm \dots

is divided by

2 a

Stretch yourself

The sum of the squares of two consecutive positive integers is 181. Find the integers.

Hint — Let the integers be $n$ and $n + 1$ . Form an equation and rearrange to zero.

Questions students ask

How do I know whether to factorise or use the formula?

If an answer is asked to a number of decimal places, or you cannot find factors quickly, use the formula. Otherwise try factorising first.

Can a quadratic have no solutions?

Yes — when

b^{2} - 4 a c < 0

, there are no real solutions and the graph does not cross the

x

-axis.

Is completing the square another method?

Yes. It also gives turning points, and is covered in the next lesson.

Key takeaways

Rearrange to $a x^{2} + b x + c = 0$ before solving.
Factorise, then set each bracket to zero.
Use $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ when it does not factorise.
The discriminant $b^{2} - 4 a c$ tells you how many solutions there are.
In context, check both solutions and reject impossible ones with a reason.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 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 42 marks

Solve $x^{2} - 9 x + 20 = 0$ .[2]

Question 2Grade 63 marks

Solve $x^{2} - 6 x + 4 = 0$ . Give your answers to 2 decimal places.[3]

Question 3Grade 84 marks

A ball's height, in metres, after $t$ seconds is $h = 30 t - 5 t^{2}$ .

Find the two times when the ball is 40 m high, and explain why there are two.[4]

Unlimited practice

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Solving Quadratic Equations

Setting up: make it equal zero

Solving by factorising

The quadratic formula

Quadratics in context

Practise it the way it is examined

Skill drill

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Related topics

Solving Quadratic Equations

Setting up: make it equal zero

Solving by factorising

The quadratic formula

Quadratics in context

Practise it the way it is examined

Skill drill

Practise this topic

Related topics