Solving Quadratic Equations

Three methods and when to use them

SPEC 2.3 • ALGEBRA AND FUNCTIONS

The Three Methods

There are three main methods for solving quadratic equations of the form

a x^{2} + b x + c = 0

:

Method 1: Factorising

Quickest when the quadratic factorises nicely into integer factors.

Method 2: Quadratic Formula

Always works for any quadratic equation - the universal method.

Method 3: Completing the Square

Useful for exact answers and finding the vertex.

Key Point

All three methods will give you the same solutions - choose the most efficient one for each problem!

Method 1: Factorising

When the quadratic factorises into nice integer factors, this is the quickest method.

Process

Write the quadratic in the form

a x^{2} + b x + c = 0

Factorise into two brackets

Set each bracket equal to zero

Solve for x

Example

x^{2} - 5 x + 6 = 0

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

x - 2 = 0

or

x - 3 = 0

Solution:

x = 2

or

x = 3

Worked Example01

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

1

Find two numbers that multiply to give

2 \times 3 = 6

and add to give

7

2

Numbers are

6

and

1

:

6 \times 1 = 6

,

6 + 1 = 7

3

Split middle term:

2 x^{2} + 6 x + x + 3 = 0

4

Factorise in pairs:

2 x (x + 3) + 1 (x + 3) = 0

5

Factor out common bracket:

(2 x + 1) (x + 3) = 0

6

Set each factor to zero:

2 x + 1 = 0

or

x + 3 = 0

Answer: $x = - \frac{1}{2}$ or $x = - 3$

Method 2: Quadratic Formula

The quadratic formula works for any quadratic equation, even when factorising is difficult or impossible.

Formula

For any quadratic equation

a x^{2} + b x + c = 0

:

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

Important

This formula is NOT given in your exam formula booklet. You must memorise it!

Common Error

Watch the signs! Remember it's

- b

, so if

b = - 5

, then

- b = 5

(positive).

Worked Example02

Solve $x^{2} - 2 x - 15 = 0$ using the quadratic formula.

1

Identify:

a = 1

,

b = - 2

,

c = - 15

2

Substitute:

x = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( 1 ) ( - 15 )}{2 ( 1 )}

3

Simplify:

x = \frac{2 \pm 4 + 60}{2} = \frac{2 \pm 64}{2}

4

Calculate:

x = \frac{2 \pm 8}{2}

5

Two solutions:

x = \frac{2 + 8}{2} = 5

or

x = \frac{2 - 8}{2} = - 3

Answer: $x = 5$ or $x = - 3$

Note: This could also be solved by factorising as

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

Worked Example03

Solve $3 x^{2} - 5 x + 1 = 0$ , giving exact answers.

1

Identify:

a = 3

,

b = - 5

,

c = 1

2

Substitute:

x = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( 3 ) ( 1 )}{2 ( 3 )}

3

Simplify:

x = \frac{5 \pm 25 - 12}{6} = \frac{5 \pm 13}{6}

4

13

cannot be simplified further

Answer: $x = \frac{5 + 13}{6}$ or $x = \frac{5 - 13}{6}$

Note: Cannot simplify further - leave as exact fractions

Method 3: Completing the Square

This method rewrites the quadratic in a form that's easy to solve by taking square roots.

Process

Write

a x^{2} + b x + c

in the form

a (x + p)^{2} + q

Set equal to zero:

a (x + p)^{2} + q = 0

Rearrange:

(x + p)^{2} = - \frac{q}{a}

Take square roots:

x + p = \pm - \frac{q}{a}

Solve for

x

Recall

From Lesson 2.05:

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

Worked Example04

Solve $2 x^{2} + 12 x + 7 = 0$ by completing the square, giving exact answers.

1

Factor out coefficient of

x^{2}

:

2 (x^{2} + 6 x) + 7 = 0

2

Complete square inside bracket:

x^{2} + 6 x = (x + 3)^{2} - 9

3

Substitute back:

2 [(x + 3)^{2} - 9] + 7 = 0

4

Expand:

2 (x + 3)^{2} - 18 + 7 = 0

5

Simplify:

2 (x + 3)^{2} - 11 = 0

6

Rearrange:

(x + 3)^{2} = \frac{11}{2}

7

Take roots:

x + 3 = \pm \frac{11}{2} = \pm \frac{22}{2}

Answer: $x = - 3 + \frac{22}{2}$ or $x = - 3 - \frac{22}{2}$

Rearranging First

Before solving, the equation must be in the form

a x^{2} + b x + c = 0

.

Common Situations

Brackets need expanding:

x (x + 3) = 10

Terms on both sides:

x^{2} + 5 = 3 x

Need to collect like terms:

2 x^{2} + x = x^{2} + 5 x - 3

Example

x (x + 3) = 10

Expand:

x^{2} + 3 x = 10

Rearrange:

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

Now solve by factorising or formula!

Worked Example05

Solve $x (2 x - 1) = 4$ , giving exact answers.

1

Expand:

2 x^{2} - x = 4

2

Rearrange:

2 x^{2} - x - 4 = 0

3

Identify:

a = 2

,

b = - 1

,

c = - 4

4

Apply formula:

x = \frac{1 \pm 1 + 32}{4} = \frac{1 \pm 33}{4}

Answer: $x = \frac{1 + 33}{4}$ or $x = \frac{1 - 33}{4}$

Quadratics in Disguise

Some equations are quadratic in a function of x. We use substitution to solve them.

Common Types

x^{4} - 5 x^{2} + 4 = 0

— quadratic in

x^{2}

e^{2 x} - 5 e^{x} + 6 = 0

— quadratic in

e^{x}

sin^{2} θ + sin θ - 2 = 0

— quadratic in

sin θ

Method

Let

u =

[the function], e.g.,

u = x^{2}

Rewrite the equation in terms of

u

Solve the quadratic in

u

Substitute back to find

x

Worked Example06

Solve $x^{4} - 5 x^{2} + 4 = 0$

1

Let

u = x^{2}

, so

x^{4} = u^{2}

2

Equation becomes:

u^{2} - 5 u + 4 = 0

3

Factorise:

(u - 1) (u - 4) = 0

4

So

u = 1

or

u = 4

5

Substitute back:

x^{2} = 1

or

x^{2} = 4

6

Solve:

x = \pm 1

or

x = \pm 2

Answer: $x = - 2, - 1, 1, 2$

Worked Example07

Solve $9^{x} - 10 (3^{x}) + 9 = 0$

1

Note that

9^{x} = (3^{2})^{x} = (3^{x})^{2}

2

Let

u = 3^{x}

, so

9^{x} = u^{2}

3

Equation becomes:

u^{2} - 10 u + 9 = 0

4

Factorise:

(u - 1) (u - 9) = 0

5

So

u = 1

or

u = 9

6

Substitute back:

3^{x} = 1

or

3^{x} = 9

7

Solve:

x = 0

or

x = 2

Answer: $x = 0$ or $x = 2$

Worked Example08

Solve $e^{2 x} - 5 e^{x} + 6 = 0$ , giving exact answers.

1

Note that

e^{2 x} = (e^{x})^{2}

2

Let

u = e^{x}

, so

e^{2 x} = u^{2}

3

Equation becomes:

u^{2} - 5 u + 6 = 0

4

Factorise:

(u - 2) (u - 3) = 0

5

So

u = 2

or

u = 3

6

Substitute back:

e^{x} = 2

or

e^{x} = 3

7

Take natural logs:

x = ln 2

or

x = ln 3

Answer: $x = ln 2$ or $x = ln 3$

Practice

Question 1: Solve

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

by factorising.

Find factors of -10 that add to 3: that's 5 and -2

(x + 5) (x - 2) = 0

x + 5 = 0

or

x - 2 = 0

Answer: $x = - 5$ or $x = 2$

Practice

Question 2: Solve

2 x^{2} - 5 x - 1 = 0

, giving exact answers in simplified form.

Use quadratic formula with

a = 2

,

b = - 5

,

c = - 1

x = \frac{5 \pm 25 + 8}{4} = \frac{5 \pm 33}{4}

33

cannot be simplified further

Answer: $x = \frac{5 + 33}{4}$ or $x = \frac{5 - 33}{4}$

Practice

Question 3: Solve

e^{2 x} - 7 e^{x} + 12 = 0

, giving exact answers.

Let

u = e^{x}

, so equation becomes

u^{2} - 7 u + 12 = 0

Factorise:

(u - 3) (u - 4) = 0

So

u = 3

or

u = 4

Substitute back:

e^{x} = 3

or

e^{x} = 4

Take natural logs:

Answer: $x = ln 3$ or $x = ln 4$

Summary

Three Methods for Solving Quadratics

MethodKey Formula/Process

Factorising

(x - p) (x - q) = 0 \Rightarrow x = p

or

x = q

Quadratic Formula

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

Completing Square

(x + p)^{2} = k \Rightarrow x = - p \pm k

Key Exam Skills

Always rearrange to

a x^{2} + b x + c = 0

first

Simplify all surds in your final answer

For "quadratics in disguise" - use substitution (let

u = ...

)

For exponentials: use logs to find

x

after substitution