Solving Quadratics

What you'll learn

How to recognise a quadratic equation.
How to factorise quadratics of the form $x^2+bx+c$ .
How to use factorised brackets to solve a quadratic equation.
How to avoid the most common sign mistakes.

1. What is a quadratic?

A quadratic is an expression or equation where the highest power of the variable is squared, such as $x^2+5x+6$ or $a^2-4a-12=0$ .

Definition

Quadratic equation

A quadratic equation is an equation where the highest power of the variable is 2, usually written in the form $ax^2+bx+c=0$ , where $\neq 0$ .

In this topic, most questions have coefficient 1 in front of the squared term, for example:

x^2+7x+12=0

These are usually solved by factorising.

Definition

Factorising

To factorise means to rewrite an expression as a product of brackets. For example, $x^2+5x+6$ factorises to $(x + 2) (x + 3)$ .

2. Why factorising helps you solve

If a quadratic is written as two brackets multiplied together, you can use a very important fact:

Key Idea

The zero product rule

If two things multiply to make zero, then at least one of them must be zero. So if $(x + 4) (x - 7) = 0$ , then either $x + 4 = 0$ or $x - 7 = 0$ .

This is why solving quadratics by factorising has two parts:

Factorise the quadratic.
Set each bracket equal to zero.

A quadratic graph can cross the x-axis twice. These crossing points are the solutions, also called roots.

A parabola crossing the x-axis at two labelled roots

Definition

Root or solution

A root or solution of a quadratic equation is a value of the variable that makes the equation true.

3. Recap: expanding double brackets

Before factorising, it helps to remember how brackets expand.

For example:

(x + 4) (x + 3)

Multiply each term in the first bracket by each term in the second bracket:

(x+4)(x+3)=x2+3x+4x+12=x2+7x+12\begin{aligned} (x+4)(x+3) &= x^2+3x+4x+12 \\ &= x^2+7x+12 \end{aligned}

So factorising goes backwards:

x^2+7x+12=(x+4)(x+3)

Tip

Think backwards

Expanding asks, “What do these brackets make?” Factorising asks, “Which brackets made this expression?”

4. Factorising $x^2+bx+c$

For a quadratic like $x^2+bx+c$ , you need two numbers that:

multiply to give the constant term $c$ ;
add to give the coefficient of $x$ , which is $b$ .

For example, in $x^2+9x+20$ :

the two numbers multiply to 20;
the two numbers add to 9.

The numbers are 4 and 5, so:

x^2+9x+20=(x+4)(x+5)

Example

Factorise then solve a positive quadratic

Factorise and solve $u^2+8u+15=0$ .

Find two numbers that multiply to 15 and add to 8.
The numbers are 3 and 5, because 3 × 5 = 15 and 3 + 5 = 8.
Write the quadratic as two brackets.
$u^2+8u+15=(u+3)(u+5)$
Now solve the equation.
$(u + 3) (u + 5) = 0$
Set each bracket equal to zero.
$\quad \text{or} \quad u+5=0$
Solve each simple equation.
$\quad \text{or} \quad u=-5$

Common Mistake

Forgetting the final solve step

If the question says “solve”, factorising alone is not enough. After writing the brackets, you must give the values of the variable.

5. When the constant is positive

If the constant term is positive, the two bracket numbers must have the same sign.

Positive middle term

For $x^2+bx+c$ , where both $b$ and $c$ are positive, both bracket numbers are positive.

Example:

x^2+11x+30=(x+5)(x+6)

Negative middle term

For $x^2-bx+c$ , where the constant is positive but the middle term is negative, both bracket numbers are negative.

Example:

x^2-9x+20=(x-4)(x-5)

Example

Solve when both bracket signs are negative

Solve $p^2-7p+12=0$ .

Find two numbers that multiply to 12 and add to -7.
The numbers are -3 and -4, because $(−3)×(−4)=12(-3)\times(-4)=12$ and $- 3 + (- 4) = - 7$ .
Factorise the quadratic.
$p^2-7p+12=(p-3)(p-4)$
Set the factorised form equal to zero.
$(p - 3) (p - 4) = 0$
Set each bracket equal to zero.
$\quad \text{or} \quad p-4=0$
Solve.
$\quad \text{or} \quad p=4$

Tip

Check the signs

If the constant is positive, your two signs should match: both plus or both minus.

6. When the constant is negative

If the constant term is negative, the two bracket numbers must have opposite signs.

For example, in $x^2+4x-21$ :

the numbers multiply to -21;
the numbers add to 4.

The pair is 7 and -3, because $7×(−3)=−217\times(-3)=-21$ and $7 + (- 3) = 4$ .

So:

x^2+4x-21=(x+7)(x-3)

Example

Solve with one plus bracket and one minus bracket

Solve $x^2+2x-48=0$ .

Find two numbers that multiply to -48 and add to 2.
List useful factor pairs of 48: 1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8.
Since the product is negative, one number is positive and one is negative.
The pair that gives a sum of 2 is 8 and -6.
Factorise the quadratic.
$x^2+2x-48=(x+8)(x-6)$
Set the brackets equal to zero.
$\quad \text{or} \quad x-6=0$
Solve each one.
$\quad \text{or} \quad x=6$

Common Mistake

Choosing the right pair but wrong signs

For $x^2+2x-48$ , the numbers 6 and 8 are useful, but the signs matter. You need +8 and -6, not +6 and -8, because the middle term must be +2.

7. Solving directly from the equation

Sometimes the question only asks you to solve, not factorise first. You still usually factorise as your method.

Example

Solve a quadratic directly

Solve $m^2+13m+36=0$ .

Look for two numbers that multiply to 36 and add to 13.
The pair is 4 and 9.
Factorise.
$m^2+13m+36=(m+4)(m+9)$
Rewrite the equation in factorised form.
$(m + 4) (m + 9) = 0$
Use the zero product rule.
$\quad \text{or} \quad m+9=0$
Solve both equations.
$\quad \text{or} \quad m=-9$

8. Quick checking by expanding

You can check your factorisation by expanding the brackets again.

Suppose you think:

x^2-5x-24=(x-8)(x+3)

Expand to check:

(x−8)(x+3)=x2+3x−8x−24=x2−5x−24\begin{aligned} (x-8)(x+3) &= x^2+3x-8x-24 \\ &= x^2-5x-24 \end{aligned}

So the factorisation is correct.

Tip

A fast mental check

For brackets like $(x + a) (x + b)$ , the middle term comes from adding $a$ and $b$ , and the final term comes from multiplying $a$ and $b$ .

9. Why there are usually two answers

A quadratic can often have two solutions because either bracket can be zero.

For example:

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

This is true when:

x - 2 = 0

or when:

x + 5 = 0

So the two solutions are:

\quad \text{or} \quad x=-5

Common Mistake

Do not divide by a bracket containing the variable

If you have $(x - 2) (x + 5) = 0$ , do not divide by $x - 2$ or $x + 5$ . One of those brackets might be zero, and division by zero is not allowed.

10. Mini method summary

For quadratics in the form $x^2+bx+c=0$ :

Find two numbers that multiply to $c$ .
Make sure they add to $b`.
Put them into brackets.
Set each bracket equal to zero.
Solve the two simple linear equations.

Common Mistake

Sign slip in the method

The add number is the coefficient of $x$ , including its sign. For $x^2-10x+16$ , the target sum is -10, not 10.

Example

Full exam-style example

Solve $z^2-4z-45=0$ .

The constant is -45, so the two numbers must have opposite signs.
Find a factor pair of 45 with a difference of 4.
The useful pair is 9 and 5. To make the sum -4, use -9 and +5.
Factorise.
$z^2-4z-45=(z-9)(z+5)$
Write the equation in factorised form.
$(z - 9) (z + 5) = 0$
Set each bracket equal to zero.
$\quad \text{or} \quad z+5=0$
Solve.
$\quad \text{or} \quad z=-5$

Exam technique

In the exam

First decide whether the constant is positive or negative; this tells you whether the bracket signs match or differ.
If you are stuck, list factor pairs of the constant and test which pair gives the middle coefficient.
After finding the two answers, substitute them back into the original equation mentally if time allows.

Self review

Check yourself

Can you explain why $(x + 6) (x - 2) = 0$ gives two possible equations to solve?
If the constant term is negative, what must be true about the signs in the two brackets?
How would you check whether $(x - 7) (x + 4)$ really expands to $x^2-3x-28$ ?

Solving Quadratics

What you'll learn

How to recognise a quadratic equation.
How to factorise quadratics of the form $x^2+bx+c$ .
How to use factorised brackets to solve a quadratic equation.
How to avoid the most common sign mistakes.

1. What is a quadratic?

A quadratic is an expression or equation where the highest power of the variable is squared, such as $x^2+5x+6$ or $a^2-4a-12=0$ .

Definition

Quadratic equation

A quadratic equation is an equation where the highest power of the variable is 2, usually written in the form $ax^2+bx+c=0$ , where $\neq 0$ .

In this topic, most questions have coefficient 1 in front of the squared term, for example:

x^2+7x+12=0

These are usually solved by factorising.

Definition

Factorising

To factorise means to rewrite an expression as a product of brackets. For example, $x^2+5x+6$ factorises to $(x + 2) (x + 3)$ .

2. Why factorising helps you solve

If a quadratic is written as two brackets multiplied together, you can use a very important fact:

Key Idea

The zero product rule

If two things multiply to make zero, then at least one of them must be zero. So if $(x + 4) (x - 7) = 0$ , then either $x + 4 = 0$ or $x - 7 = 0$ .

This is why solving quadratics by factorising has two parts:

Factorise the quadratic.
Set each bracket equal to zero.

A quadratic graph can cross the x-axis twice. These crossing points are the solutions, also called roots.

A parabola crossing the x-axis at two labelled roots

Definition

Root or solution

A root or solution of a quadratic equation is a value of the variable that makes the equation true.

3. Recap: expanding double brackets

Before factorising, it helps to remember how brackets expand.

For example:

(x + 4) (x + 3)

Multiply each term in the first bracket by each term in the second bracket:

(x+4)(x+3)=x2+3x+4x+12=x2+7x+12\begin{aligned} (x+4)(x+3) &= x^2+3x+4x+12 \\ &= x^2+7x+12 \end{aligned}

So factorising goes backwards:

x^2+7x+12=(x+4)(x+3)

Tip

Think backwards

Expanding asks, “What do these brackets make?” Factorising asks, “Which brackets made this expression?”

4. Factorising $x^2+bx+c$

For a quadratic like $x^2+bx+c$ , you need two numbers that:

multiply to give the constant term $c$ ;
add to give the coefficient of $x$ , which is $b$ .

For example, in $x^2+9x+20$ :

the two numbers multiply to 20;
the two numbers add to 9.

The numbers are 4 and 5, so:

x^2+9x+20=(x+4)(x+5)

Example

Factorise then solve a positive quadratic

Factorise and solve $u^2+8u+15=0$ .

Find two numbers that multiply to 15 and add to 8.
The numbers are 3 and 5, because 3 × 5 = 15 and 3 + 5 = 8.
Write the quadratic as two brackets.
$u^2+8u+15=(u+3)(u+5)$
Now solve the equation.
$(u + 3) (u + 5) = 0$
Set each bracket equal to zero.
$\quad \text{or} \quad u+5=0$
Solve each simple equation.
$\quad \text{or} \quad u=-5$

Common Mistake

Forgetting the final solve step

If the question says “solve”, factorising alone is not enough. After writing the brackets, you must give the values of the variable.

5. When the constant is positive

If the constant term is positive, the two bracket numbers must have the same sign.

Positive middle term

For $x^2+bx+c$ , where both $b$ and $c$ are positive, both bracket numbers are positive.

Example:

x^2+11x+30=(x+5)(x+6)

Negative middle term

For $x^2-bx+c$ , where the constant is positive but the middle term is negative, both bracket numbers are negative.

Example:

x^2-9x+20=(x-4)(x-5)

Example

Solve when both bracket signs are negative

Solve $p^2-7p+12=0$ .

Find two numbers that multiply to 12 and add to -7.
The numbers are -3 and -4, because $(−3)×(−4)=12(-3)\times(-4)=12$ and $- 3 + (- 4) = - 7$ .
Factorise the quadratic.
$p^2-7p+12=(p-3)(p-4)$
Set the factorised form equal to zero.
$(p - 3) (p - 4) = 0$
Set each bracket equal to zero.
$\quad \text{or} \quad p-4=0$
Solve.
$\quad \text{or} \quad p=4$

Tip

Check the signs

If the constant is positive, your two signs should match: both plus or both minus.

6. When the constant is negative

If the constant term is negative, the two bracket numbers must have opposite signs.

For example, in $x^2+4x-21$ :

the numbers multiply to -21;
the numbers add to 4.

The pair is 7 and -3, because $7×(−3)=−217\times(-3)=-21$ and $7 + (- 3) = 4$ .

So:

x^2+4x-21=(x+7)(x-3)

Example

Solve with one plus bracket and one minus bracket

Solve $x^2+2x-48=0$ .

Find two numbers that multiply to -48 and add to 2.
List useful factor pairs of 48: 1 and 48, 2 and 24, 3 and 16, 4 and 12, 6 and 8.
Since the product is negative, one number is positive and one is negative.
The pair that gives a sum of 2 is 8 and -6.
Factorise the quadratic.
$x^2+2x-48=(x+8)(x-6)$
Set the brackets equal to zero.
$\quad \text{or} \quad x-6=0$
Solve each one.
$\quad \text{or} \quad x=6$

Common Mistake

Choosing the right pair but wrong signs

For $x^2+2x-48$ , the numbers 6 and 8 are useful, but the signs matter. You need +8 and -6, not +6 and -8, because the middle term must be +2.

7. Solving directly from the equation

Sometimes the question only asks you to solve, not factorise first. You still usually factorise as your method.

Example

Solve a quadratic directly

Solve $m^2+13m+36=0$ .

Look for two numbers that multiply to 36 and add to 13.
The pair is 4 and 9.
Factorise.
$m^2+13m+36=(m+4)(m+9)$
Rewrite the equation in factorised form.
$(m + 4) (m + 9) = 0$
Use the zero product rule.
$\quad \text{or} \quad m+9=0$
Solve both equations.
$\quad \text{or} \quad m=-9$

8. Quick checking by expanding

You can check your factorisation by expanding the brackets again.

Suppose you think:

x^2-5x-24=(x-8)(x+3)

Expand to check:

(x−8)(x+3)=x2+3x−8x−24=x2−5x−24\begin{aligned} (x-8)(x+3) &= x^2+3x-8x-24 \\ &= x^2-5x-24 \end{aligned}

So the factorisation is correct.

Tip

A fast mental check

For brackets like $(x + a) (x + b)$ , the middle term comes from adding $a$ and $b$ , and the final term comes from multiplying $a$ and $b$ .

9. Why there are usually two answers

A quadratic can often have two solutions because either bracket can be zero.

For example:

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

This is true when:

x - 2 = 0

or when:

x + 5 = 0

So the two solutions are:

\quad \text{or} \quad x=-5

Common Mistake

Do not divide by a bracket containing the variable

If you have $(x - 2) (x + 5) = 0$ , do not divide by $x - 2$ or $x + 5$ . One of those brackets might be zero, and division by zero is not allowed.

10. Mini method summary

For quadratics in the form $x^2+bx+c=0$ :

Find two numbers that multiply to $c$ .
Make sure they add to $b`.
Put them into brackets.
Set each bracket equal to zero.
Solve the two simple linear equations.

Common Mistake

Sign slip in the method

The add number is the coefficient of $x$ , including its sign. For $x^2-10x+16$ , the target sum is -10, not 10.

Example

Full exam-style example

Solve $z^2-4z-45=0$ .

The constant is -45, so the two numbers must have opposite signs.
Find a factor pair of 45 with a difference of 4.
The useful pair is 9 and 5. To make the sum -4, use -9 and +5.
Factorise.
$z^2-4z-45=(z-9)(z+5)$
Write the equation in factorised form.
$(z - 9) (z + 5) = 0$
Set each bracket equal to zero.
$\quad \text{or} \quad z+5=0$
Solve.
$\quad \text{or} \quad z=-5$

Exam technique

In the exam

First decide whether the constant is positive or negative; this tells you whether the bracket signs match or differ.
If you are stuck, list factor pairs of the constant and test which pair gives the middle coefficient.
After finding the two answers, substitute them back into the original equation mentally if time allows.

Self review

Check yourself

Can you explain why $(x + 6) (x - 2) = 0$ gives two possible equations to solve?
If the constant term is negative, what must be true about the signs in the two brackets?
How would you check whether $(x - 7) (x + 4)$ really expands to $x^2-3x-28$ ?

Solving Quadratics

What you'll learn

1. What is a quadratic?

2. Why factorising helps you solve

3. Recap: expanding double brackets

4. Factorising $x^2+bx+c$

5. When the constant is positive

Positive middle term

Negative middle term

6. When the constant is negative

7. Solving directly from the equation

8. Quick checking by expanding

9. Why there are usually two answers

10. Mini method summary

Recap questions

You've reached the end

Solving Quadratics

What you'll learn

1. What is a quadratic?

2. Why factorising helps you solve

3. Recap: expanding double brackets

4. Factorising $x^2+bx+c$

5. When the constant is positive

Positive middle term

Negative middle term

6. When the constant is negative

7. Solving directly from the equation

8. Quick checking by expanding

9. Why there are usually two answers

10. Mini method summary

Recap questions

You've reached the end

Solving Quadratics

What you'll learn

1. What is a quadratic?

2. Why factorising helps you solve

3. Recap: expanding double brackets

4. Factorising x2+bx+cx^2+bx+cx2+bx+c

5. When the constant is positive

Positive middle term

Negative middle term

6. When the constant is negative

7. Solving directly from the equation

8. Quick checking by expanding

9. Why there are usually two answers

10. Mini method summary

Recap questions

You've reached the end

Solving Quadratics Revision Guide

Solving Quadratics

What you'll learn

1. What is a quadratic?

2. Why factorising helps you solve

3. Recap: expanding double brackets

4. Factorising x2+bx+cx^2+bx+cx2+bx+c

5. When the constant is positive

Positive middle term

Negative middle term

6. When the constant is negative

7. Solving directly from the equation

8. Quick checking by expanding

9. Why there are usually two answers

10. Mini method summary

Recap questions

You've reached the end

Solving Quadratics Revision Guide

4. Factorising $x^2+bx+c$

4. Factorising $x^2+bx+c$