GCSE Maths Algebra Inequalities

Here we will learn about quadratic inequalities including how to solve quadratic inequalities, identify solution sets using inequality notation and represent solutions on a number line.

There are also quadratic inequalities worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

Quadratic inequalities are similar to quadratic equations and when plotted they display a parabola. We can solve quadratic inequalities to give a range of solutions.

For example,

The quadratic equation x^{2}+ 6x +5 = 0 has two solutions.

This is shown on the graph below where the parabola crosses the x axis.

We could solve this by factorising: (x + 1)(x + 5) = 0 .

The two values of x that equate this equation to zero are x = - 1 and x = - 5 .

For example,

The quadratic inequality x^{2}+ 6x +5  \leq 0 can be solved by factorising but instead of two solutions, there are a range of solutions.

x^{2}+ 6x +5 \leq 0 means that the we need to know the x values that when the graph is less than 0 .

This corresponds to where the curve is below the x axis.

We can see the from the graph that the x values need to be between -1 and -5 .

Therefore the solution to this inequality can be written using inequality notation,

-5\leq x \leq-1 .

For example,

The quadratic inequality x^{2} + 6x +5 > 0 is similar to the previous inequality and has a range of solutions.

x^{2}+ 6x +5 > 0 means that we need to know the x values when the graph is greater than 0

This corresponds to where the curve is above the x axis.

We can see the from the graph that the x values need to be greater than -1 and less than -5 .

Therefore the solution to this inequality can be written using inequality notation as two inequalities  x > -1 or x < -5 .

## How to solve quadratic inequalities

In order to solve quadratic inequalities by factorising:

2. Find the values of x that make each bracket equal zero.
3. Write the solution using inequality notation.

1. Identify values of a, b and c to substitute into the quadratic formula.
3. Simplify to calculate the solutions of the inequality.
4. Write the solution using inequality notation.

## Related lessons on inequalities

Quadratic inequalities is part of our series of lessons to support revision on inequalities. You may find it helpful to start with the main inequalities lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

### Example 1: solving by factorising

Solve x ^{2}+7x+10 < 0 .

(x + 2)(x + 5) < 0

2Find the values of x that make each bracket equal zero.

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

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

3Write the solution using inequality notation.

We need the values of x that produce a graph that is less than 0 and so below the x axis.

The inequality is given by -5 < x < -2 .

4Represent the solution set on a number line.

-5 and -2 are not included in the solution set so these values are indicated with open circles.

### Example 2: solving by factorising

Solve x^{2}+3x+10\leq0 .

Find the values of x that make each bracket equal zero.

Write the solution using inequality notation.

Represent the solution set on a number line.

### Example 3: solving by factorising

Solve x^{2}-6x+8\geq0 .

Find the values of x that make each bracket equal zero.

Write the solution using inequality notation.

Represent the solution set on a number line.

### Example 4: solving by factorising

List the integers that satisfy 2x^{2} + 9x + 4\leq0

Find the values of x that make each bracket equal zero.

Write the solution using inequality notation.

List the integer values that are included in the solution set.

### Example 5: solving by factorising

Solve x^{2}-9>0 .

Find the values of x that make each bracket equal zero.

Write the solution using inequality notation.

Represent the solution set on a number line.

### Example 6: solving by factorising and rearranging

List the integers that satisfy x^{2}<16 .

Rearrange the inequality.

Factorise the inequality.

Find the values of x that make each bracket equal zero.

Write the solution using inequality notation.

List the integer values that are included in the solution set.

### Example 7: solving by using the quadratic formula

Solve 5x^{2}+6x-12\leq0 .

Identify values of a, b and c to substitute into the quadratic formula.

Substitute the values into the quadratic formula.

Simplify to calculate the solutions of the inequality.

Write the solution using inequality notation.

### Example 8: solving by using the quadratic formula

Solve 2x^{2}-3x- 4 < 0 .

Identify values of a, b and c to substitute into the quadratic formula.

Substitute the values into the quadratic formula.

Simplify to calculate the solutions of the inequality.

Write the solution using inequality notation.

### Example 9: one solution

Solve x^{2}+6x+9\leq0 .

Factorise the inequality.

Find the values of x that make each bracket equal zero.

Write the solution using inequality notation.

### Common misconceptions

• Square root of a number has two solutions

A common error when calculating the square root of a number is only writing the positive solution. A negative number squared will also give a positive solution.

For example,

When solving x^2=16, x=\pm \sqrt{16} =\pm 4 .

• Not maintaining the original inequality sign

A common error is to accidentally change the direction of the inequality sign or not writing an inequality sign in the solution.

• Writing a double inequality when the solution set is split

When solving  x^{2} + 10x +16  < 0 there is one solution set which is below the x axis. Therefore the solution is -8 < x < -2 .

When solving  x^{2} + 10x +16  > 0 there is one solution set which is above the x axis. Therefore the solution is x<-8 and x > -2 .

1. Solve x^{2}+13x+30<0 .

-10 <x< -3

10 <x<3

-10 and -3

-10 >x> -3
x^{2} + 13x +30 < 0

Factorise the quadratic, (x + 3)(x +10) < 0 .

Find the values of x that make the bracket equal zero, x = -10 and x = -3 .

Write the solution using interval notation, -10 < x < -3 .

2. Solve x^{2}+3x-10\leq0 .

5\leq x \leq-2

-5\leq x \leq 2

-5 and -2

-5\leq x \leq -2
x^{2}+3x-10 \leq 0

Factorise the quadratic, (x + 5)(x -2) \leq 0 .

Find the values of x that make the bracket equal zero, x = -5 and x = 2 .

Write the solution using interval notation, -5\leq x \leq2 .

3. Solve x^{2}-7x+6\geq0 .

1 \leq x < 6

2 \geq x \geq 3

1 \geq x \leq 6

x \leq 1 and x\geq 6

x^{2}-7x+6\geq 0

Factorise the quadratic, (x-6)(x-1)\geq 0 .

Find the values of x that make the bracket equal zero, x = 6 and x =1 .

Write the solution using interval notation, x \leq 1 and x\geq6 .

4. Solve 2x^{2}-11x+ 5 \leq 0 .

-5\leq x \leq -\frac{1}{2}

1\leq x \leq 5

\frac{1}{2}\leq x \leq 5

-\frac{1}{2}\leq x \leq 5
2x^{2}-11x+ 5 \leq 0

Factorise the quadratic, (2x-1)(x-5) \leq 0 .

Find the values of x that make the bracket equal zero, x =\frac{1}{2} and x =5 .

Write the solution using interval notation, \frac{1}{2} \leq x\leq 5 .

5. Solve x^{2}-25>0 .

-5\leq x \leq 5

x = -5 and x =5

-5 > x > 5

x > 5
x^{2} – 25 > 0

Factorise the quadratic, (x + 5)(x -5) > 0 .

Find the values of x that make the bracket equal zero, x = -5 and x =5 .

Write the solution using interval notation, x < -5 and x > 5 .

6. Solve   x^{2}<1 .

-1 < x < 1

-1 > x > 1

x < -1

-1\leq x \leq 1
x^{2} < 1

Rearrange to make the inequality less than zero, x^{2} – 1<0 .

Factorise the quadratic, (x + 1)(x -5) < 0 .

Find the values of x that make the bracket equal zero, x = -1 and x =1 .

Write the solution using interval notation, -1 < x < 1 .

7. Solve 3x^{2}+4x-2\leq 0 .

-0.39\leq x \leq 1.72

-1.72\geq x \leq 0.39

-1.72\leq x \leq 0.39

-3\leq x \leq 4

3x^{2} + 4x – 2 \leq 0 .

Identify a, b and c from the form ax^{2} +bx + c \leq 0 , a = 3, b = 4, c = -2 .

Substitute the values in to the quadratic formula, x = \frac{-4 \pm \sqrt{4^{2}-4 \times 3 \times-2}}{2 \times 3} .

Simplify the inequality, x = \frac{-4 \pm \sqrt{16–24}}{6} .

Simplify further, x=\frac{-4 \pm \sqrt{40}}{6} .

You could simplify to surd form, x = \frac{-4 \pm 2 \sqrt{10}}{6} .

Simplify the fraction. You could leave in surd form, x = \frac{-2 \pm \sqrt{10}}{3} .

Correct to 2 decimal places, x \leq 0.39 and x \geq -1.72 .

Written in surd form written in interval notation, \frac{-2+\sqrt{10}}{3} \leq x \leq \frac{-2+\sqrt{10}}{3} .

Written as an inequality with values rounded to 2 decimal places, -1.72 \leq x \leq 0.39 .

8. Solve x^{2}-10x+25 \leq 0 .

x = 5

-5\leq x \leq -\frac{1}{2}

-5 >x > 5

x\leq5

x^{2} -10x +25 \leq 0 .

Factorise the quadratic, (x – 5)(x -5) \leq 0 .

Find the values of x that make the bracket equal zero, x =5 .

The quadratic only has one root at 5, so is never less than zero but equal to zero at 5, x = 5 .

1. Solve the inequality x^{2}-8x+15\leq 0 .

(4 marks)

(x-3)(x-5) \leq 0

Factorise the inequality. 1 mark if there are one or two errors.

(2)

Roots found x = 3 and x = 5

(1)

3\leq x \leq 5

(1)

2. (a) Find the interval for which x^{2}+3x – 10 < 0 .

(b) Represent your solution to part (a) on the number line below.

(6 marks)

(a)

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

1 mark if there are one or two errors.

(2)

Roots found x =2 and x = -5

(1)

-5 < x < 2

(1)

(b)

-5 and 2 indicated.

(1)

Correct solution with open circles.

(1)

3. Here is a rectangle.

All measurements are in centimetres.

x is an integer.

The total area of the rectangle is less than 21cm^{2} .

Show that 4 <x < 7 .

x^{2} – 4x <21 .

(1)

x^{2} – 4x – 21< 0 .

(1)

(x – 7)(x + 3) < 0 .

(1)

7 and -3

(1)

4<x < 7 as you cannot have a negative value for distance.

(1)

## Learning checklist

You have now learned how to: