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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 .

In order to solve quadratic inequalities by factorising:

**Factorise the quadratic expression.****Find the values of**x**that make each bracket equal zero.****Write the solution using inequality notation.**

In order to solve quadratic inequalities by using the quadratic formula:

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

Get your free quadratic inequalities worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free quadratic inequalities worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**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:

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

**Factorise the quadratic expression.**

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

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

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

3**Write 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 .

4**Represent 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.

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

**Factorise the quadratic expression.**

(x-2)(x+5)\leq0

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

(x - 2) = 0 when x = 2

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

**Write the solution using inequality notation.**

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

The inequality is given by -5\leq x \leq2 .

**Represent the solution set on a number line.**

-5 and 2 are included in the solution set so these values are indicated with closed circles.

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

**Factorise the quadratic expression.**

(x-2)(x-4)\geq0

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

(x -2) = 0 when x = 2

(x -4 ) = 0 when x = 4

**Write the solution using inequality notation.**

We need the values of x that produce a graph that is** greater than or equal to ** 0 and so **above the ** x ** axis.**

There will be two separate inequalities as there are two separate regions indicated on the graph.

x \geq4 and x \leq2 .

**Represent the solution set on a number line.**

2 and 4 are included in the solution set so these values are indicated with closed circles.

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

**Factorise the quadratic expression.**

(2x+1)(x+4)\leq0

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

(2x + 1) = 0 when x = -\frac{1}{2}

(x +4 ) = 0 when x = -4

**Write the solution using inequality notation.**

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

The inequality is given by -4\leq x \leq-\frac{1}{2} .

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

-4, -3, -2, -1

x is any value greater than or equal to -4 and less than or equal to -\frac{1}{2} . The highest integer below -\frac{1}{2} is -1 .

Solve x^{2}-9>0 .

**Factorise the quadratic expression.**

(x -3)(x + 3) > 0

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

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

(x -3 ) = 0 when x = 3

**Write the solution using inequality notation.**

We need the values of x that produce a graph that is** greater than ** 0 and so **above the ** x ** axis.**

There will be two separate inequalities as there are two separate regions indicated on the graph.

x>3 and x < -3 .

**Represent the solution set on a number line.**

-3 and 3 are not included in the solution set so these values are indicated with closed circles.

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

**Rearrange the inequality.**

Rearrange the inequality so that it is less than 0 by subtracting 16 from both sides.

x^{2}-16< 0**Factorise the inequality.**

(x - 4)(x + 4) = 0

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

(x +4) = 0 when x = -4

(x -4 ) = 0 when x = 4

**Write 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 -4 < x < 4 .

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

-3, -2, -1, 0, 1, 2, 3

x is any value greater than -4 and less than 4 .

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

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

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

A quadratic expression is in the form ax^{2} + bx + c .

In this inequality a = 5, b = 6 and c = -12 .

**Substitute the values into the quadratic formula.**

The quadratic formula is x = \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} .

Substituting the values a = 5, b = 6 and c = -12 gives x = \frac{-6 \pm \sqrt{6^{2}-4 \times 5 \times-12}}{2 \times 5} .

We need to use the quadratic formula to solve rather than factorising as the discriminant is not a square number.

**Simplify to calculate the solutions of the inequality.**

\[x = \frac{-6 \pm \sqrt{6^{2}-4 \times 5 \times-12}}{2 \times 5} \\
x = \frac{-6 \pm \sqrt{36–240}}{10} \\
x = \frac{-6 \pm \sqrt{276}}{10}\]

\[x= \frac{-6+2\sqrt{69}}{10} \quad and \quad x=\frac{-6-2\sqrt{69}}{10}\\
x= \frac{-3+\sqrt{69}}{5} \quad and \quad x=\frac{-3-\sqrt{69}}{5}\]

In this case the answer is given in surd form but you may be asked to leave the answers rounded to a degree of accuracy.

**Write the solution using inequality notation.**

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

The inequality is given by \frac{-3- \sqrt{69}}{5} \leq x \leq \frac{-3+ \sqrt{69}}{5} .

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

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

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

A quadratic expression is in the form ax^{2}+bx+c .

In this inequality a = 2, b = -3 and c = -4 .

**Substitute the values into the quadratic formula.**

The quadratic formula is x = \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} .

Substituting the values a = 2, b =-36 and c = -4 gives

x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4 \times 2 \times-4}}{2 \times 2} .

**Simplify to calculate the solutions of the inequality.**

\[x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4 \times 2 \times-4}}{2 \times 2} \\
x=\frac{3 \pm \sqrt{9–32}}{4} \\
x=\frac{3 \pm \sqrt{41}}{4}\]

\[x=\frac{3+6.4}{4} \quad and \quad x=\frac{3-6.4}{4}\\
x = 2.35 \quad and \quad x = -0.85\]

In this case the answer is being given to two decimal places but you could leave the solution in surd form.

**Write 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 -0.85 < x< 2.35 .

Rounded to two decimal places.

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

**Factorise the inequality.**

(x + 3)(x + 3) \leq 0 or (x+3)^{2}

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

As both brackets are equal, there is only one solution.

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

**Write the solution using inequality notation.**

We need the values of x that produce a graph that is** less than or equal to ** 0 and so **below the** x ** axis. **This only happens when, x = -3 .

**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\leq0 .

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)**

Show answer

(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)**

Correct answer.

**(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)**

Show answer

(a)

(x -2)(x + 5) < 01 mark if there are one or two errors.

**(2)**

Roots found x =2 and x = -5

**(1)**

**(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 .

Show answer

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)**

You have now learned how to:

- Solve quadratic inequalities
- Use inequality notation and represent solutions on a number line
- List integers that are in the solution set

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