**The Quadratic Formula:**

The quadratic formula is a formula that you can substitute values into in order to **find the solutions to any quadratic equation**. This is just one method of solving quadratics, you will encounter more throughout the course.

Make sure you are happy with the following topics before continuing:

**The Quadratic Formula**

The quadratic formula is a formula that you can substitute values into in order to get the solutions to any quadratic equation.

x=\dfrac{-\textcolor{red}{b}\pm\sqrt{\textcolor{red}{b}^2-4\textcolor{blue}{a}\textcolor{Orange}{c}}}{2\textcolor{blue}{a}}

The \textcolor{blue}{a}, \textcolor{red}{b} and \textcolor{Orange}{c} values correspond to values in a quadratic equation, shown below

\textcolor{blue}{a}x^2+\textcolor{red}{b}x+\textcolor{Orange}{c}=0

**Take note:**

Notice that there is a “plus or minus” symbol in there {\textcolor{blue}{(\pm)}}.

This is because a quadratic has up to** two real solutions** – putting a plus sign there will give you one solution and putting a minus sign there will give you the other. In other words, the two solutions are

x=\dfrac{-b\textcolor{blue}{+}\sqrt{b^2-4ac}}{2a} \,\,\,\,\,\,\, and \,\,\,\,\,\,\, x=\dfrac{-b\textcolor{blue}{-}\sqrt{b^2-4ac}}{2a}

You are **not given **this formula in an exam so you do have to **memorise it**.

**Example 1: Quadratic where a=1**

Use the **quadratic formula** the following quadratic equation:

x^2+2x-35=0

**[2 marks]**

Firstly, we have to identify what a,b, and c are:

a=1, b=2, c=-35

Next we need to **substitute** these into the formula:

x=\dfrac{-2\pm\sqrt{2^2-4\times1\times(-35)}}{2}

Simplifying this we get

x=\dfrac{-2+\sqrt{144}}{2}, x=\dfrac{-2-\sqrt{144}}{2}

We know \sqrt{144}=12, so the two final solutions are

x_1=\dfrac{-2+12}{2}=5, x_2=\dfrac{-2-12}{2}=-7

**Example 2: Quadratic where a>1**

Use the **quadratic formula** to solve the following quadratic equation:

2x^2-6x+3=0

Give your answer to 2 decimal places.

**[2 marks]**

a=2, b=-6, c=3

Putting these into the formula, we get

x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4\times2\times3}}{2\times2}

so, the solutions are

x=\dfrac{6+\sqrt{12}}{4} and x=\dfrac{6-\sqrt{12}}{4}

12 is not a square number, which is how we know this won’t give us a nice answer.

The question asks for 2dp, so putting these into the calculator, we get

x_1=2.366...=2.37 (2dp) and x_2=0.6339...=0.63 (2dp)

**Note:** You can put the first quadratic formula straight into the calculator without any simplifying and use the + and – to get your two answers.

## GCSE Maths Predicted Papers

(104 Reviews) £4.99### Example Questions

**Question 1:** Use the quadratic formula to find the solutions to x^2+11x+16=0 to 3 significant figures.

**[2 marks]**

Firstly, the quadratic formula is

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

Then, we can identify that here, a=1, b=11, and c=16. Putting these values into the formula, we get

x=\dfrac{-11\pm\sqrt{11^2-4\times1\times16}}{2}

The part inside the square root is

11^2-4\times1\times16=121-64=57

So, the solutions become

x=\dfrac{-11\pm\sqrt{57}}{2}

Putting these into a calculator (one with +, one with -), we get the final solutions

x=-1.7250...= -1.73\text{ (3sf), and }\,x=-9.2749...=-9.27\text{ (3sf)}

**Question 2:** Use the quadratic formula to find the solutions to x^2-2x=44 to 3 significant figures.

**[2 marks]**

Firstly, the quadratic formula is

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

Then, we can identify that here, a=1, b=-2, and c=-44. Putting these values into the formula, we get

x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4\times1\times(-44)}}{2}

The part inside the square root is

(-2)^2-4\times1\times(-44)=4+176=180

So, the solutions become

x=\dfrac{2\pm\sqrt{180}}{2}

Putting these into a calculator (one with +, one with -), we get the final solutions

x=7.7082...=7.71\text{ (3sf), and }\,x=-5.7082...=-5.71\text{ (3sf)}

**Question 3:** Use the quadratic formula to find the solutions to 4x^2+7x-1=0 to 2 decimal places.

**[2 marks]**

Firstly, the quadratic formula is

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

Then, we can identify that here, a=4, b=7, and c=-1. Putting these values into the formula, we get

x=\dfrac{-7\pm\sqrt{7^2-4\times4\times(-1)}}{2\times4}

The part inside the square root is

7^2-4\times4\times(-1)=49+16=65

So, the solutions become

x=\dfrac{-7\pm\sqrt{65}}{8}

Putting these into a calculator (one with +, one with -), we get the final solutions

x=0.1327...=0.13\text{ (2dp), and }\,x=-1.8827...=-1.88\text{ (2dp)}

**Question 4:** Use the quadratic formula to find the solutions to x^2+8x=-13 to 2 decimal places.

**[2 marks]**

Firstly, the quadratic formula is

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

Then, we can identify that here, a=1, b=8, and c=13. Putting these values into the formula, we get

x=\dfrac{-8\pm\sqrt{8^2-4\times1\times13}}{2\times1}

The part inside the square root is

8^2-4\times8\times13=64-52=12

So, the solutions become

x=\dfrac{-8\pm\sqrt{12}}{2}

Putting these into a calculator (one with +, one with -), we get the final solutions

x=-2.2679...=-2.27\text{ (2dp), and }\,x=-5.73205...=-5.73\text{ (2dp)}

**Question 5:** Use the quadratic formula to find the solutions to 25x^2-30x+7=0, giving your answer in surd form.

**[3 marks]**

Firstly, the quadratic formula is

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

Then, we can identify that here, a=25, b=-30, and c=7. Putting these values into the formula, we get

x=\dfrac{-(-30)\pm\sqrt{(-30)^2-4\times25\times7}}{2\times25}

The part inside the square root is

30^2-4\times25\times7=900-700=200

So, the solutions become

x=\dfrac{30\pm\sqrt{200}}{50}

Putting these into a calculator (one with +, one with -), we get the final solutions,

x=\dfrac{3\pm\sqrt{2}}{5}

### Worksheets and Exam Questions

#### (NEW) The Quadratic Formula Exam Style Questions - MME

Level 6-7 New Official MME### Drill Questions

#### Quadratic Formula (1) - Drill Questions

#### Quadratic Formula (2) - Drill Questions

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