Quadratic Formula Questions, Worksheets and Revision

Quadratic Formula Questions, Worksheets and Revision

GCSE 6 - 7AQAEdexcelOCRWJECHigherEdexcel 2022WJEC 2022

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:

Level 6-7 GCSE AQA Edexcel OCR WJEC

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

Level 6-7 GCSE AQA Edexcel OCR WJEC

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.

Level 6-7 GCSE AQA Edexcel OCR WJEC

Example 1: Quadratic where a=1

Use the quadratic formula to solve 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}=5x_2=\dfrac{-2-12}{2}=-7

Level 6-7 GCSE AQA Edexcel OCR WJEC

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.

Level 6-7 GCSE AQA Edexcel OCR WJEC

Example Questions

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)}

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)}

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)}

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)}

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}

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