Quadratic Formula Questions | Worksheets and Revision | MME

# Quadratic Formula Questions, Worksheets and Revision

Level 6 Level 7

## What you need to know

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:

$x^2+2x-35=0$

[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

$x=\dfrac{-2+\sqrt{144}}{2}$ and $x=\dfrac{-2-\sqrt{144}}{2}$

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

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

### Example 2:

$2x^2-6x+3=0$

[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 $2$dp, 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.

### Example Questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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,

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

Level 6-7

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Level 6-7

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