# How do you solve 5x^2 = 60?

Nov 6, 2015

$x = \pm 2 \sqrt{3}$
The explanation is a bit long but this is because I have shown the thinking process in detail.

#### Explanation:

When you are required to find the value of something (in your case $x$) your objective is to get only 1 of your target ( your $x$) on the left of the equals sign and everything else on the other. Mathematically you are saying that one of "whatever it is" is of value "everything else on the others side".

The first step of isolating $x$ is to strip off the 5.

As the association between $x$ and 5 is that of multiplication we do this by changing it to 1 as 1 times anything has no effect. The same thing goes for division as dividing by 1 does not change anything.

Divide both sides by 5

$\left(5 {x}^{2}\right) \div i \mathrm{de} 5 = \left(60\right) \div i \mathrm{de} 5$
The brackets are there only to show you the original equation.

This is the same as:

$\left(5 {x}^{2}\right) \times \frac{1}{5} = \left(60\right) \times \frac{1}{5}$

This gives us

$\frac{5}{5} \times {x}^{2} = \frac{60}{5}$

${x}^{2} = 12$

but we need only one $x$ on the left side
Known that $\sqrt{{x}^{2}} = x$

Taking square root of both sides gives

$x = \sqrt{12}$ which is not quite correct. We have missed a value!

To take this extra value into account we must write:

$x = \pm \sqrt{12}$

as $\left(- \sqrt{12}\right) \times \left(- \sqrt{12}\right) = x = \left(+ \sqrt{12}\right) \times \left(+ \sqrt{12}\right)$

But $12 = 3 \times 4$ and $4 = {2}^{2}$

$x = \pm \sqrt{3 \times {2}^{2}}$

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