# How do you solve 5x^2 - 125 = 0?

Mar 13, 2018

Start by factorizing.

#### Explanation:

Start by factorizing

$5 \left({x}^{2} - 25\right) = 0$

then you can divide both sides by 5 and since $\frac{0}{5} = 0$ you get the expression:

${x}^{2} - 25 = 0$

rearrange so that 25 is on the right hand side

${x}^{2} = 25$

x=± sqrt25

x=± 5

Mar 13, 2018

$x = \pm 5$

#### Explanation:

To make progress, an attempt should be made to have just $x$ on one side of the equation to see what it is equal to (on the other side of the equation).

By inspection, both $5$ and $- 125$ are divisible by $5$. Zero is also divisible by $5$ in the sense $\frac{0}{5} = 0$

So, dividing both sides of the equation by $5$ (also called "dividing through by $5$)

$5 {x}^{2} - 125 = 0$

implies

$\frac{5 {x}^{2}}{5} - \frac{125}{5} = \frac{0}{5}$

that is

${x}^{2} - 25 = 0$

Now $25$ may be added to both sides to give

${x}^{2} - 25 + 25 = 0 + 25$

that is

${x}^{2} = 25$

You will recognise $25$ as a perfect square so finding a solution should be easy but take care! Remember square numbers have two roots, a positive one and a negative one. So

${x}^{2} = 25$

implies

$\sqrt{{x}^{2}} = \sqrt{25}$

that is

$x = 5$
or
$x = - 5$