# How do you solve x^2-75=0?

Aug 17, 2016

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

#### Explanation:

Treat ${x}^{2} - 75$ as the difference of squares

We know that ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

So replacing $a$ with $x$
and $b$ with $5 \sqrt{3}$ (so that ${b}^{2} = 75$)
we have
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 75 = 0$
is equivalent to
$\textcolor{w h i t e}{\text{XXX}} \left(x - 5 \sqrt{3}\right) \left(x + 5 \sqrt{3}\right) = 0$

which implies
either $\left(x - 5 \sqrt{3}\right) = 0 \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} \left(x + 5 \sqrt{3}\right) = 0$
$\textcolor{w h i t e}{\text{XX")rarr x=5sqrt(3)color(white)("XXXXXXXX}} \rightarrow x = - 5 \sqrt{3}$