# How do you solve x^2-4=0?

Aug 1, 2015

${x}_{1 , 2} = \pm 2$

#### Explanation:

The easiest way to solve this quadratic equation is to isolate ${x}^{2}$ on one side of the equation, then take the square root of both sides.

To do that, add $4$ to both sides of the equation

${x}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} = 4$

${x}^{2} = 4$

Taking the square roots of both sides of the equation will get you

$\sqrt{\left({x}^{2}\right)} = \sqrt{4}$

$= \pm 2 = \left\{\begin{matrix}{x}_{1} = \textcolor{g r e e n}{- 2} \\ {x}_{2} = \textcolor{g r e e n}{2}\end{matrix}\right.$

Alternatively, you could recognize that you're dealing with the difference of two squares, for which you know that

color(blue)(a^2 - b^2 = (a-b)(a+b)

In your case, you would get

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

$\left(x - 2\right) \left(x + 2\right) = 0$

This equation equals zero if $\left(x - 2\right) = 0$ or if $\left(x + 2\right) = 0$, which means that you have

$x + 2 = 0 \implies {x}_{1} = - 2$

and

$x - 2 = 0 \implies {x}_{2} = 2$