# How do you solve x^2/(x^2-4) = x/(x+2)-2/(2-x)?

Aug 28, 2017

$x = 2$

#### Explanation:

Write as:

${x}^{2} / \left({x}^{2} - {2}^{2}\right) = \frac{x}{x + 2} - \frac{2}{2 - x}$

Using the principle that $\textcolor{w h i t e}{\text{s}} {a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

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$\textcolor{b r o w n}{\text{The sort of cheating bit - not really!}}$
Consider $- \frac{2}{2 - x}$.$\textcolor{w h i t e}{c}$ We need to change this such that the
$\textcolor{w h i t e}{}$
denominator is $x - 2$ if we whish to use the context of
$\textcolor{w h i t e}{}$
$\textcolor{w h i t e}{\text{s}} {a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

Suppose we had $+ \frac{2}{x - 2}$ which is the format we wish to have. If we use this is it another form of what was originally written. That is; does it have the same value?

color(green)([2/(x-2)color(red)(xx1)] ->[2/(x-2)color(red)(xx(-1)/(-1))] = -2/((2-x))

So $+ \frac{2}{x - 2}$ has the same value as $- \frac{2}{2 - x}$
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Write as:

${x}^{2} / \left({x}^{2} - {2}^{2}\right) = \frac{x}{x + 2} + \frac{2}{x - 2}$

This is the same as:

${x}^{2} / \left({x}^{2} - {2}^{2}\right) = \frac{x \times 2}{\left(x + 2\right) \left(x - 2\right)}$

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

As the denominators are the same we can just consider the numerators.

${x}^{2} = 2 x$

Divide both sides by x

$x = 2$