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

Dec 29, 2015

${x}^{2} / \left({x}^{2} - 4\right) = \frac{x}{x} ^ 2 - \frac{2 x}{2 - x}$

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

Here we can see that $L . C . M$ is $x \left(x - 2\right) \left(x + 2\right)$
Multiply both sides by $L . C . M$ i.e. by $x \left(x - 2\right) \left(x + 2\right)$

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

$\implies {x}^{3} = \left(x - 2\right) \left(x + 2\right) + 2 {x}^{2} \left(x + 2\right)$
$\implies {x}^{3} = {x}^{2} - 4 + 2 {x}^{3} + 4 {x}^{2}$
$\implies {x}^{3} + 5 {x}^{2} - 4 = 0$
$\implies {x}^{3} + {x}^{2} + 4 {x}^{2} + 4 x - 4 x - 4 = 0$
$\implies {x}^{2} \left(x + 1\right) + 4 x \left(x + 1\right) - 4 \left(x + 1\right) = 0$
$\implies \left(x + 1\right) \left({x}^{2} + 4 x - 4\right) = 0$
$\implies$ either $x + 1 = 0 \mathmr{and} {x}^{2} + 4 x - 4 = 0$

$\implies x = - 1 \mathmr{and} x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \left(- 4\right)}}{2 \cdot 1}$

$\implies x = - 1 \mathmr{and} x = \frac{- 4 \pm \sqrt{16 + 16}}{2}$

$\implies x = - 1 \mathmr{and} x = \frac{- 4 \pm 4 \sqrt{1 + 1}}{2}$

$\implies x = - 1 \mathmr{and} x = \frac{2 \left(- 2 \pm 2 \sqrt{1 + 1}\right)}{2}$

$\implies x = - 1 \mathmr{and} x = - 2 \pm 2 \sqrt{2}$

Hence $x$ can have three values which are $x = - 1 , - 2 + 2 \sqrt{2} , - 2 - 2 \sqrt{2}$