# How do you solve the rational equation x/(x-2) + (2x)/(4-x^2) = 5/(x+2)?

Feb 13, 2016

$x = \frac{5 \pm \sqrt{15} i}{2}$

#### Explanation:

$\frac{x}{x - 2} + \frac{2 x}{4 - {x}^{2}} = \frac{5}{x + 2}$

First put all the denominatores is standard for by multiplying $\frac{2 x}{4 - {x}^{2}}$by (-1/-1)

$\frac{x}{x - 2} - \frac{2 x}{{x}^{2} - 4} = \frac{5}{x + 2}$

${x}^{2} - 4 = \left(x + 2\right) \left(x - 2\right)$, so we can easily put the fractions with the same denominator:

$\frac{x \left(x + 2\right)}{{x}^{2} - 4} - \frac{2 x}{{x}^{2} - 4} = \frac{5 \left(x - 2\right)}{{x}^{2} - 4}$

No we can eliminate the denominators. The equivalent of multiplying both sides of equation by ${x}^{2} - 4$, but we must keep in mind that x!=+-2, because these values woul put ${x}^{2} - 4 = 0$

$x \left(x + 2\right) - \left(2 x\right) = 5 \left(x - 2\right)$

${x}^{2} + 2 x - 2 x = 5 x - 10$

${x}^{2} - 5 x + 10 = 0$

Now we must use the expression for solving a second degree polynom:

$x = \frac{5 \pm \sqrt{{5}^{2} - 4 \cdot 10}}{2}$

$x = \frac{5 \pm \sqrt{- 15}}{2}$

$x = \frac{5 \pm \sqrt{15} i}{2}$