# How do you solve the quadratic 2/(x+5)-x/(x-5)=1 using any method?

Jan 5, 2017

Stipulate that $x \ne \pm 5$. Multiply both sides by $\left(x - 5\right) \left(x + 5\right)$. Collect all the terms on one side. Factor or use the quadratic formula. Discard any restricted roots.

#### Explanation:

Stipulate that $x \ne \pm 5$:

2/(x + 5) - x/(x - 5) = 1; x !=+-5

Multiply both sides by $\left(x - 5\right) \left(x + 5\right)$:

2(x - 5) - x(x + 5) = x^2 - 25; x !=+-5

2x - 10 - x^2 - 5x = x^2 - 25; x !=+-5

Collect all of the terms to one side:

2x^2 + 3x - 15 = 0;x !=+-5

Check the discriminant:

$d = {b}^{2} - 4 \left(a\right) \left(c\right) = {3}^{2} - 4 \left(2\right) \left(- 15\right) = 129$

Drop the restriction, because, with 129 under the radical, there is no way to obtain a root equal to $\pm 5$.

$x = \frac{- b \pm \sqrt{d}}{2 a}$
$x = \frac{- 3 \pm \sqrt{129}}{2 \left(2\right)}$
$x = \left\{\begin{matrix}\frac{- 3 + \sqrt{129}}{4} \\ \frac{- 3 - \sqrt{129}}{4}\end{matrix}\right.$