# How do you solve 2/(x-2) + 7/(x^2-4) = 5/x?

May 23, 2016

$x = 5 \text{ and } - \frac{4}{3}$

#### Explanation:

Known: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

So ${x}^{2} - 4 \to {x}^{2} - {2}^{2} \to \left(x - 2\right) \left(x + 2\right)$

Write the given equation as:

$\frac{2}{x - 2} + \frac{7}{\left(x - 2\right) \left(x + 2\right)} = \frac{5}{x}$

$\implies \frac{2 \left(x + 2\right) + 7}{\left(x - 2\right) \left(x + 2\right)} = \frac{5}{x}$

$\implies \frac{2 x + 4 + 7}{\left(x - 2\right) \left(x + 2\right)} = \frac{5}{x}$

Cross multiplying

$x \left(2 x + 11\right) = 5 \left(x - 2\right) \left(x + 2\right)$

$x \left(2 x + 11\right) = 5 \left({x}^{2} - 4\right)$

$2 {x}^{2} + 11 x = 5 {x}^{2} - 20$

$3 {x}^{2} - 11 x - 20 = 0$

not all the solutions are whole numbers

Using: $y = a {x}^{2} + b x + c$ where a=3; b= -11; c=-20

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\implies x = \frac{11 \pm \sqrt{{11}^{2} - 4 \left(3\right) \left(- 20\right)}}{2 \left(3\right)}$

$x = \frac{11 \pm \sqrt{121 + 240}}{6}$

$x = \frac{11 \pm 19}{6}$

$x = 5 \text{ and } - \frac{4}{3}$