# How do you solve 15x^(-2)+7x^(-1)-2=0?

Nov 6, 2015

Multiply through by ${x}^{2}$ to turn it into a standard quadratic equation, and then solve with the quadratic formula

#### Explanation:

First, we multiply each side of the equation by ${x}^{2}$.

${x}^{2} \left(15 {x}^{-} 2 + 7 {x}^{-} 1 - 2\right) = 0 {x}^{2}$
$\implies 15 + 7 x - 2 {x}^{2} = 0$ (by applying the rule ${x}^{a} \cdot {x}^{b} = {x}^{a + b}$)

Rearranging the terms gives us a familiar quadratic form.

$- 2 {x}^{2} + 7 x + 15 = 0$

Finally, we apply the quadratic formula $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ where $a = - 2$, $b = 7$, and $c = 15$

This gives

$x = - \frac{3}{2}$ or $x = 5$.