Answer
$x = \frac{-1±\sqrt{21}}{2}$
Work Step by Step
$\frac{1}{x^2 - 3x + 2} = \frac{1}{x+2} + \frac{5}{x^2-4}$
Simplify by multiplying both sides by the least common multiple $(x-2)(x+2)(x-1)$. Thus, the equation becomes:
$\frac{1}{(x-2)(x-1)}⋅ (x-2)(x+2)(x-1)= \frac{1}{x+2}⋅(x-2)(x+2)(x-1) + \frac{5}{(x+2)(x-2)}⋅(x-2)(x+2)(x-1)$
$x+2=(x-2)(x-1) + 5(x-1)$
$x+2=x^2 - 3x + 2 + 5x-5$
$x+2=x^2 + 2x-3$
$x^2 + x-5=0$
Solve by using the quadratic formula:
$a=1$, $b=1$, $c=-5$
$x = \frac{-b±\sqrt{b^2-4ac}}{2a}$
$x = \frac{-1±\sqrt{(-1)^2-(4⋅1⋅-5)}}{2(1)}$
$x = \frac{-1±\sqrt{1-(-20)}}{2}$
$x = \frac{-1±\sqrt{21}}{2}$
$x = \frac{-1±\sqrt{21}}{2}$
