Answer
The solution set is {${-5,-4}$}.
Work Step by Step
$20x^{-2} + 9x^{-1} + 1 = 0$
Let $u=x^{-1}$
Rewrite the given equation as a quadratic equation in $u$ and solve for $u$.
$20u^2 + 9u + 1 = 0$
Get the factors.
$20u^2 + 9u + 1 = 0$: $(5u+1)(4u+1)=0$
Thus,
$5u+1=0$ or $4u+1=0$
$u=-\frac{1}{5}$ or $u=-\frac{1}{4}$
Solve for $u$
$-\frac{1}{5}=x^{-1}$ and $-\frac{1}{4}=x^{-1}$
$-\frac{1}{5}=\frac{1}{x}$ and $-\frac{1}{4}=\frac{1}{x}$
Cross-multiply.
$x=-5$ and $x=-4$
The solution set is {${-5,-4}$}.
