Answer
The solution set is {$\frac{2-\sqrt{7}}{3},\frac{2+\sqrt{7}}{3}$}.
Work Step by Step
$x^{-2} - 4x^{-1} = 3$
Let $u=x^{−1}$
Rewrite the given equation as a quadratic equation in $u$ and solve for $u$.
$x^{-2} - 4x^{-1} = 3$: $u^2 - 4u = 3$
Subtract $3$ on both sides.
$u^2 - 4u-3 = 3-3$
$u^2 - 4u-3 = 0$
Solve using the quadratic formula: $u = \frac{-b±\sqrt{b^2-4ac}}{2a}$
$a=1$, $b=-4$, $c=-3$
$u = \frac{-(-4)±\sqrt{(-4)^2-(4⋅1⋅-3)}}{2⋅1}$
$u = \frac{4±\sqrt{16-(-12)}}{2}$
$u = \frac{4±\sqrt{28}}{2}$
$u = \frac{4±\sqrt{4⋅7}}{2}$
$u = \frac{4±2\sqrt{7}}{2}$
$u = 2±\sqrt{7}$
Solve for $u$.
$2±\sqrt{7} = x^{-1}$
$2±\sqrt{7} = \frac{1}{x}$
$x = \frac{1}{2±\sqrt{7}}$
$x = \frac{1}{2+\sqrt{7}}$ or $x = \frac{1}{2-\sqrt{7}}$
$x = \frac{1}{2+\sqrt{7}}⋅\frac{2-\sqrt{7}}{2-\sqrt{7}}$ or $x = \frac{1}{2-\sqrt{7}} ⋅\frac{2+\sqrt{7}}{2+\sqrt{7}}$
$x=\frac{2-\sqrt{7}}{3}$ or $x=\frac{2+\sqrt{7}}{3}$
The solution set is {$\frac{2-\sqrt{7}}{3},\frac{2+\sqrt{7}}{3}$}.
