## Intermediate Algebra for College Students (7th Edition)

The solution set is {$\frac{2-\sqrt{7}}{3},\frac{2+\sqrt{7}}{3}$}.
$x^{-2} - 4x^{-1} = 3$ 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 {$\frac{2-\sqrt{7}}{3},\frac{2+\sqrt{7}}{3}$}.