## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$$x=-\frac{4+\sqrt{19}}{3},\:x=\frac{\sqrt{19}-4}{3}$$
Solving using the quadratic formula, we find: $$\frac{1}{x^2}x^2-3x^2=\frac{8}{x}x^2\\ 1-3x^2=8x \\ -3x^2-8x+1=0\\ x_{1,\:2}=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\left(-3\right)1}}{2\left(-3\right)}\\ x=-\frac{4+\sqrt{19}}{3},\:x=\frac{\sqrt{19}-4}{3}$$