# How do you solve x^3-2x>1?

May 10, 2015

First subtract 1 from both sides to get ${x}^{3} - 2 x - 1 > 0$.

Notice that $x = - 1$ is a solution of ${x}^{3} - 2 x - 1 = 0$, so $\left(x + 1\right)$ is a factor of ${x}^{3} - 2 x - 1$. The other factor is ${x}^{2} - x - 1$.

${x}^{3} - 2 x - 1 = \left({x}^{2} - x - 1\right) \left(x + 1\right)$.

Solve ${x}^{2} - x - 1 = 0$ using the quadratic formula

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot \left(- 1\right)}}{2 \cdot 1}$

$= \frac{1 \pm \sqrt{5}}{2}$

So ${x}^{3} - 2 x - 1$ will cross the $x$-axis at $x = - 1$, $x = \frac{1 - \sqrt{5}}{2}$ and $x = \frac{1 + \sqrt{5}}{2}$.

Since the dominant term is ${x}^{3}$, the function ${x}^{3} - 2 x - 1$ will be negative for $x < - 1$, positive for $- 1 < x < \frac{1 - \sqrt{5}}{2}$, negative for $\frac{1 - \sqrt{5}}{2} < x < \frac{1 + \sqrt{5}}{2}$ and positive for $x > \frac{1 + \sqrt{5}}{2}$.

So the solution to the original problem is:

$- 1 < x < \frac{1 - \sqrt{5}}{2}$ or $x > \frac{1 + \sqrt{5}}{2}$.