How do you solve and graph x^3>2x^2+x?

May 31, 2018

The answers are: $1 - \sqrt{2} < x < 0$ and $x > 1 + \sqrt{2}$

Explanation:

${x}^{3} > 2 {x}^{2} + x$

${x}^{3} - 2 {x}^{2} - x > 0$

$x \left({x}^{2} - 2 x - 1\right) > 0$

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

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

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

$x = \frac{2 \pm \sqrt{8}}{2}$

$x = \frac{2 \pm 2 \sqrt{2}}{2}$

$x = 1 \pm \sqrt{2}$

So:
$x \left({x}^{2} - 2 x - 1\right) > 0$

$x \left(x - \left(1 + \sqrt{2}\right)\right) \left(x - \left(1 - \sqrt{2}\right)\right) = 0$ is the graph below

graph{x(x-(1+sqrt2))(x-(1-sqrt2)) [-10, 10, -5, 5]}

$x \left(x - \left(1 + \sqrt{2}\right)\right) \left(x - \left(1 - \sqrt{2}\right)\right) > 0$ means that you need to find the parts of the graph that is above the line $y = 0$

Therefore, the answers are: $1 - \sqrt{2} < x < 0$ and $x > 1 + \sqrt{2}$