# How do you solve x^3 + 4x^2 - x using the quadratic formula?

Oct 25, 2015

$x = \left\{0 , - 2 \pm \sqrt{5}\right\}$

#### Explanation:

${x}^{3} + 4 {x}^{2} - x = 0$ is a cubic equation. You first have to factor $x$ out.

$\left[1\right] \text{ } {x}^{3} + 4 {x}^{2} - x = 0$

$\left[2\right] \text{ } x \left({x}^{2} + 4 x - 1\right) = 0$

The first root is $x = 0$ (from the $x$ you factored out). You can use the quadratic formula to find the other two roots from ${x}^{2} + 4 x - 1$.

$a = 1$
$b = 4$
$c = - 1$

$\left[3\right] \text{ } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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

$\left[5\right] \text{ } x = \frac{- 4 \pm \sqrt{20}}{2}$

$\left[6\right] \text{ } x = \frac{- 4 \pm 2 \sqrt{5}}{2}$

$\left[7\right] \text{ } \textcolor{b l u e}{x = - 2 \pm \sqrt{5}}$

So the roots of the equation are:

$x = \left\{0 , - 2 \pm \sqrt{5}\right\}$