# How do you solve x^3<=4x^2+3x using a sign chart?

Feb 9, 2017

The answer is x in ]-oo,2-sqrt7]uu [0,2+sqrt7]

#### Explanation:

Let's rewrite and factorise the inequality

${x}^{3} - 4 {x}^{2} - 3 x \le 0$

x(x^2-4x-3))<=0

We calculate the roots of the equation ${x}^{2} - 4 x - 3 = 0$

$\Delta = {\left(- 4\right)}^{2} - 4 \cdot 1 \cdot \left(- 3\right) = 16 + 12 = 28$

${x}_{1} = \frac{4 + \sqrt{28}}{2} = \frac{4 + 2 \sqrt{7}}{2} = 2 + \sqrt{7}$

${x}_{2} = \frac{4 - \sqrt{28}}{2} = \frac{4 - 2 \sqrt{7}}{2} = 2 - \sqrt{7}$

Let $f \left(x\right) = x \left(x - \left(2 + \sqrt{7}\right)\right) \left(x - \left(2 - \sqrt{7}\right)\right)$

Now, we build the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$2 - \sqrt{7}$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$2 + \sqrt{7}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - {x}_{2}$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - {x}_{1}$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$+$

Therefore,

$f \left(x\right) \le 0$ when x in ]-oo,2-sqrt7]uu [0,2+sqrt7]