# How do you solve the inequality 5x+4<=3x^2 and write your answer in interval notation?

Dec 11, 2017

The solution is $x \in \left(- \infty , - 0.59\right] \cup \left[2.26 , + \infty\right)$

#### Explanation:

Let's rewrite the inequality

$5 x + 4 \le 3 {x}^{2}$

$3 {x}^{2} - 5 x - 4 \ge 0$

Let $f \left(x\right) = 3 {x}^{2} - 5 x - 4$

The roots of the quadratic equation $3 {x}^{2} - 5 x - 4 = 0$, are

$x = \frac{5 \pm \sqrt{{\left(- 5\right)}^{2} - \left(4\right) \cdot \left(3\right) \cdot \left(- 4\right)}}{6} = \frac{5 \pm \sqrt{73}}{6}$

${x}_{1} = \frac{5 - \sqrt{73}}{6} = - 0.59$

${x}_{2} = \frac{5 + \sqrt{73}}{6} = 2.26$

Let's 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}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$${x}_{1}$$\textcolor{w h i t e}{a a a a a a}$${x}_{2}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - {x}_{1}$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - {x}_{2}$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(white)(aaaa)-$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{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}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a}$$+$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , {x}_{1}\right] \cup \left[{x}_{2} , + \infty\right)$

graph{3x^2-5x-4 [-11.39, 11.11, -6.615, 4.635]}