# How do you solve -x ^ { 2} - 7x - 15\leq - 5?

Jun 2, 2018

$x \le - 5$ or $x \ge - 2$

#### Explanation:

Writing
$- {x}^{2} - 7 x - 15 \le - 5$

$- {x}^{2} - 7 x - 10 \le 0$
${x}^{2} + 7 x + 10 \ge 0$
so

${\left(x + \frac{7}{2}\right)}^{2} - \frac{9}{4} \ge 0$
$\left(x + \frac{7}{2} - \frac{3}{2}\right) \left(x + \frac{7}{2} + \frac{3}{2}\right) \ge 0$
$\left(x + 2\right) \left(x + 5\right) \ge 0$
so
$x \ge - 2$ or $x \le - 5$

Jun 2, 2018

The solutions are $x \in \left(- \infty , - 5\right] \cup \left[- 2 , + \infty\right)$

#### Explanation:

The inequality is

$- {x}^{2} - 7 x - 15 \le - 5$

${x}^{2} + 7 x - 5 + 15 \ge 0$

${x}^{2} + 7 x + 10 \ge 0$

$\left(x + 5\right) \left(x + 2\right) \ge 0$

Let $f \left(x\right) = \left(x + 5\right) \left(x + 2\right)$

Build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- 5$$\textcolor{w h i t e}{a a a a a a}$$- 2$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 5$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x + 2$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$color(white)(aaaaa)-$\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}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a 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 $x \in \left(- \infty , - 5\right] \cup \left[- 2 , + \infty\right)$

graph{x^2+7x+10 [-12.875, 7.125, -3.24, 6.76]}