# How do you solve -x^2-9x-16<=-2?

Sep 2, 2017

$x \ge - 2 \mathmr{and} x \ge - 7$

#### Explanation:

$- {x}^{2} - 9 x - 16 \le - 2$

According to the law of inequality, when you multiply through with minus $\left(-\right)$ sign, the inequality sign changes..

$- {x}^{2} - 9 x - 16 \le - 2$

Multiply through by $\left(-\right)$

$- \left(- {x}^{2} - 9 x - 16\right) \le - \left(- 2\right)$

Note $- \times - = +$

${x}^{2} + 9 x + 16 \ge 2$, Note that the inequality sign changes..

Collect like terms

${x}^{2} + 9 x + 16 - 2 \ge 0$

${x}^{2} + 9 x + 14 \ge 0$

**Solving the Quadratic Equation, we have $7 \mathmr{and} 2$ as the factors..

Where $\to 7 \times 2 = 14 , 7 x + 2 x = 9 x$

Hence we have..

${x}^{2} + 9 x + 14 \ge 0 \Rightarrow {x}^{2} + 2 x + 7 x + 14 \ge 0$

${x}^{2} + 2 x + 7 x + 14 \ge 0$

$\left({x}^{2} + 2 x\right) \left(+ 7 x + 14\right) \ge 0$

$x \left(x + 2\right) + 7 \left(x + 2\right) \ge 0$

$\left(x + 2\right) \left(x + 7\right) \ge 0$

$x + 2 \ge 0 \mathmr{and} x + 7 \ge 0$

$x \ge - 2 \mathmr{and} x \ge - 7$