# Question #fc972

Oct 18, 2016

$x \in \left[- 5 , - 4\right] \cup \left[- 2 , - 1\right]$

#### Explanation:

$- 2 x - 4 \le {\left(x + 2\right)}^{2} \le - 2 x - 1$ Adding $2 x$ to each inequality term

$- 4 \le {\left(x + 2\right)}^{2} + 2 x \le - 1$ so the solution set is for all $x$ that simultaneously

1)
$- 4 \le {\left(x + 2\right)}^{2} + 2 x \to 0 \le {\left(x + 2\right)}^{2} + 2 x + 4$

and

2)
${\left(x + 2\right)}^{2} + 2 x \le - 1 \to {\left(x + 2\right)}^{2} + 2 x + 1 \le 0$

Solving 1)

the roots for

${x}^{2} + 6 x + 8 = 0$

are

$- 4 , - 2$

so for

$x \le - 4$ and $x \ge - 2$ we have ${x}^{2} + 6 x + 8 \ge 0$

Solving 2)

the roots for

${\left(x + 2\right)}^{2} + 2 x + 1 = 0$

are

$- 5 , - 1$

so for

$- 5 \le x \le - 1$ we have ${\left(x + 2\right)}^{2} + 2 x + 1 \le 0$

and the intersection set is

$- 5 \le x \le - 4$ and $- 2 \le x \le - 1$

so the solution set is

$x \in \left[- 5 , - 4\right] \cup \left[- 2 , - 1\right]$