Question #fc972

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Answer 1 · 2016-10-18T15:42:24

$x in [-5,-4] uu [-2,-1]$

$-2x-4<=(x+2)^2<=-2x-1$ Adding $2x$ to each inequality term

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

1)
$-4 le (x+2)^2+2x->0 le (x+2)^2+2x+4$

and

2)
$(x+2)^2+2x le -1-> (x+2)^2+2x+1 le 0$

Solving 1)

the roots for

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

are

$-4,-2$

so for

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

Solving 2)

the roots for

$(x+2)^2+2x+1=0$

are

$-5,-1$

so for

$-5 le x le-1$ we have $(x+2)^2+2x+1 le 0$

and the intersection set is

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

so the solution set is

$x in [-5,-4] uu [-2,-1]$