Question

If $-7\le x^2+8x+5\le 14$, then the interval(s) in which $x$ lies is/are

• A. $(-8,-6)$
• B. $(-5,-2)$
• C. $(-1,1)$
• D. $(1,2)$

Answer 1

Answer: (A), (C)

The correct options are
A $(-8,-6)$
C $(-1,1)$

$-7\le x^2+8x+5\le 14$
$\Rightarrow -7\le x^2+8x+16-11\le 14$
$\Rightarrow 4\le (x+4{)}^2\le 25$

When $4\le (x+4{)}^2$
$\Rightarrow x+4\in (\infty -2]\cup [2,\infty )\Rightarrow x\in (\infty -6]\cup [-2,\infty )\cdots \left(1\right)$

When $(x+4{)}^2\le 25$
$\Rightarrow -5\le x+4\le 5$
$\Rightarrow -9\le x\le 1\Rightarrow x\in [-9,1]\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get
$x\in [-9,-6]\cup [-2,1]$

Hence, from the given options $(-1,1)$ and $(-8,-6)$ are correct.