if $s=\sqrt{x^2+6x+9} +\sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that... 
if $s=\sqrt{x^2+6x+9} + \sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that $s=9$

After noticing that the first radical is $x+3$ and the second one is  $x+12$ all I get is $s=2x+15$, which btw. is suggested as an alternative answer in the exam.
So, how to go about this?
 A: Note that $$x^2+6x+9=(x+3)^2$$ and $$x^2+24x+144=(x+12)^2$$, so you will get
$$s=|x+3|+|x+12|$$, now consider the inequality $$-10<x<-8$$
A: It is not true that $\sqrt{x^2 + 6x + 9} = \sqrt{(x + 3)^2} = x + 3$ since $x + 3 < 0$ if $x < -3$.  Remember that $\sqrt{x}$ is the principal (nonnegative) square root of $x$.    Actually,
$$\sqrt{x^2 + 6x + 9} = |x + 3|$$
Similarly,
$$\sqrt{x^2 + 24x + 144} = |x + 12|$$
Since $x + 3 \geq 0$ if $x \geq -3$ and $x + 3 < 0$ if $x < -3$,
$$
|x + 3| = \begin{cases}
x + 3 & \text{if $x \geq -3$}\\
-x - 3 & \text{if $x < -3$}
\end{cases}
$$ 
Since $x + 12 \geq 0$ if $x \geq -12$ and $x + 12 < 0$ if $x < -12$
$$
|x + 12| = \begin{cases}
x + 12 & \text{if $x \geq -12$}\\
-x - 12 & \text{if $x < -12$}
\end{cases}
$$ 
Thus, if $-10 < x < -8$,
\begin{align*}
s & = \sqrt{x^2 + 6x + 9} + \sqrt{x^2 + 12x + 24}\\
  & = |x + 3| + |x + 12|\\
  & = -x - 3 + x + 12\\
  & = 9 
\end{align*}
A: $$\sqrt{x^2+6x+9} = \pm(x+3)$$
And 
$$\sqrt{x^2+24x+144} = \pm(x+12)$$
Hence $$S = \pm (x+3) \pm (x+12)$$
There are $4$ different cases :
$$S = 2x+15$$
$$S = -9$$
$$S = 9$$
And 
$$S = -2x-15$$
