<s>[INST] 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?
 [/INST] $$\sqrt{x^2+6x+9} = \pm(x+3)$$
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 
$$S = -2x-15$$
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