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

Question

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?

Answer 1

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*}

Answer 2

$$\sqrt{x^2+6x+9} = \pm(x+3)$$

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 $$S = -2x-15$$

Answer 3

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 $$-10<x<-8$$