Solve (x^2 + 4x - 12) * sqrt(x + 5) = 0 and give the sum of its roots.

Step-by-step solution

1. 1

   Set the domain first. The square root requires $x+5\ge 0$, so $x\ge -5$. This restriction is what makes the problem more than a routine factorisation: one root of the quadratic will fall outside it.

2. 2

   Split the product into two cases. The product vanishes when $x^{2}+4x-12=0$ or when $\sqrt{x+5}=0$.

3. 3

   Solve the quadratic factor. The discriminant is $D=4^{2}-4(1)(-12)=16+48=64$, so $x=\frac{-4\pm 8}{2},$ giving $x_1=2$ and $x_2=-6$. Since $-6<-5$, the value $x=-6$ is outside the domain and must be discarded; only $x=2$ remains.

4. 4

   Solve the radical factor. $\sqrt{x+5}=0$ forces $x+5=0$, so $x=-5$, which is exactly the left endpoint of the domain and therefore admissible.

5. 5

   Add the surviving roots. The solution set is $\{-5,2\}$, so the sum is $-5+2=-3$. Checking: at $x=2$ the quadratic factor is $4+8-12=0$, and at $x=-5$ the radical is $0$, so both make the product vanish.