Solve the equation (x squared plus 4x minus 12) times the square root of (x plus 5) equals zero, and give the sum of its roots.

Step-by-step solution

1. 1

   Find the domain first, not last. A square root is only defined when its radicand is non-negative, so before solving anything:

   $x+5\ge 0 \quad\Longrightarrow\quad x\ge -5$

   Doing this first is what prevents an extraneous root from sneaking into the final sum.

2. 2

   Apply the zero-product property. A product is zero exactly when one of its factors is zero, so solve $x^2+4x-12=0$ and $\sqrt{x+5}=0$ separately. Beginners often forget the second branch, because a radical that equals zero does not look like an equation to solve.

3. 3

   Solve the quadratic factor.

   $x^2+4x-12=(x+6)(x-2)=0 \quad\Longrightarrow\quad x=-6 \ \text{ or } \ x=2$

   But $x=-6$ fails the domain $x\ge-5$, so it is rejected. Only $x=2$ survives.

4. 4

   Solve the radical factor.

   $\sqrt{x+5}=0 \quad\Longrightarrow\quad x+5=0 \quad\Longrightarrow\quad x=-5$

   This value sits exactly on the boundary of the domain, so it is legitimate: at $x=-5$ the radical is defined and equals $0$.

5. 5

   Collect the roots and add them. The solution set is $\{-5,\ 2\}$, so

   $-5+2=-3$

6. 6

   Verify both roots in the original equation. At $x=2$: $(4+8-12)\sqrt{7}=0\cdot\sqrt7=0$. At $x=-5$: $(25-20-12)\sqrt{0}=(-7)(0)=0$. Both give $0$, and $x=-6$ is not testable at all since $\sqrt{-1}$ is undefined over the reals.