2011 AMC 10B Problem 19

Both sides are nonnegative square roots, so the equation holds exactly when the radicands are equal and nonnegative:
$$5|x| + 8 = x^2 - 16.$$

Since $x^2 = |x|^2$, substitute $u = |x| \ge 0$:
$$u^2 - 5u - 24 = 0 \quad\Longrightarrow\quad (u - 8)(u + 3) = 0.$$
The root $u = -3$ is impossible for an absolute value, so $|x| = 8$ and $x = \pm 8$.

Check: $5 \cdot 8 + 8 = 48$ and $64 - 16 = 48$, so both roots are genuine.

The product of the roots is $8 \cdot (-8) = -64$, so the answer is $\boxed{\textbf{(A)}\ -64}$.

Because the equation depends on $x$ only through $|x|$ and $x^2$, it is symmetric under $x \to -x$; solving for $|x|$ and then unfolding the sign is cleaner than splitting into $x \ge 0$ and $x < 0$ cases. After squaring, always confirm the radicands are nonnegative, though here $u \ge 0$ handles it automatically.

• Applying Vieta to $x^2 - 5x - 24 = 0$ and answering $-24$, ignoring that the equation involves $|x|$ and has roots symmetric about $0$.
• Keeping $u = -3$ as a root and getting $576$ or $24$ from products involving $\pm 3$.