2011 AMC 10B Problem 19

1. Remove the square roots. Two square roots are equal exactly when the numbers inside them are equal (and not negative). So we need
$$5|x| + 8 = x^2 - 16.$$
2. Use only $|x|$. Squaring a number gives the same result as squaring its absolute value: $x^2 = |x|^2$. For example $(-8)^2 = 8^2 = 64$. So call $u = |x|$, a number that is $0$ or more. The equation becomes
$$u^2 - 16 = 5u + 8, \quad\text{or}\quad u^2 - 5u - 24 = 0.$$
3. Factor. Look for two numbers that multiply to $-24$ and add to $-5$: they are $-8$ and $3$. So $(u - 8)(u + 3) = 0$, and $u = 8$ or $u = -3$.
4. An absolute value cannot be negative, so $u = -3$ is thrown out. That leaves $|x| = 8$, so $x = 8$ or $x = -8$.
5. Check. For $x = 8$: $5 \times 8 + 8 = 48$ and $64 - 16 = 48$. Both are equal and positive. The same numbers appear for $x = -8$. So both roots work.
6. The product of the roots is $8 \times (-8) = -64$.

The answer is $\boxed{\textbf{(A)}\ -64}$.

The equation only uses $|x|$ and $x^2$, and neither changes when you switch $x$ to $-x$. So every positive root comes with a matching negative root. Solving for $|x|$ first, then adding the $\pm$ at the end, is cleaner than splitting into "$x$ positive" and "$x$ negative" cases. Always plug roots back in after removing square roots.