2020 AMC 10A Problem 5

An absolute value equals $2$ when the number inside is $2$ or $-2$. So split into two ordinary equations.

1. Inside equals $2$. Then $x^2 - 12x + 34 = 2$, so $x^2 - 12x + 32 = 0$. We need two numbers that multiply to $32$ and add to $12$: they are $4$ and $8$. So $(x - 4)(x - 8) = 0$, which gives $x = 4$ or $x = 8$.
2. Inside equals $-2$. Then $x^2 - 12x + 34 = -2$, so $x^2 - 12x + 36 = 0$. This is a perfect square: $(x - 6)^2 = 0$. The only solution is $x = 6$.
3. Check one: $x = 4$ gives $16 - 48 + 34 = 2$. Good. $x = 6$ gives $36 - 72 + 34 = -2$, and $|-2| = 2$. Good.
4. The solutions are $4$, $6$ and $8$. Their sum is $4 + 6 + 8 = 18$.

The answer is $\boxed{\textbf{(C)}\ 18}$.

An absolute-value equation is really two equations, one for each sign. Be careful when a case turns out to be a perfect square: it gives only one number, not two. Here $x^2 - 12x + 34 = (x - 6)^2 - 2$ is never smaller than $-2$, so the second case just touches that lowest value once.

The absolute value equals $2$ when the inside is $2$ or $-2$.

Case 1: $x^2 - 12x + 34 = 2$, i.e. $x^2 - 12x + 32 = 0$, which factors as $(x-4)(x-8) = 0$. Roots $4$ and $8$.

Case 2: $x^2 - 12x + 34 = -2$, i.e. $x^2 - 12x + 36 = 0$, which is $(x-6)^2 = 0$. The single root is $6$.

The real numbers satisfying the equation are $4$, $6$, $8$, and their sum is $18$.

The answer is $\boxed{\textbf{(C)}\ 18}$.