# 2020 AMC 10A Problems/Problem 5

## Problem

What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$ $\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$

## Solution 1

Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.

Case 1:

The equation yields $x^2-12x+34=2$, which is equal to $(x-4)(x-8)=0$. Therefore, the two values for the positive case is $4$ and $8$.

Case 2:

Similarly, taking the nonpositive case for the value inside the absolute value notation yields $-x^2+12x-34=2$. Factoring and simplifying gives $(x-6)^2=0$, so the only value for this case is $6$.

Summing all the values results in $4+8+6=\boxed{\textbf{(C) }18}$.

## Solution 2

We have the equations $x^2-12x+32=0$ and $x^2-12x+36=0$.

Notice that the second is a perfect square with a double root at $x=6$, and the first has real roots. By Vieta's, the sum of the roots of the first equation is $12$. $12+6=\boxed{\textbf{(C) }18}$.

## Video Solution

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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 