# 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 (Casework and Factoring)

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 (Casework and Vieta)

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 two distinct real roots. By Vieta's, the sum of the roots of the first equation is $-(-12)$ or $12$. $12+6=\boxed{\textbf{(C) }18}$.

## Solution 3 (Casework and Graphing)

Completing the square gives \begin{align*} \left|(x-6)^2-2\right|&=2 \\ (x-6)^2-2&=\pm2. \hspace{15mm}(\bigstar) \end{align*} Note that the graph of $y=(x-6)^2-2$ is an upward parabola with the vertex $(6,-2)$ and the axis of symmetry $x=6;$ the graphs of $y=\pm2$ are horizontal lines.

We apply casework to $(\bigstar):$

1. $(x-6)^2-2=2$
2. The line $y=2$ intersects the parabola $y=(x-6)^2-2$ at two points that are symmetric about the line $x=6.$

In this case, the average of the solutions is $6,$ so the sum of the solutions is $12.$

3. $(x-6)^2-2=-2$
4. The line $y=-2$ intersects the parabola $y=(x-6)^2-2$ at one point: the vertex of the parabola.

In this case, the only solution is $x=6.$

Finally, the sum of all solutions is $12+6=\boxed{\textbf{(C) } 18}.$

~MRENTHUSIASM

## Video Solution 2

Education, The Study Of Everything

~IceMatrix

~bobthefam

~savannahsolver

## Video Solution 5

~ pi_is_3.14

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