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

https://youtu.be/E7zjQkZl59E

Education, The Study Of Everything

https://youtu.be/WUcbVNy2uv0

~IceMatrix

https://www.youtube.com/watch?v=7-3sl1pSojc

~bobthefam

https://youtu.be/TlIrYXcEuws

~savannahsolver

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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