Difference between revisions of "2020 AMC 10A Problems/Problem 5"

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   <li><math>(x-6)^2-2=2</math></li><p>
 
   <li><math>(x-6)^2-2=2</math></li><p>
The line <math>y=2</math> intersects the parabola <math>y=(x-6)^2-2</math> at two points. Since these two points are symmetric about the line <math>x=6,</math> the average of their <math>x</math>-coordinates is <math>6.</math> <p>
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The line <math>y=2</math> intersects the parabola <math>y=(x-6)^2-2</math> at two points, which are symmetric about the line <math>x=6.</math><p>
 
In this case, the average of the solutions is <math>6,</math> so the sum of the solutions is <math>12.</math>
 
In this case, the average of the solutions is <math>6,</math> so the sum of the solutions is <math>12.</math>
 
   <li><math>(x-6)^2-2=-2</math></li><p>
 
   <li><math>(x-6)^2-2=-2</math></li><p>

Revision as of 00:17, 18 June 2021

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, which 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, which is the vertex.

    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 1

https://youtu.be/E7zjQkZl59E

Video Solution 2

Education, The Study Of Everything

https://youtu.be/WUcbVNy2uv0

~IceMatrix

Video Solution 3

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

~bobthefam

Video Solution 4

https://youtu.be/TlIrYXcEuws

~savannahsolver

Video Solution 5

https://youtu.be/3dfbWzOfJAI?t=1544

~ pi_is_3.14

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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