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

Revision as of 22:30, 6 October 2022

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 $x^2-12x+36=2$ , 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 $-x^2+12x+36=-2$ . 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 $(\bigstar):$

$(x-6)^2-2=2$

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 $12.$

$(x-6)^2-2=-2$

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 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

@@ Line 9: / Line 9: @@
 Case 1:
-The equation yields <math>x^2-12x+34=2</math>, which is equal to <math>(x-4)(x-8)=0</math>. Therefore, the two values for the positive case is <math>4</math> and <math>8</math>.
+The equation yields <math>x^2-12x+36=2</math>, which is equal to <math>(x-4)(x-8)=0</math>. Therefore, the two values for the positive case is <math>4</math> and <math>8</math>.
 Case 2:
-Similarly, taking the nonpositive case for the value inside the absolute value notation yields <math>-x^2+12x+36=0</math>. Factoring and simplifying gives <math>(x-6)^2=0</math>, so the only value for this case is <math>6</math>.
+Similarly, taking the nonpositive case for the value inside the absolute value notation yields <math>-x^2+12x+36=-2</math>. Factoring and simplifying gives <math>(x-6)^2=0</math>, so the only value for this case is <math>6</math>.
 Summing all the values results in <math>4+8+6=\boxed{\textbf{(C) }18}</math>.

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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