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

Revision as of 12:46, 4 May 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 $(\bigstar):$

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

The line $y=2$ intersects the parabola $y=(x-6)^2-2$ at two points. Since these two points are symmetric about the line $x=6,$ the average of their $x$ -coordinates is $6.$

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, which is the vertex.

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

WILL FINISH BY TOMORROW. NO EDITING PLEASE ...

~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 28: / Line 28: @@
 (x-6)^2-2&=\pm2. \hspace{15mm}(\bigstar)
 \end{align*}</cmath>
-Note that the graph of <math>y=(x-6)^2-2</math> is an upward parabola with vertex <math>(6,-2).</math> We apply casework to <math>(\bigstar):</math>
+Note that the graph of <math>y=(x-6)^2-2</math> is an upward parabola with the vertex <math>(6,-2)</math> and the axis of symmetry <math>x=6;</math> the graphs of <math>y=\pm2</math> are horizontal lines.
+We apply casework to <math>(\bigstar):</math>
+<ol style="margin-left: 1.5em;">
+  <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>
+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>
+The line <math>y=-2</math> intersects the parabola <math>y=(x-6)^2-2</math> at one point, which is the vertex.<p>
+In this case, the only solution is <math>x=6.</math>
+</ol>
 <b>WILL FINISH BY TOMORROW. NO EDITING PLEASE ...</b>

2020 AMC 10A (Problems • Answer Key • Resources)
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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