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

(→‎Solution 1)
(→‎Solution 1 (Casework and Factoring): (there was a extra negative sign in x^2-12x+34=-2. I believe it is a typo.))
 
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==Problem 5==
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==Problem==
 
What is the sum of all real numbers <math>x</math> for which <math>|x^2-12x+34|=2?</math>
 
What is the sum of all real numbers <math>x</math> for which <math>|x^2-12x+34|=2?</math>
  
 
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25</math>
 
<math>\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25</math>
  
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== Solution 1==  
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== Solution 1 (Casework and Factoring)==  
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Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
 
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
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Case 1:
 
Case 1:
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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+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>.
  
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Case 2:
 +
 +
Similarly, taking the nonpositive case for the value inside the absolute value notation yields <math>x^2-12x+34=-2</math>. Factoring and simplifying gives <math>(x-6)^2=0</math>, so the only value for this case is <math>6</math>.
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 +
Summing all the values results in <math>4+8+6=\boxed{\textbf{(C) }18}</math>.
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== Solution 2 (Casework and Vieta)==
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We have the equations <math>x^2-12x+32=0</math> and <math>x^2-12x+36=0</math>.
 +
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Notice that the second is a perfect square with a double root at <math>x=6</math>, and the first has two distinct real roots. By Vieta's, the sum of the roots of the first equation is <math>-(-12)</math> or <math>12</math>. <math>12+6=\boxed{\textbf{(C) }18}</math>.
  
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Case 2:
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==Solution 3 (Casework and Graphing)==
 +
Completing the square gives
 +
<cmath>\begin{align*}
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\left|(x-6)^2-2\right|&=2 \\
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(x-6)^2-2&=\pm2. \hspace{15mm}(\bigstar)
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\end{align*}</cmath>
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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.
  
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Similarly, taking the nonpositive case for the value inside the absolute value notation yields <math>-x^2+12x-34=2</math>. Factoring and simplifying gives <math>(x-6)^2=0</math>, so the only value for this case is <math>6</math>.
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We apply casework to <math>(\bigstar):</math>
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<ol style="margin-left: 1.5em;">
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  <li><math>(x-6)^2-2=2</math></li><p>
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The line <math>y=2</math> intersects the parabola <math>y=(x-6)^2-2</math> at two points that are symmetric about the line <math>x=6.</math><p>
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In this case, the average of the solutions is <math>6,</math> so the sum of the solutions is <math>12.</math>
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  <li><math>(x-6)^2-2=-2</math></li><p>
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The line <math>y=-2</math> intersects the parabola <math>y=(x-6)^2-2</math> at one point: the vertex of the parabola.<p>
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In this case, the only solution is <math>x=6.</math>
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</ol>
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Finally, the sum of all solutions is <math>12+6=\boxed{\textbf{(C) } 18}.</math>
  
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~MRENTHUSIASM
  
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Summing all the values results in <math>4+8+6=\boxed{\text{(C) }18}</math>.
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==Video Solution 1==
  
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== Solution 2==
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https://youtu.be/E7zjQkZl59E
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We have the equations <math>x^2-12x+32=0</math> and <math>x^2-12x+36=0</math>.
 
  
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Notice that the second is a perfect square with a double root at <math>x=6</math>, and the first has real roots. By Vieta's, the sum of the roots of the first equation is <math>12</math>. <math>12+6=\boxed{\text{(C) }18}</math>.
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==Video Solution 2==
 +
Education, The Study Of Everything
  
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==Video Solution==
 
 
https://youtu.be/WUcbVNy2uv0
 
https://youtu.be/WUcbVNy2uv0
  
 
~IceMatrix
 
~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==
 
==See Also==

Latest revision as of 17:38, 12 December 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 $x^2-12x+34=2$, 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 $12.$

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

    3. 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

See Also

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

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