Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(→‎Solution 1 (Casework and Factoring): (there was a extra negative sign in x^2-12x+34=-2. I believe it is a typo.))
 
(22 intermediate revisions by 11 users not shown)
Line 1: Line 1:
−
==Problem 5==
+
==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>
  
−
== Solution 1==  
+
== Solution 1 (Casework and Factoring)==  
 +
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
 +
 
 +
Case 1:
  
−
Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.
+
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 first case 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>.
+
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>.
+
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>.
  
−
Summing all the values results in <math>4+8+6=\boxed{\text{(C) }18}</math>.
+
Summing all the values results in <math>4+8+6=\boxed{\textbf{(C) }18}</math>.
  
−
== Solution 2==  
+
== Solution 2 (Casework and Vieta)==  
 
We have the equations <math>x^2-12x+32=0</math> and <math>x^2-12x+36=0</math>.
 
We have the equations <math>x^2-12x+32=0</math> and <math>x^2-12x+36=0</math>.
  
−
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>.
+
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>.
 +
 
 +
==Solution 3 (Casework and Graphing)==
 +
Completing the square gives
 +
<cmath>\begin{align*}
 +
\left|(x-6)^2-2\right|&=2 \\
 +
(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 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 that 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>
 +
  <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: the vertex of the parabola.<p>
 +
In this case, the only solution is <math>x=6.</math>
 +
</ol>
 +
Finally, the sum of all solutions is <math>12+6=\boxed{\textbf{(C) } 18}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Video Solution 1==
 +
 
 +
https://youtu.be/E7zjQkZl59E
 +
 
 +
==Video Solution 2==
 +
Education, The Study Of Everything
  
−
==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)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_5&oldid=183904"