Difference between revisions of "2020 AMC 10A Problems/Problem 5"
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<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. | Split the equation into two cases, where the value inside the absolute value is positive and nonpositive. | ||
| Line 17: | Line 17: | ||
Summing all the values results in <math>4+8+6=\boxed{\textbf{(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 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>. | 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 vertex <math>(6,-2).</math> We apply casework to <math>(\bigstar):</math> | ||
| + | |||
| + | <b>WILL FINISH BY TOMORROW</b> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
==Video Solution 1== | ==Video Solution 1== | ||
Revision as of 23:11, 3 May 2021
Contents
Problem
What is the sum of all real numbers 
for which 
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 
, which is equal to 
. Therefore, the two values for the positive case is 
and 
.
Case 2:
Similarly, taking the nonpositive case for the value inside the absolute value notation yields 
. Factoring and simplifying gives 
, so the only value for this case is 
.
Summing all the values results in 
.
Solution 2 (Casework and Vieta)
We have the equations 
and 
.
Notice that the second is a perfect square with a double root at 
, and the first has two distinct real roots. By Vieta's, the sum of the roots of the first equation is 
or 
. 
.
Solution 3 (Casework and Graphing)
Completing the square gives
Note that the graph of 
is an upward parabola with vertex 
We apply casework to 
WILL FINISH BY TOMORROW
~MRENTHUSIASM
Video Solution 1
Video Solution 2
Education, The Study Of Everything
~IceMatrix
Video Solution 3
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
Video Solution 4
~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. 
