Difference between revisions of "2020 AMC 10A Problems/Problem 5"
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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{\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 real roots. By Vieta's, the sum of the roots of the first equation is <math>12</math>. <math>12+6=\boxed{\textbf{(C) }18}</math>. | ||
| − | ==Video Solution== | + | ==Video Solution 1== |
https://youtu.be/E7zjQkZl59E | https://youtu.be/E7zjQkZl59E | ||
| + | ==Video Solution 2== | ||
Education, The Study Of Everything | Education, The Study Of Everything | ||
| Line 36: | Line 37: | ||
~IceMatrix | ~IceMatrix | ||
| + | ==Video Solution 3== | ||
https://www.youtube.com/watch?v=7-3sl1pSojc | https://www.youtube.com/watch?v=7-3sl1pSojc | ||
~bobthefam | ~bobthefam | ||
| + | ==Video Solution 4== | ||
https://youtu.be/TlIrYXcEuws | https://youtu.be/TlIrYXcEuws | ||
Revision as of 13:32, 22 December 2020
Contents
Problem
What is the sum of all real numbers 
for which 
Solution 1
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
We have the equations 
and 
.
Notice that the second is a perfect square with a double root at 
, and the first has real roots. By Vieta's, the sum of the roots of the first equation is 
. 
.
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
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. 
