2010 AMC 10B Problems/Problem 13
Problem
What is the sum of all the solutions of ?
Solution 1
We evaluate this in cases:
Case 1
When we are going to have . When we are going to have and when we are going to have . Therefore we have .
Subcase 1
When we are going to have . When this happens, we can express as . Therefore we get . We check if is in the domain of the numbers that we put into this subcase, and it is, since . Therefore is one possible solution.
Subcase 2
When we are going to have , therefore can be expressed in the form . We have the equation . Since is less than , is another possible solution.
Case 2 :
When , . When we can express this in the form . Therefore we have . This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have
We have now evaluated all the cases, and found the solution to be which have a sum of
Solution 2
From the equation , we have , or . Therefore, , or . From here we have four possible cases:
1. ; this simplifies to , so .
2. ; this simplifies to .
3. ; this simplifies to , so .
4. ; this simplifies to , so . However, this solution is extraneous because the absolute value of cannot be negative.
The sum of all of the solutions of is
Video Solution
https://youtu.be/vYXz4wStBUU?t=272
~IceMatrix
~savannahsolver
See Also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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