Difference between revisions of "1973 Canadian MO Problems/Problem 2"
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We can break this up into cases based upon if <math>x+3</math> and <math>x-1</math> are positive or negative. | We can break this up into cases based upon if <math>x+3</math> and <math>x-1</math> are positive or negative. |
Latest revision as of 23:07, 29 December 2015
Problem
Find all real numbers that satisfy the equation . (Note: if if .)
Solution
We can break this up into cases based upon if and are positive or negative.
In this case . Then we have .
In this case we have that . Thus, .
There are obviously no solutions here since and is a contradiction.
In this case we have . Thus, .
Thus all solutions to this are and
See also
1973 Canadian MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 | Followed by Problem 3 |