Difference between revisions of "1959 IMO Problems/Problem 2"
Phoenixfire (talk | contribs) (→Solution) |
|||
Line 23: | Line 23: | ||
Since the term inside the square root is a perfect square, and by factoring 2 out, we get | Since the term inside the square root is a perfect square, and by factoring 2 out, we get | ||
<cmath>A^2 = 2(x + \sqrt{(x-1)^2})</cmath> | <cmath>A^2 = 2(x + \sqrt{(x-1)^2})</cmath> | ||
− | Use the property that <math>\sqrt{x^2}=x</math> to get | + | Use the property that <math>\sqrt{x^2}=|x|</math> to get |
<cmath>A^2 = 2(x+|x-1|)</cmath> | <cmath>A^2 = 2(x+|x-1|)</cmath> | ||
Revision as of 05:35, 22 March 2020
Problem
For what real values of is
given (a) , (b) , (c) , where only non-negative real numbers are admitted for square roots?
Solution
The square roots imply that .
Square both sides of the given equation:
Add the first and the last terms to get:
Multiply the middle terms, and use to get:
Since the term inside the square root is a perfect square, and by factoring 2 out, we get Use the property that to get
Case I: If , then , and the equation reduces to . This is precisely part (a) of the question, for which the valid interval is now
Case II: If , then and we have which simplifies to
This tells there that there is no solution for (b), since we must have
For (c), we have , which means that , so the only solution is .
~flamewavelight (Expanded)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1959 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |