Difference between revisions of "1959 IMO Problems/Problem 2"
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== Solution == | == Solution == | ||
− | + | The square roots imply that <math>x\ge \frac{1}{2}</math>. | |
Square both sides of the given equation: <cmath>A^2 = \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big) </cmath> | Square both sides of the given equation: <cmath>A^2 = \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big) </cmath> |
Revision as of 04:28, 17 December 2019
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 |