Difference between revisions of "1959 IMO Problems/Problem 2"
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Firstly, the square roots imply that a valid domain for x is <math>x\ge \frac{1}{2}</math>. | Firstly, the square roots imply that a valid domain for x is <math>x\ge \frac{1}{2}</math>. | ||
− | Square both sides of the given equation | + | Square both sides of the given equation: <cmath> \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big) = A^2</cmath>and simplify to obtain <cmath>A^2 = 2(x+|x-1|)</cmath> |
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+ | Add the first and the last terms to get <cmath>2x + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} = A^2</cmath> | ||
If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>. Otherwise, we have | If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>. Otherwise, we have |
Revision as of 14:28, 15 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
Firstly, the square roots imply that a valid domain for x is .
Square both sides of the given equation: and simplify to obtain
Add the first and the last terms to get
If , then we must clearly have
. Otherwise, we have
Hence for (a) the solution is , for (b) there is no solution, since we must have
, and for (c), the only solution is
. Q.E.D.
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 |