Difference between revisions of "1959 IMO Problems/Problem 2"
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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> | 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> | ||
− | 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> | + | 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> |
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+ | Multiply the middle terms, and use <math>a^2 - b^2 = (a + b)(a - b)</math> to get:<cmath>2x + 2 \sqrt{x^2 + 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 | ||
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Hence for (a) the solution is <math> x \in \left[ \frac{1}{2}, 1 \right]</math>, for (b) there is no solution, since we must have <math>A^2 \ge 2</math>, and for (c), the only solution is <math> x=\frac{3}{2}</math>. Q.E.D. | Hence for (a) the solution is <math> x \in \left[ \frac{1}{2}, 1 \right]</math>, for (b) there is no solution, since we must have <math>A^2 \ge 2</math>, and for (c), the only solution is <math> x=\frac{3}{2}</math>. Q.E.D. | ||
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+ | ~flamewavelight (Expanded) | ||
{{Alternate solutions}} | {{Alternate solutions}} |
Revision as of 13:31, 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
Multiply the middle terms, and use 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.
~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 |