Difference between revisions of "1959 IMO Problems/Problem 2"
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<cmath>A^2 > 2 </cmath> | <cmath>A^2 > 2 </cmath> | ||
− | + | This tells there that there is no solution for (b), since we must have <math>A^2 \ge 2</math> | |
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+ | For (c), we have <math>A = 2</math>, which means that <math>A^2 = 4</math>, so the only solution is <math> x=\frac{3}{2}</math>. Q.E.D. | ||
~flamewavelight (Expanded) | ~flamewavelight (Expanded) |
Revision as of 15:02, 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:
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
. 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 |