Difference between revisions of "1963 IMO Problems/Problem 1"
(New page: ==Problem== Find all real roots of the equation <center><math>\sqrt{x^2-p}+2\sqrt{x^2-1}=x</math>,</center>where <math>p</math> is a real parameter. ==Solution== {{solution}} ==See Also...) |
Brut3Forc3 (talk | contribs) (→Solution: --Woot Team) |
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==Solution== | ==Solution== | ||
− | {{solution}} | + | Assuming <math>x \geq 0</math>, square the equation, obtaining <math>4\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2</math>. If we have <math>p + 4 \geq 4x^2</math>, we can square again, obtaining <math>x^2 = \frac {(p - 4)^2}{4(4 - 2p)}</math>, or <math>x = \pm\frac {p - 4}{2\sqrt {4 - 2p}}</math>. We must have <math>4 - 2p > 0 \iff p < 2</math>, so we have <math>x = \frac {4 - p}{2\sqrt {4 - 2p}}</math>. However, this is only a solution when <math>p + 4 \geq 4x^2 = \frac {(p - 4)^2}{4 - 2p} \iff (p + 4)(4 - 2p)\leq(p - 4)^2 \iff 0\leq p(3p - 4)</math>, so we have <math>p\leq 0</math> or <math>p \geq \frac {4}{3}</math>. But if <math>p < 0</math>, then <math>\sqrt {x^2 - p} > x</math>, contradiction. So we have <math>x = \frac {4 - p}{2\sqrt {4 - 2p}}</math> for <math>p = 0, \frac {4}{3}\leq p < 2</math>. |
==See Also== | ==See Also== | ||
{{IMO box|year=1963|before=First Question|num-a=2}} | {{IMO box|year=1963|before=First Question|num-a=2}} |
Revision as of 21:05, 16 February 2009
Problem
Find all real roots of the equation
where is a real parameter.
Solution
Assuming , square the equation, obtaining . If we have , we can square again, obtaining , or . We must have , so we have . However, this is only a solution when , so we have or . But if , then , contradiction. So we have for .
See Also
1963 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |