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: <cmath> \Big( x + \sqrt{2x - 1}\Big) | + | 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> | ||
− | 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> | ||
− | Multiply the middle terms, and use <math>(a + b)(a - b) = a^2 - b^2</math> to get:< | + | Multiply the middle terms, and use <math>(a + b)(a - b) = a^2 - b^2</math> to get: |
+ | <math>2x + 2 \sqrt{x^2 - 2x + 1} = A^2<cmath> | ||
− | Since the term inside the square root is a perfect square, and by factoring 2 out, we get <cmath>A^2 = 2(x+|x-1|)</ | + | Since the term inside the square root is a perfect square, and by factoring 2 out, we get |
+ | </cmath>2(x + \sqrt{(x-1)^2}) = A^2<cmath> | ||
+ | Use the property that </cmath>\sqrt{x^2}=|x|<cmath> to get | ||
+ | </cmath>A^2 = 2(x+|x-1|)</math><math> | ||
− | If <math>x \le 1< | + | If </math>x \le 1<math>, then we must clearly have </math>A^2 =2<math>. Otherwise, we have |
<cmath>x = \frac{A^2 + 2}{4} > 1,</cmath> | <cmath>x = \frac{A^2 + 2}{4} > 1,</cmath> | ||
<cmath>A^2 > 2 </cmath> | <cmath>A^2 > 2 </cmath> | ||
− | Hence for (a) the solution is <math> x \in \left[ \frac{1}{2}, 1 \right]< | + | 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}$. Q.E.D. |
~flamewavelight (Expanded) | ~flamewavelight (Expanded) |
Revision as of 14:41, 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:
$2x + 2 \sqrt{x^2 - 2x + 1} = A^2<cmath>
Since the term inside the square root is a perfect square, and by factoring 2 out, we get
</cmath>2(x + \sqrt{(x-1)^2}) = A^2<cmath>
Use the property that </cmath>\sqrt{x^2}=|x|<cmath> to get
</cmath>A^2 = 2(x+|x-1|)$ (Error compiling LaTeX. Unknown error_msg)x \le 1
A^2 =2$. Otherwise, we have
<cmath>x = \frac{A^2 + 2}{4} > 1,</cmath> <cmath>A^2 > 2 </cmath>
Hence for (a) the solution is$ (Error compiling LaTeX. Unknown error_msg) x \in \left[ \frac{1}{2}, 1 \right]A^2 \ge 2
x=\frac{3}{2}$. 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 |