# 1959 IMO Problems/Problem 2

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## Problem

For what real values of $x$ is

$\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = A,$

given (a) $A=\sqrt{2}$, (b) $A=1$, (c) $A=2$, where only non-negative real numbers are admitted for square roots?

## Solution

The square roots imply that $x\ge \frac{1}{2}$.

Square both sides of the given equation: $$A^2 = \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big)$$

Add the first and the last terms to get: $$A^2 = 2x + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}}$$

Multiply the middle terms, and use $(a + b)(a - b) = a^2 - b^2$ to get: $$A^2 = 2x + 2 \sqrt{x^2 - 2x + 1}$$

Since the term inside the square root is a perfect square, and by factoring 2 out, we get $$A^2 = 2(x + \sqrt{(x-1)^2})$$ Use the property that $\sqrt{x^2}=|x|$ to get $$A^2 = 2(x+|x-1|)$$

Case I: If $x \le 1$, then $|x-1| = 1 - x$, and the equation reduces to $A^2 = 2$. This is precisely part (a) of the question, for which the valid interval is now $x \in \left[ \frac{1}{2}, 1 \right]$

Case II: If $x > 1$, then $|x-1| = x - 1$ and we have $$x = \frac{A^2 + 2}{4} > 1$$ which simplifies to $$A^2 > 2$$

This tells there that there is no solution for (b), since we must have $A^2 \ge 2$

For (c), we have $A = 2$, which means that $A^2 = 4$, so the only solution is $x=\frac{3}{2}$.

~flamewavelight (Expanded)

~phoenixfire (edited)