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

I am puzzling over the following problem, which involves an equation of the form:

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

The problem involves finding real values of x corresponding to A = $\sqrt{2}$, A = 1, and A = 2, where the roots must be of non-negative real numbers.

So far, I have found that A = $\sqrt{2}$ when x = 1:

$$\sqrt{1 + \sqrt{2(1) -1}} + \sqrt{x- \sqrt{2(1)-1}} = \sqrt{1 + 1} + \sqrt{1 - 1} = \sqrt{2}$$

But it is not so obvious to me how to go about obtaining values of x when A = 1, or A = 2, for example. And how can I be certain that I have found all values of x, if there is more than one? And how can I be absolutely sure that there is only one solution?

And if we think about it the other way around- in general, for which values of A are there solutions?

Some thoughts on any of the above would be much appreciated!

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This is quite a famous question, and there are various parts you need to be careful with (e.g. what terms must be positive, can $x=0, x = 0.5$?). As a start, you could square the equation and see what you get. Lots of things will cancel. –  Calvin Lin Jun 21 '13 at 21:55
@Seraphina: Reduce the number of square roots on both side by squaring both sides, isolate the remaining radical, square again, algebra, solve. Be careful to find solutions that satisfy after all of this! –  Amzoti Jun 21 '13 at 21:59

Hint: If you square the equation, you get that

$$2x + 2 |1-x| = A^2$$

Question: Does every solution to the above equation become a solution to the original equation? Why, or why not?

Consider $x= 0$, $x=0.5$, $x=1$, $x=\sqrt{2}$, $x = \sqrt{2}$, $x = \frac{e}{\pi}$ etc...

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We look for real solutions. Let $y=2x-1$. Note that $x\ge \frac{1}{2}$. We have $x=\frac{y+1}{2}$. Then $$x+\sqrt{2x-1}=\frac{y+1}{2}+\sqrt{y}=\frac{1}{2}(1+\sqrt{y})^2.$$ Taking the square root, we find that $$\sqrt{x+\sqrt{2x-1}}=\frac{1}{\sqrt{2}}(1+\sqrt{y}).$$ Almost similarly, we find that $$\sqrt{x-\sqrt{2x-1}}=\frac{1}{\sqrt{2}}|1-\sqrt{y}|.$$ Adding, we end up with the equation $$(1+\sqrt{y})+|1-\sqrt{y}|=\sqrt{2}A.$$ If $\sqrt{y}\le 1$, we get $A=\sqrt{2}$, with no other conditions on $y$. That gives us an interval of solutions when $A=\sqrt{2}$.

If $\sqrt{y}\gt 1$, we arrive after a little manipulation at $y=\frac{A^2}{2}$. So there is no solution if $A=1$, and indeed if $A\lt \sqrt{2}$. There is a unique solution if $A\gt \sqrt{2}$.

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