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

Question

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!

Answer 1

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...

Answer 2

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 $A\gt \sqrt{2}$.