Why this has a solution

Question

Given: $$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A\tag{1}$$ In general, given $A$, we can solve for $x$ as: $$a=\sqrt{2x-1}$$ $$\sqrt{x+a}+\sqrt{x-a}=A\Rightarrow2x+2\sqrt{x^2-a^2}=A^2\Rightarrow2x-A^2=-2\sqrt{x^2-a^2}$$ $$4x^2+A^4-4A^2x=4x^2-4a^2\Rightarrow A^4-4=x(4A^2-8)$$ $$x=\frac{A^4-4}{4A^2-8}=\frac14(A^2+2)\tag{2}$$ But when the original equation is set to equal $\sqrt{2}$: $$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\tag{3}$$ There is no solution for $x$, but suppose we set $A=\sqrt{2}$ in eq. $(2)$, we get: $$x=\frac14\big((\sqrt{2})^2+2\big)\Rightarrow x=1$$

How is this possible? Have I made a mistake along the way?

Answer 1

There is a little mistake in your solution in terms of $\rm A$.

You wrote $$x=\frac{A^4-4}{4A^2-8}=\frac14(A^2+2)$$

You cancelled the factor $A^2-2$, which isn't allowed, if $A=\sqrt 2$.

Answer 2

Note that $x\ge\frac{1}{2}$. \begin{align*} A^2&=x+\sqrt{2x-1}+2\sqrt{x^2-(2x-1)}+x-\sqrt{2x-1}\\ &=2x+2\sqrt{(x-1)^2} \end{align*}

If $x\ge 1$, $A^2=2x+2(x-1)=4x-2$ and hence $A=\sqrt{4x-2}$.

If $\frac{1}{2}<x<1$, $A^2=2x+2(1-x)=2$ and hence $A=\sqrt{2}$.

You missed the case when $A=\sqrt{2}$ in $(2)$.

Answer 3

Hint:

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