Rewriting $x-2\mp\sqrt{2x-5}$ as $a^2\mp2ab+b^2$

Question

I am trying to solve this equation $$\sqrt{x-2-\sqrt{2x-5}}+\sqrt{x+2-3\sqrt{2x-5}}=\sqrt{2}$$ and so far no success. I am trying to rewrite the expressions under the radicals as $a^2\mp2ab+b^2$ but I keep getting to the same point and getting stuck.

I do it like this:

Since we have $x-2$ then $a^2+b^2$ must be equal to $x-2$. Then I have $2ab=\sqrt{2x-5}$ and from that I can express $a$ or $b$. I go with $a$ and get $a=\sqrt{2x-5}/2b$ and then I put that into $a^2+b^2$ so i can get what $a$ and $b$ will be equal to. But here i just keep getting into a bigger mess.

I'm currently trying to just raise to square both sides until I can get the answer, but it is as long and cumbersome as my previous method. Is there any method I can use to make the rewriting easier or do I keep doing it the brute force way and just go until I see an answer.

Answer 1

It's $$\sqrt{2x-4-2\sqrt{2x-5}}+\sqrt{2x+4-6\sqrt{2x-5}}=2$$ or $$\sqrt{\left(\sqrt{2x-5}-1\right)^2}+\sqrt{\left(\sqrt{2x-5}-3\right)^2}=2$$ or $$|\sqrt{2x-5}-1|+|\sqrt{2x-5}-3|=2.$$ But by the triangle inequality $$2=|\sqrt{2x-5}-1|+|\sqrt{2x-5}-3|\geq|\sqrt{2x-5}-1+3-\sqrt{2x-5}|=2.$$ Thus, $$1\leq\sqrt{2x-5}\leq3$$ or $$3\leq x\leq7.$$

Answer 2

Let $\sqrt{2x-5}=a$

For real $a, 2x-5\ge0$ and for real $x, a\ge0 ,x=\dfrac{a^2+5}2$

$$x-2-\sqrt{2x-5}=\dfrac{a^2+1}2-a=\dfrac{(a-1)^2}2$$

Similarly, $$x+2-3\sqrt{2x-5}=\dfrac{a^2+9}2-3a=\dfrac{(a-3)^2}2$$

So, we have $$2=|a-1|+|a-3|$$

For $a<1,$ $$2=1-a+3-a, a=1\text{ which is untenable}$$

For $1\le a\le3,$ $$2=a-1+3-a\text{ which is an identity}\implies1\le a^2\le9\iff\dfrac{1+5}2\le x\le\dfrac{9+5}2$$

What if $a>3?$