Roots addition.

Question

$$\sqrt{\frac{a+x^2}{x}-2\sqrt{a}}+\sqrt{\frac{a+x^2}{x}+2\sqrt{a}}=Q $$ One is expected to find $Q$ respecting $a>0$, $x>\sqrt{a}$ .

I'd like to have my solution checked; namely the correct answer is $2\sqrt{x}$ but I simply fail to see where I made a mistake.
And it'd be nice to hear if there is any smoother solution, or any other way to solve this.

I introduced two subsequent substitutions.

1. $\frac{a+x^2}{x}=A, \;\;\;2\sqrt{a}=B \; \Rightarrow \; \sqrt{A-B}+\sqrt{A+B}=Q$
2. $\sqrt{A-B}=k, \;\;\; \sqrt{A+B}=n$

Now I have: $ \;\;\;$ $k+n=Q, \;\;\; k^2+n^2=2A, \;\;\; kn=\sqrt{A^2-B^2}$
$$k^2+n^2+2kn=\left ( k+n \right )^2=Q^2$$ $$Q^2=2A+2\sqrt{A^2-B^2}=\frac{2(a+x^2)}{x}+2\sqrt{\frac{\left (a+x^2 \right )^2}{x^2}-\left ( 2\sqrt{a} \right )^2}=$$ $$=\frac{2(a+x^2)}{x}+2\sqrt{\frac{a^2+2ax^2+x^4-4ax^2}{x^2}}=$$ $$=\frac{2(a+x^2)}{x}+2\sqrt{\frac{a^2-2ax^2+x^4}{x^2}}=$$ $$=\frac{2(a+x^2)}{x}+2\sqrt{\frac{\left ( a-x^2 \right )^2}{x^2}}=\frac{2(a+x^2)}{x}+\frac{2\left (a-x^2 \right)}{x}=\frac{4a}{x}$$ And as a result: $$Q=2\sqrt{\frac{a}{x}}$$

Answer 1

Since we have $x\gt \sqrt a\gt 0$, we have $$x^2\gt a\Rightarrow a-x^2\lt 0.$$ Hence, note that we have $$\sqrt{\frac{(a-x^2)^2}{x^2}}=\frac{\sqrt{(a-x^2)^2}}{\sqrt{x^2}}=\frac{|a-x^2|}{|x|}=\frac{\color{red}{-}(a-x^2)}{x}.$$

Answer 2

Set $\sqrt{a}=b$ so the first summand is $$ \sqrt{\frac{b^2+x^2}{x}-2b}= \sqrt{\frac{(x-b)^2}{x}}=\frac{x-b}{\sqrt{x}} $$ because, by assumption, $x>b$ (and so also positive).

Similarly $$ \sqrt{\frac{b^2+x^2}{x}+2b}= \sqrt{\frac{(x+b)^2}{x}}=\frac{x+b}{\sqrt{x}} $$

Thus your expression is $$ Q=\frac{x-b}{\sqrt{x}}+\frac{x+b}{\sqrt{x}}= \frac{2x}{\sqrt{x}}=2\sqrt{x} $$ Your error is in $$ \sqrt{\frac{(a-x^2)^2}{x^2}}=\frac{a-x^2}{x} $$ because the right-hand side is positive, while the right-hand side is negative.