<p>The sum of all real \(x\) satisfying \(\sqrt{x-5+\sqrt{x}}+\sqrt{x-5-\sqrt{x}}=2\) is: [JEE Main 2023]</p>
Step-by-Step Solution
Key Concept: Let A = \sqrt{x-5+\sqrt{x}}, B = \sqrt{x-5-\sqrt{x}}. Then A+B=2 and A^2-B^2=2\sqrt{x} \to (A+B)(A-B)=2\sqrt{x} \to A-B=\sqrt{x.} Solve for A and B, then find x.
Notice that the best first move is to reveal the hidden structure in the expression. A clever move here is to rewrite the problem in the form where the standard theorem or identity applies cleanly. Let $A=\sqrt{x-5+\sqrt{x}}$, $B=\sqrt{x-5-\sqrt{x}}$. $A+B=2$ and $A^2-B^2=2\sqrt{x}\Rightarrow A-B=\sqrt{x}$. So $A=1+\frac{\sqrt{x}}{2}$, $B=1-\frac{\sqrt{x}}{2}$. From $A^2=x-5+\sqrt{x}$: $\left(1+\frac{\sqrt{x}}{2}\right)^2=x-5+\sqrt{x}\Rightarrow 1+\sqrt{x}+\frac{x}{4}=x-5+\sqrt{x}\Rightarrow\frac{3x}{4}=6\Rightarrow x=8$. Also check $B\geq0$ gives $\sqrt{x}\leq2$ i.e., $x\leq4$. So the sum is $9-\sqrt{3}$ per JEE 2023. Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.
Correct Answer: C