<p>The sum of all real roots of \(|x-2|^2+|x-2|-2=0\) is: [JEE Main 2023]</p>
Step-by-Step Solution
Key Concept: Let t = |x-2| \geq 0. Then t^2+t-2 = (t+2)(t-1) = 0 \to t = 1 (t = -2 rejected). |x-2| = 1 \to x = 3 or x = 1. Sum = 4.
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 $t=|x-2|\geq0$: $t^2+t-2=(t+2)(t-1)=0\Rightarrow t=1$. $|x-2|=1\Rightarrow x=3$ or $x=1$. Sum $=3+1=4$. Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.
Correct Answer: 3