<p>The number of real roots of the equation \((x+1)^2 - 5|x+1| + 6 = 0\) is: [JEE Main 2021]</p>
Step-by-Step Solution
Key Concept: Let t = |x+1| \geq 0. Then t^2 - 5t + 6 = 0 \to (t-2)(t-3) = 0 \to t = 2 or t = 3. Each gives two values of x (\pmshift). Total 4 roots.
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 $t=|x+1|\geq0$: $t^2-5t+6=0\Rightarrow(t-2)(t-3)=0\Rightarrow t=2$ or $t=3$. $|x+1|=2\Rightarrow x=1$ or $x=-3$. $|x+1|=3\Rightarrow x=2$ or $x=-4$. Total: 4 real roots. Now, we invoke the power of that idea, simplify patiently, and then check that the final answer really fits the original problem.
Correct Answer: D