Animated Solution for Mathematics - Quadratic Equations: The sum of the squares of the roots of $|x+2{|}^2+|x-2|-2=0$ and the squares of the roots of $x^2-2|x-3|-5=0$, is

Welcome, future engineer. Today, we are not just solving an algebra problem; we are learning the art of mathematical discipline. In the JEE Advanced arena, problems involving absolute values are designed to test your patience and your ability to handle logical branches.

Let us dissect this problem, which asks us to find the sum of the squares of the roots of two distinct equations. Take a deep breath, and let us begin.

Look at the first equation: $|x-2{|}^2+|x-2|-2=0$. Many students immediately try to open the modulus by considering cases for $x-2$. While that works, it is the long, winding road.

Notice the structure: we have a term $|x-2|$ appearing as a square and as a linear term. This is a classic quadratic pattern in disguise. Let us introduce a substitution: $t=|x-2|$.

Suddenly, the equation transforms into the beautiful, simple quadratic:

Factoring this is trivial: $(t+2)(t-1)=0$. This gives us $t=-2$ or $t=1$. But here is where the discipline comes in.

We defined $t=|x-2|$. Since an absolute value represents a distance, it must be non-negative ($t\ge 0$). Thus, we must ruthlessly reject $t=-2$.

We are left with $t=1$, which implies $|x-2|=1$. This splits into two simple linear equations: $x-2=1$ and $x-2=-1$, yielding $x=3$ and $x=1$. The sum of their squares is $1^2+3^2=10$. Keep this number safe; we will need it later.

Now, we turn to the second equation: $x^2-2|x-3|-5=0$. Here, substitution fails because of the $x^2$ term. We must confront the modulus directly.

The expression $|x-3|$ changes its behavior at the critical point $x=3$. This is our fork in the road. We must analyze two distinct universes:

1. The Universe where $x\ge 3$: Here, $|x-3|=x-3$. The equation becomes $x^2-2(x-3)-5=0$. Expanding this, we get $x^2-2x+6-5=0$, which simplifies to:

This is a perfect square: $(x-1{)}^2=0$, giving $x=1$. But wait! We assumed $x\ge 3$. Since $1<3$, this root is a ghost—it does not exist in this universe. We reject it.

2. The Universe where $x<3$: Here, $|x-3|=-(x-3)$. The equation becomes $x^2-2(-(x-3))-5=0$. This simplifies to:

We have arrived at $x^2+2x-11=0$. We could use the quadratic formula to find the roots, but why do the heavy lifting when we can use the elegance of Vieta's formulas?

Let the roots be $\alpha$ and $\beta$. We know that $\alpha +\beta =-2$ and $\alpha \beta =-11$. The problem asks for the sum of the squares of the roots, which is ${\alpha }^2+{\beta }^2$.

Using the algebraic identity ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$, we substitute our values:

We have conquered both parts of this problem. From the first equation, we obtained a sum of squares equal to $10$. From the second equation, we obtained a sum of squares equal to $26$.

Adding these together, we get:

This problem was not just about finding roots; it was about managing constraints, recognizing patterns, and choosing the most efficient path to the solution. You have done well. Keep this logical rigor in your toolkit, and no JEE problem will ever be able to stand against you. The final answer is 36.