Animated Solution for Mathematics - Quadratic Equations: Let p, q and r be real numbers ($p\ne q,r\ne 0$), such that the roots of the equation $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

Welcome, my dear student. Today, we are not just solving an equation; we are embarking on a journey into the heart of algebraic symmetry.

When you first look at the equation $\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$, it might seem like a daunting rational expression. It is easy to feel intimidated by the variables $p$, $q$, and $r$ floating around.

But I want you to take a deep breath. In the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

Many students rush to solve for $x$ immediately. They try to isolate $x$, and they get lost in a sea of fractions. Do not do that.

Instead, look at the structure. We have a rational equation, but we know that any rational equation of this form can be transformed into a quadratic equation. Our first mission is to clear the denominators.

We start by combining the terms on the left-hand side. By taking the common denominator $(x+p)(x+q)$, we transform the left side into:

Look at that numerator: $2x+p+q$. It is simple, clean, and linear. Now, let us expand the denominator: $(x+p)(x+q)=x^2+(p+q)x+pq$.

Suddenly, the equation is taking the shape of a standard quadratic equation:

Now, we cross-multiply. This is where the magic happens. We get:

Rearranging this into the standard form $A{x}^2+Bx+C=0$, we get:

Here is where you must pause. The problem tells us the roots are equal in magnitude but opposite in sign. Let the roots be $\alpha$ and $-\alpha$.

If you were to plot these on a number line, they are perfectly balanced around the origin. This is the Symmetry I mentioned. If the roots are $\alpha$ and $-\alpha$, their sum must be zero.

Using Vieta's formulas, the sum of the roots is $-\frac{B}{A}$. In our equation, $A=1$ and $B=(p+q-2r)$. Therefore, the sum of the roots is $-(p+q-2r)$.

Since the sum is zero, we have:

This is a massive breakthrough! We have found that $r=\frac{p+q}{2}$. We have unlocked the value of $r$ without even knowing what $x$ is.

We are not done yet. The question asks for the sum of the squares of the roots. The roots are $\alpha$ and $-\alpha$, so the sum of their squares is ${\alpha }^2+(-\alpha {)}^2=2{\alpha }^2$.

We know from Vieta's formulas that the product of the roots is $\frac{C}{A}$. Here, the product is $\alpha \cdot (-\alpha )=-{\alpha }^2$. The constant term $C$ is $(pq-r(p+q))$. So:

Multiply by $-1$ to get ${\alpha }^2=r(p+q)-pq$. Now, substitute our value of $r=\frac{p+q}{2}$:

Expanding $(p+q{)}^2$ gives us $p^2+q^2+2pq$. So:

Taking the common denominator of $2$:

Look at that! The $2pq$ and $-2pq$ cancel out perfectly. We are left with ${\alpha }^2=\frac{p^2+q^2}{2}$.

Finally, the sum of the squares of the roots is $2{\alpha }^2=2\cdot \left(\frac{p^2+q^2}{2}\right)=p^2+q^2$.

We started with a complex-looking rational equation and ended with a beautifully simple result: $p^2+q^2$. This is the essence of JEE Advanced mathematics.

It is not about brute force; it is about identifying the underlying structure, respecting the symmetry, and letting the algebra guide you to the truth. You have done well today. Keep this mindset, and no problem will ever be too difficult for you.