Animated Solution for Mathematics - Quadratic Equations: If $\alpha$ and $\beta$ are the roots of the equation $x^2+px+2=0$ and $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ are the roots of the equation $2x^2+2qx+1=0$, then $(\alpha -\frac{1}{\alpha })(\beta -\frac{1}{\beta })(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })$ is equal to :

Welcome, fellow traveler on the JEE journey. Today, we are not just solving a quadratic equation; we are uncovering the hidden symmetry within roots.

When you look at a problem like this, it is easy to feel overwhelmed by the sheer number of variables. But take a deep breath. The beauty of these problems lies in the fact that you rarely need to know the exact values of the roots; you only need to know how they behave together.

The Foundation of Vieta's

Let us start with the first equation: $x^2+px+2=0$. We are told the roots are $\alpha$ and $\beta$.

Immediately, your mind should jump to Vieta's formulas. We know the sum of the roots is $\alpha +\beta =-p$ and the product is $\alpha \beta =2$. Keep these two treasures safe; they are the keys to the entire kingdom.

Now, look at the second equation: $2x^2+2qx+1=0$. The roots are given as $\frac{1}{\alpha }$ and $\frac{1}{\beta }$.

Again, apply Vieta's. The sum is $\frac{1}{\alpha }+\frac{1}{\beta }=-\frac{2q}{2}=-q$. The product is $\left(\frac{1}{\alpha }\right)\left(\frac{1}{\beta }\right)=\frac{1}{2}$, which simplifies to $\alpha \beta =2$. This confirms our consistency!

The Strategic Grouping

Now, we face the target expression:

Do not panic at the four brackets. Instead, let us be strategic. We will split this into two parts, $P_1$ and $P_2$.

Let $P_1=\left(\alpha -\frac{1}{\alpha }\right)\left(\beta -\frac{1}{\beta }\right)$ and $P_2=\left(\alpha +\frac{1}{\beta }\right)\left(\beta +\frac{1}{\alpha }\right)$.

The Algebraic Dance

Let us tackle $P_2$ first, as it is the friendlier of the two. Expanding the brackets, we get:

Substituting $\alpha \beta =2$, we get:

Now for $P_1$. This one requires a bit more care with the signs. Expanding it, we get:

Grouping the terms, we have:

We know ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =(-p{)}^2-2(2)=p^2-4$. Substituting this back in:

Finally, we bring it all together. The total expression $E$ is simply $P_1\times P_2$:

Look at that! The complexity has vanished, leaving behind a clean, elegant result. This is the power of algebraic manipulation.

You didn't need to find $\alpha$ or $\beta$; you just needed to trust the structure of the math. Keep practicing this, and you will find that even the most intimidating JEE problems are just puzzles waiting to be solved.