Animated Solution for Mathematics - Quadratic Equations: If $\alpha ,\beta$ are the roots of $x^2+px+q=0$ and $\gamma ,\delta$ are the roots of $x^2+rx+s=0$, evaluate $(\alpha -\gamma )(\alpha -\delta )(\beta -\gamma )(\beta -\delta )$ in terms of $p,q,r$ and $s$. Deduce the condition that the equations have a common root.

Welcome, fellow JEE warrior. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of variables.

We have two quadratic equations, $x^2+px+q=0$ with roots $\alpha$ and $\beta$, and $x^2+rx+s=0$ with roots $\gamma$ and $\delta$. We are tasked with evaluating the product $E=(\alpha -\gamma )(\alpha -\delta )(\beta -\gamma )(\beta -\delta )$.

Many students would immediately reach for the quadratic formula, trying to find explicit expressions for $\alpha ,\beta ,\gamma ,$ and $\delta$. But stop! That is the trap. In the world of JEE Advanced, we do not fight the math; we dance with it. Let us find the rhythm of this problem.

The secret to this problem lies in the Factor Theorem. We know that if $\gamma$ and $\delta$ are the roots of the second quadratic equation, then the expression $x^2+rx+s$ can be factored perfectly as $(x-\gamma )(x-\delta )$.

This is not just a coincidence; it is an identity that holds true for any value of $x$. Now, look at our target expression $E$. It contains the term $(\alpha -\gamma )(\alpha -\delta )$.

If we set $x=\alpha$ in our identity, we get:

Suddenly, the product of two differences has transformed into a simple quadratic expression. This is the 'Aha!' moment. We have reduced the complexity of the problem by shifting our perspective.

We are not done yet. We still have an ${\alpha }^2$ term, which feels a bit heavy. But remember, $\alpha$ is a root of the first equation, $x^2+px+q=0$.

This means ${\alpha }^2+p\alpha +q=0$, or more usefully, ${\alpha }^2=-p\alpha -q$. By substituting this into our expression, we replace the quadratic term with a linear one:

Look at that! We have taken a product of two terms and turned it into a clean, linear expression.

Now, we apply the principle of symmetry. Since $\beta$ is also a root of the first equation, the exact same logic applies. We do not need to repeat the derivation.

We can confidently state that:

Our total product $E$ is now simply the product of these two linear expressions:

We have successfully reduced a product of four terms into a product of two.

Let us expand this product. Multiplying these terms out, we get:

This is where the beauty of Vieta's formulas shines. We know that for the first equation, the sum of roots $\alpha +\beta =-p$ and the product of roots $\alpha \beta =q$.

Substituting these values, we get:

With a little algebraic housekeeping—factoring out $(r-p)$ and simplifying—we arrive at the elegant result:

Finally, consider the condition for a common root. If the equations share a root, then one of the differences must be zero, forcing $E=0$.

Thus, the condition is $(q-s{)}^2-(p-r)(ps-qr)=0$. You see? We didn't need to calculate the roots. We used the structure of the polynomials themselves to reveal the answer. Keep this mindset, and no problem will ever be too daunting.