Animated Solution for Mathematics - Quadratic Equations: Let $\alpha ,\beta$ be the roots of $x^2-x+p=0$ and $\gamma ,\delta$ be the roots of $x^2-4x+q=0$. If $\alpha ,\beta ,\gamma ,\delta$ are in G.P., then the integral values of $p$ and $q$ respectively, are

Imagine you are standing at the threshold of a classic JEE Advanced challenge. You are presented with two quadratic equations, $x^2-x+p=0$ and $x^2-4x+q=0$, and told that their roots, $\alpha ,\beta ,\gamma ,\delta$, dance together in a perfect Geometric Progression (G.P.).

To begin, we must translate the language of G.P. into the language of algebra. If our roots are in G.P., we can define them as $\alpha =a$, $\beta =ar$, $\gamma =a{r}^2$, and $\delta =a{r}^3$, where $a$ is the first term and $r$ is the common ratio.

We turn to the power of Vieta's formulas. For the first equation, $x^2-x+p=0$, the sum of the roots is $\alpha +\beta =1$. Substituting our G.P. terms, we get $a+ar=1$, which simplifies to:

For the second equation, $x^2-4x+q=0$, the sum of the roots is $\gamma +\delta =4$. Substituting our terms, we get $a{r}^2+a{r}^3=4$, which factors into:

We have two equations and two unknowns. If we divide Equation 2 by Equation 1, we get:

Notice how the terms $a$ and $(1+r)$ vanish, leaving us with the simple, powerful result:

This gives us two potential paths: $r=2$ or $r=-2$.

The problem demands that $p$ and $q$ be integers. If we test $r=2$, we find $a(1+2)=1$, so $a=\frac{1}{3}$.

Calculating $p=\alpha \beta =a^{2}r=(\frac{1}{9})(2)=\frac{2}{9}$. Since this is not an integer, we must reject $r=2$.

However, if we test $r=-2$, we find $a(1-2)=1$, so $-a=1$, which means $a=-1$. This yields integer-friendly values.

With $a=-1$ and $r=-2$, we can now find our targets. For $p$, the product of the roots is:

For $q$, the product of the roots is:

We have arrived at the final solution. The values are $p=-2$ and $q=-32$.