Animated Solution for Mathematics - Quadratic Equations: Let $S=\left\{\alpha :{\log }_2\left(9^{2\alpha -4}+13\right)-{\log }_2\left(\frac{5}{2}\cdot 3^{2\alpha -4}+1\right)=2\right\}$. Then the maximum value of $\beta$ for which the equation $x^2-2{\left({\sum }_{\alpha \in S}\alpha \right)}^{2}x+{\sum }_{\alpha \in S}(\alpha +1{)}^2\beta =0$ has real roots, is

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving an equation; we are peeling back the layers of a mathematical onion.

At first glance, this problem might look like a chaotic mess of exponents and logarithms, 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 strip away that mask together.

The Logarithmic Transformation

We begin with the equation:

When you see two logarithms with the same base being subtracted, your mind should immediately jump to the quotient rule: ${\log }_{b}A-{\log }_{b}B={\log }_b(A/B)$. By applying this, we condense the entire left side into a single, manageable fraction:

Now, we invoke the definition of a logarithm. If ${\log }_2(X)=2$, then $X=2^2=4$. Suddenly, the logarithmic fog clears, and we are left with a rational equation:

The Art of Substitution

Here is where the 'Aha!' moment happens. Look at the terms $9^{2\alpha -4}$ and $3^{2\alpha -4}$. Since $9=3^2$, we can write $9^{2\alpha -4}$ as $(3^{2\alpha -4}{)}^2$.

Let us perform a substitution: let $y=3^{2\alpha -4}$. Our equation transforms into a beautiful, simple quadratic:

Cross-multiplying gives us $y^2+13=4(\frac{5}{2}y+1)$, which simplifies to $y^2-10y+9=0$. Factoring this is a breeze: $(y-9)(y-1)=0$. We have found our values for $y$: $y=1$ and $y=9$.

Unveiling the Set $S$

We must return to our original variable, $\alpha$. If $y=1$, then $3^{2\alpha -4}=3^0$, which implies $2\alpha -4=0$, so $\alpha =2$.

If $y=9$, then $3^{2\alpha -4}=3^2$, which implies $2\alpha -4=2$, so $\alpha =3$. Thus, our set $S$ is simply $\{2,3\}$.

The complexity has vanished, leaving behind two humble integers.

The Final Challenge

Now, we move to the second act. We are given the quadratic equation $x^2-2{\left({\sum }_{\alpha \in S}\alpha \right)}^{2}x+{\sum }_{\alpha \in S}(\alpha +1{)}^2\beta =0$.

First, let us calculate the sums. The sum of elements in $S$ is $2+3=5$. The sum of $(\alpha +1{)}^2$ is $(2+1{)}^2+(3+1{)}^2=9+16=25$.

Substituting these into our equation, we get:

For this quadratic to have real roots, the discriminant $D=b^2-4ac$ must be greater than or equal to zero. Plugging in our values:

And there it is! The maximum value of $\beta$ is 25. You navigated the logs, mastered the substitution, and conquered the discriminant. This is the essence of JEE Advanced—not just calculating, but seeing the structure of the problem and guiding it to its inevitable conclusion.