Let `S ={alpha:log_2(9^(2alpha-4)+13)-log_2(5/2.3^(2alpha-4)+1)=2}`Then the maximum value of `beta` for which the equation `x^2-2 (sum_(a in s)a)^2 x+sum_(a in s)(alpha+1)^2beta=0` has real roots, is ______ .

Answer

To solve the given problem step by step, we will follow the instructions provided in the video transcript and clarify the calculations.

### Step 1: Simplifying the Logarithmic Equation
We start with the equation given in the problem:

\[ \log_2(9^{2\alpha - 4} + 13) - \log_2\left(\frac{5}{2} \cdot 3^{2\alpha - 4} + 1\right) = 2 \]

Using the property of logarithms that states \(\log_a(b) - \log_a(c) = \log_a\left(\frac{b}{c}\right)\), we can rewrite the equation as:

\[ \log_2\left(\frac{9^{2\alpha - 4} + 13}{\frac{5}{2} \cdot 3^{2\alpha - 4} + 1}\right) = 2 \]

### Step 2: Exponentiating Both Sides
Exponentiating both sides with base 2 gives us:

\[ \frac{9^{2\alpha - 4} + 13}{\frac{5}{2} \cdot 3^{2\alpha - 4} + 1} = 2^2 = 4 \]

### Step 3: Cross-Multiplying
Cross-multiplying results in:

\[ 9^{2\alpha - 4} + 13 = 4\left(\frac{5}{2} \cdot 3^{2\alpha - 4} + 1\right) \]

This simplifies to:

\[ 9^{2\alpha - 4} + 13 = 10 \cdot 3^{2\alpha - 4} + 4 \]

### Step 4: Rearranging the Equation
Rearranging gives:

\[ 9^{2\alpha - 4} - 10 \cdot 3^{2\alpha - 4} + 9 = 0 \]

### Step 5: Substituting Variables
Let \(y = 3^{2\alpha - 4}\). Then, \(9^{2\alpha - 4} = y^2\). The equation becomes:

\[ y^2 - 10y + 9 = 0 \]

### Step 6: Solving the Quadratic Equation
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

Here, \(a = 1\), \(b = -10\), and \(c = 9\):

\[ y = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2} \]

Thus, we find:

\[ y = 9 \quad \text{or} \quad y = 1 \]

### Step 7: Finding Corresponding Values of \(\alpha\)
1. If \(y = 9\):
\[ 3^{2\alpha - 4} = 9 \implies 2\alpha - 4 = 2 \implies 2\alpha = 6 \implies \alpha = 3 \]

2. If \(y = 1\):
\[ 3^{2\alpha - 4} = 1 \implies 2\alpha - 4 = 0 \implies 2\alpha = 4 \implies \alpha = 2 \]

Thus, the values of \(\alpha\) are \(2\) and \(3\).

### Step 8: Summing the Values of \(\alpha\)
The sum of the values of \(\alpha\) in set \(S\) is:

\[ \sum_{a \in S} a = 2 + 3 = 5 \]

### Step 9: Formulating the Quadratic Equation
Now we substitute this sum into the quadratic equation:

\[ x^2 - 2 \cdot 5^2 x + \sum_{a \in S} (\alpha + 1)^2 \beta = 0 \]

Calculating \(\sum_{a \in S} (\alpha + 1)^2\):

\[ (2 + 1)^2 + (3 + 1)^2 = 3^2 + 4^2 = 9 + 16 = 25 \]

Thus, the equation becomes:

\[ x^2 - 50x + 25\beta = 0 \]

### Step 10: Condition for Real Roots
For the quadratic equation to have real roots, the discriminant must be non-negative:

\[ D = b^2 - 4ac \geq 0 \]

Calculating the discriminant:

\[ 50^2 - 4 \cdot 1 \cdot 25\beta \geq 0 \]

This simplifies to:

\[ 2500 - 100\beta \geq 0 \implies 2500 \geq 100\beta \implies \beta \leq 25 \]

### Conclusion
The maximum value of \(\beta\) for which the equation has real roots is:

\[ \boxed{25} \]