If `alpha, beta` ar the roots of the quadratic equation `x ^(2) -(3+ 2 ^(sqrt(log _(2)3))-3 ^(sqrt(log _(3)2)))x-2 (3 ^(log _(3)2)-2^(log _(z)3))=0, ` then the value of `alpha ^(2) + alpha beta ` is equal to :

Options

A. 3
B. 5
C. 7
D. 11

Answer

To solve the given quadratic equation and find the value of \( \alpha^2 + \alpha\beta \), we will follow these steps:

### Step 1: Identify the quadratic equation
The given quadratic equation is:
\[ x^2 - (3 + 2^{\sqrt{\log_2 3}} - 3^{\sqrt{\log_3 2}})x - 2(3^{\log_3 2} - 2^{\log_2 3}) = 0 \]

### Step 2: Simplify the coefficients
We need to simplify the coefficients of the quadratic equation.

1. **Simplifying the coefficient of \( x \)**:
- The coefficient of \( x \) is \( 3 + 2^{\sqrt{\log_2 3}} - 3^{\sqrt{\log_3 2}} \).
- Using the change of base formula, we can express \( \sqrt{\log_3 2} \) as \( \frac{\sqrt{\log_2 2}}{\sqrt{\log_2 3}} \).
- Thus, we can rewrite \( 3^{\sqrt{\log_3 2}} \) as \( 3^{\frac{\sqrt{\log_2 2}}{\sqrt{\log_2 3}}} = 2^{\sqrt{\log_2 3}} \).
- Therefore, the coefficient simplifies to \( 3 + 2^{\sqrt{\log_2 3}} - 2^{\sqrt{\log_2 3}} = 3 \).

2. **Simplifying the constant term**:
- The constant term is \( -2(3^{\log_3 2} - 2^{\log_2 3}) \).
- Using the property \( a^{\log_a b} = b \), we have \( 3^{\log_3 2} = 2 \) and \( 2^{\log_2 3} = 3 \).
- Therefore, the constant term simplifies to \( -2(2 - 3) = -2(-1) = 2 \).

### Step 3: Write the simplified quadratic equation
After simplification, the quadratic equation becomes:
\[ x^2 - 3x + 2 = 0 \]

### Step 4: Find the roots of the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1, b = -3, c = 2 \).
- The discriminant \( D = b^2 - 4ac = (-3)^2 - 4(1)(2) = 9 - 8 = 1 \).
- The roots are:
\[ x = \frac{3 \pm \sqrt{1}}{2} = \frac{3 \pm 1}{2} \]
Thus, the roots are \( x = 2 \) and \( x = 1 \). Therefore, \( \alpha = 2 \) and \( \beta = 1 \).

### Step 5: Calculate \( \alpha^2 + \alpha\beta \)
Now we can find \( \alpha^2 + \alpha\beta \):
\[ \alpha^2 + \alpha\beta = 2^2 + (2)(1) = 4 + 2 = 6 \]

### Final Answer
The value of \( \alpha^2 + \alpha\beta \) is:
\[ \boxed{6} \]