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 +beta^2` is equal to :

Options

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

Answer

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

### Step 1: Write down 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: Identify the coefficients
From the standard form \( ax^2 + bx + c = 0 \), we can identify:
- \( a = 1 \)
- \( b = -(3 + 2^{\sqrt{\log_2 3}} - 3^{\sqrt{\log_3 2}}) \)
- \( c = -2(3^{\log_3 2} - 2^{\log_2 3}) \)

### Step 3: Calculate the product of the roots \( \alpha\beta \)
Using Vieta's formulas, the product of the roots \( \alpha\beta \) is given by:
\[ \alpha\beta = \frac{c}{a} = -2(3^{\log_3 2} - 2^{\log_2 3}) \]
Using the property \( a^{\log_b c} = c^{\log_b a} \), we can simplify:
\[ 3^{\log_3 2} = 2 \quad \text{and} \quad 2^{\log_2 3} = 3 \]
Thus,
\[ \alpha\beta = -2(2 - 3) = -2(-1) = 2 \]

### Step 4: Calculate the sum of the roots \( \alpha + \beta \)
The sum of the roots \( \alpha + \beta \) is given by:
\[ \alpha + \beta = -\frac{b}{a} = 3 + 2^{\sqrt{\log_2 3}} - 3^{\sqrt{\log_3 2}} \]
Using the same logarithmic properties:
\[ 2^{\sqrt{\log_2 3}} = 3^{\sqrt{\log_3 2}} \]
Thus,
\[ \alpha + \beta = 3 + 2^{\sqrt{\log_2 3}} - 2^{\sqrt{\log_2 3}} = 3 \]

### Step 5: Calculate \( \alpha^2 + \alpha\beta + \beta^2 \)
We can express \( \alpha^2 + \alpha\beta + \beta^2 \) in terms of \( \alpha + \beta \) and \( \alpha\beta \):
\[ \alpha^2 + \alpha\beta + \beta^2 = (\alpha + \beta)^2 - \alpha\beta \]
Substituting the values we found:
\[ \alpha^2 + \alpha\beta + \beta^2 = (3)^2 - 2 = 9 - 2 = 7 \]

### Final Answer
Thus, the value of \( \alpha^2 + \alpha\beta + \beta^2 \) is:
\[ \boxed{7} \]