If yadi `x + y = sqrt3 and x - y = sqrt2,` then the value of `8xy (x ^(2) + y ^(2))` is

Options

A. 6
B. `sqrt6`
C. `5`
D. `sqrt5`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
\[ x + y = \sqrt{3} \quad \text{(1)} \]
\[ x - y = \sqrt{2} \quad \text{(2)} \]

2. **Add the Two Equations**:
Adding equations (1) and (2):
\[ (x + y) + (x - y) = \sqrt{3} + \sqrt{2} \]
This simplifies to:
\[ 2x = \sqrt{3} + \sqrt{2} \]
Therefore, we can solve for \(x\):
\[ x = \frac{\sqrt{3} + \sqrt{2}}{2} \quad \text{(3)} \]

3. **Subtract the Two Equations**:
Now, subtract equation (2) from equation (1):
\[ (x + y) - (x - y) = \sqrt{3} - \sqrt{2} \]
This simplifies to:
\[ 2y = \sqrt{3} - \sqrt{2} \]
Therefore, we can solve for \(y\):
\[ y = \frac{\sqrt{3} - \sqrt{2}}{2} \quad \text{(4)} \]

4. **Calculate \(xy\)**:
Now we need to calculate \(xy\):
\[ xy = \left(\frac{\sqrt{3} + \sqrt{2}}{2}\right) \left(\frac{\sqrt{3} - \sqrt{2}}{2}\right) \]
Using the difference of squares:
\[ xy = \frac{(\sqrt{3})^2 - (\sqrt{2})^2}{4} = \frac{3 - 2}{4} = \frac{1}{4} \]

5. **Calculate \(x^2 + y^2\)**:
We can use the identity \(x^2 + y^2 = (x+y)^2 - 2xy\):
\[ x^2 + y^2 = (\sqrt{3})^2 - 2 \cdot \frac{1}{4} = 3 - \frac{1}{2} = \frac{6}{2} - \frac{1}{2} = \frac{5}{2} \]

6. **Calculate \(8xy(x^2 + y^2)\)**:
Now we can find \(8xy(x^2 + y^2)\):
\[ 8xy(x^2 + y^2) = 8 \cdot \frac{1}{4} \cdot \frac{5}{2} = 2 \cdot \frac{5}{2} = 5 \]

Thus, the final value of \(8xy(x^2 + y^2)\) is:
\[ \boxed{5} \]