If `alpha, beta` are the rootsof `6x^(2)-4sqrt2 x-3=0`, then `alpha^(2)beta+alpha beta^(2)` is

Options

A. `(2)/(3)`
B. `(-2)/(3)`
C. `(sqrt2)/(3)`
D. `(-sqrt2)/(3)`

Answer

To solve the problem, we need to find the value of \( \alpha^2 \beta + \alpha \beta^2 \) given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( 6x^2 - 4\sqrt{2}x - 3 = 0 \).

### Step 1: Identify coefficients
The given quadratic equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 6 \)
- \( b = -4\sqrt{2} \)
- \( c = -3 \)

### Step 2: Calculate \( \alpha + \beta \) and \( \alpha \beta \)
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

Calculating \( \alpha + \beta \):
\[ \alpha + \beta = -\frac{-4\sqrt{2}}{6} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3} \]

Calculating \( \alpha \beta \):
\[ \alpha \beta = \frac{-3}{6} = -\frac{1}{2} \]

### Step 3: Find \( \alpha^2 \beta + \alpha \beta^2 \)
We can factor \( \alpha^2 \beta + \alpha \beta^2 \) as follows:
\[ \alpha^2 \beta + \alpha \beta^2 = \alpha \beta (\alpha + \beta) \]

### Step 4: Substitute the values
Now substituting the values we found:
\[ \alpha^2 \beta + \alpha \beta^2 = \alpha \beta (\alpha + \beta) = \left(-\frac{1}{2}\right) \left(\frac{2\sqrt{2}}{3}\right) \]

### Step 5: Calculate the result
Now, performing the multiplication:
\[ \alpha^2 \beta + \alpha \beta^2 = -\frac{1}{2} \cdot \frac{2\sqrt{2}}{3} = -\frac{2\sqrt{2}}{6} = -\frac{\sqrt{2}}{3} \]

### Final Answer
Thus, the value of \( \alpha^2 \beta + \alpha \beta^2 \) is:
\[ -\frac{\sqrt{2}}{3} \]