If one of roots of `2x^(2) - ax + 32= 0` is twice the other root, then the value of a is____

Options

A. `-2 sqrt2`
B. `8 sqrt2`
C. `12 sqrt2`
D. `-2 sqrt2`

Answer

To solve the problem, we need to find the value of \( a \) in the quadratic equation \( 2x^2 - ax + 32 = 0 \) given that one root is twice the other root.

### Step-by-Step Solution:

1. **Let the Roots be Defined**:
- Let one root be \( \alpha \) and the other root be \( \beta \).
- According to the problem, \( \beta = 2\alpha \).

2. **Sum of Roots**:
- The sum of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by the formula:
\[ \alpha + \beta = -\frac{b}{a} \]
- For our equation, \( a = 2 \), \( b = -a \), and \( c = 32 \).
- Therefore, we can write:
\[ \alpha + 2\alpha = -\frac{-a}{2} \]
- This simplifies to:
\[ 3\alpha = \frac{a}{2} \]
- Thus, we can express \( a \) in terms of \( \alpha \):
\[ a = 6\alpha \quad \text{(Equation 1)} \]

3. **Product of Roots**:
- The product of the roots is given by:
\[ \alpha \cdot \beta = \frac{c}{a} \]
- Substituting \( \beta = 2\alpha \):
\[ \alpha \cdot 2\alpha = \frac{32}{2} \]
- This simplifies to:
\[ 2\alpha^2 = 16 \]
- Dividing both sides by 2:
\[ \alpha^2 = 8 \]
- Taking the square root:
\[ \alpha = 2\sqrt{2} \quad \text{(Equation 2)} \]

4. **Substituting Back to Find \( a \)**:
- Now substitute \( \alpha = 2\sqrt{2} \) back into Equation 1:
\[ a = 6(2\sqrt{2}) = 12\sqrt{2} \]

5. **Final Answer**:
- Therefore, the value of \( a \) is:
\[ \boxed{12\sqrt{2}} \]