If `1/(2-sqrt(-2))` is one of the roots of `ax^(2)+bx+c = 0`, where a,b, c are real, then what are the values of a,b,c respectively ?

Options

A. `6, -4, 1`
B. ` 4, 6, -1`
C. `3,-2,1`
D. `6,4,1`

Explanation

To find the values of \( a, b, c \) in the quadratic equation \( ax^2 + bx + c = 0 \) given that \( \frac{1}{2 - \sqrt{-2}} \) is one of the roots, we will follow these steps:

### Step 1: Simplify the given root
The given root is \( \frac{1}{2 - \sqrt{-2}} \). We can simplify this by rationalizing the denominator.

\[ \sqrt{-2} = i\sqrt{2} \quad \text{(where \( i \) is the imaginary unit)} \]
Thus, we rewrite the root as:
\[ \frac{1}{2 - i\sqrt{2}} \]

To rationalize, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{1 \cdot (2 + i\sqrt{2})}{(2 - i\sqrt{2})(2 + i\sqrt{2})} \]

Calculating the denominator:
\[ (2 - i\sqrt{2})(2 + i\sqrt{2}) = 2^2 - (i\sqrt{2})^2 = 4 - (-2) = 4 + 2 = 6 \]

Thus, the root simplifies to:
\[ \frac{2 + i\sqrt{2}}{6} = \frac{1}{3} + \frac{i\sqrt{2}}{6} \]

### Step 2: Identify the other root
Since the coefficients \( a, b, c \) are real, the roots must come in conjugate pairs. Therefore, the other root is:
\[ \frac{1}{3} - \frac{i\sqrt{2}}{6} \]

### Step 3: Calculate the sum and product of the roots
Let \( r_1 = \frac{1}{3} + \frac{i\sqrt{2}}{6} \) and \( r_2 = \frac{1}{3} - \frac{i\sqrt{2}}{6} \).

**Sum of the roots**:
\[ r_1 + r_2 = \left(\frac{1}{3} + \frac{i\sqrt{2}}{6}\right) + \left(\frac{1}{3} - \frac{i\sqrt{2}}{6}\right) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

**Product of the roots**:
\[ r_1 \cdot r_2 = \left(\frac{1}{3} + \frac{i\sqrt{2}}{6}\right)\left(\frac{1}{3} - \frac{i\sqrt{2}}{6}\right) = \left(\frac{1}{3}\right)^2 - \left(\frac{i\sqrt{2}}{6}\right)^2 \]
Calculating:
\[ = \frac{1}{9} - \left(-\frac{2}{36}\right) = \frac{1}{9} + \frac{1}{18} = \frac{2}{18} + \frac{1}{18} = \frac{3}{18} = \frac{1}{6} \]

### Step 4: Form the quadratic equation
Using the sum and product of the roots, we can form the quadratic equation:
\[ x^2 - \left(\text{sum of roots}\right)x + \text{product of roots} = 0 \]
Substituting the values:
\[ x^2 - \frac{2}{3}x + \frac{1}{6} = 0 \]

To eliminate the fractions, multiply the entire equation by 6:
\[ 6x^2 - 4x + 1 = 0 \]

### Step 5: Identify \( a, b, c \)
From the equation \( 6x^2 - 4x + 1 = 0 \), we identify:
\[ a = 6, \quad b = -4, \quad c = 1 \]

### Final Answer
The values of \( a, b, c \) are:
\[ \boxed{(6, -4, 1)} \]