If a and b are the roots of the equation `x^2-6x + 6=0`, find the value of `2(a^(2) + b^(2))`.

Options

A. 40
B. 42
C. 48
D. 46

Answer

To solve the problem, we need to find the value of \(2(a^2 + b^2)\) where \(a\) and \(b\) are the roots of the quadratic equation \(x^2 - 6x + 6 = 0\).

### Step-by-Step Solution:

1. **Identify the coefficients of the quadratic equation**:
The given quadratic equation is \(x^2 - 6x + 6 = 0\). Here, we can identify:
- \(A = 1\) (coefficient of \(x^2\))
- \(B = -6\) (coefficient of \(x\))
- \(C = 6\) (constant term)

2. **Use Vieta's formulas to find the sum and product of the roots**:
According to Vieta's formulas:
- The sum of the roots \(a + b = -\frac{B}{A} = -\frac{-6}{1} = 6\)
- The product of the roots \(ab = \frac{C}{A} = \frac{6}{1} = 6\)

3. **Find \(a^2 + b^2\)**:
We can use the identity:
\[ a^2 + b^2 = (a + b)^2 - 2ab \]
Substituting the values we found:
\[ a^2 + b^2 = (6)^2 - 2(6) = 36 - 12 = 24 \]

4. **Calculate \(2(a^2 + b^2)\)**:
Now, we need to find \(2(a^2 + b^2)\):
\[ 2(a^2 + b^2) = 2 \times 24 = 48 \]

### Final Answer:
Thus, the value of \(2(a^2 + b^2)\) is \(48\).

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