If `alpha` is a root of the equation `4x^(2)+2x-1=0` and `f(x)=4x^(2)-3x+1`, then `2(f(alpha)+(alpha))` is equal to

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to find the value of \( 2(f(\alpha) + \alpha) \) where \( \alpha \) is a root of the equation \( 4x^2 + 2x - 1 = 0 \) and \( f(x) = 4x^2 - 3x + 1 \).

### Step 1: Find the roots of the equation \( 4x^2 + 2x - 1 = 0 \)

We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 4 \), \( b = 2 \), and \( c = -1 \).

Calculating the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 4 \cdot (-1) = 4 + 16 = 20 \]

Now substituting into the quadratic formula:
\[ x = \frac{-2 \pm \sqrt{20}}{2 \cdot 4} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4} \]

Thus, the roots are:
\[ \alpha = \frac{-1 + \sqrt{5}}{4} \quad \text{(we can take this as } \alpha \text{)} \]
\[ \text{or } \alpha = \frac{-1 - \sqrt{5}}{4} \]

### Step 2: Calculate \( f(\alpha) \)

Now we need to calculate \( f(\alpha) \):
\[ f(x) = 4x^2 - 3x + 1 \]
Substituting \( \alpha \):
\[ f(\alpha) = 4\left(\frac{-1 + \sqrt{5}}{4}\right)^2 - 3\left(\frac{-1 + \sqrt{5}}{4}\right) + 1 \]

Calculating \( \alpha^2 \):
\[ \alpha^2 = \left(\frac{-1 + \sqrt{5}}{4}\right)^2 = \frac{(-1 + \sqrt{5})^2}{16} = \frac{1 - 2\sqrt{5} + 5}{16} = \frac{6 - 2\sqrt{5}}{16} = \frac{3 - \sqrt{5}}{8} \]

Now substituting back into \( f(\alpha) \):
\[ f(\alpha) = 4 \cdot \frac{3 - \sqrt{5}}{8} - 3 \cdot \frac{-1 + \sqrt{5}}{4} + 1 \]
\[ = \frac{12 - 4\sqrt{5}}{8} + \frac{3 - 3\sqrt{5}}{4} + 1 \]
\[ = \frac{12 - 4\sqrt{5}}{8} + \frac{6 - 6\sqrt{5}}{8} + \frac{8}{8} \]
\[ = \frac{12 - 4\sqrt{5} + 6 - 6\sqrt{5} + 8}{8} = \frac{26 - 10\sqrt{5}}{8} = \frac{13 - 5\sqrt{5}}{4} \]

### Step 3: Calculate \( 2(f(\alpha) + \alpha) \)

Now we need to calculate \( 2(f(\alpha) + \alpha) \):
\[ f(\alpha) + \alpha = \frac{13 - 5\sqrt{5}}{4} + \frac{-1 + \sqrt{5}}{4} \]
\[ = \frac{13 - 5\sqrt{5} - 1 + \sqrt{5}}{4} = \frac{12 - 4\sqrt{5}}{4} = 3 - \sqrt{5} \]

Now multiplying by 2:
\[ 2(f(\alpha) + \alpha) = 2(3 - \sqrt{5}) = 6 - 2\sqrt{5} \]

### Final Answer
Thus, the value of \( 2(f(\alpha) + \alpha) \) is:
\[ \boxed{6 - 2\sqrt{5}} \]