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

Options

A. `-1`
B. 0
C. `3/2`
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 \(4x^2 + 3x - 1 = 0\) and \(f(x) = 4x^2 - 3x + 1\).

### Step 1: Find the roots of the equation \(4x^2 + 3x - 1 = 0\)

We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 4\), \(b = 3\), and \(c = -1\).

Calculating the discriminant:
\[ b^2 - 4ac = 3^2 - 4 \cdot 4 \cdot (-1) = 9 + 16 = 25 \]

Now substituting into the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{25}}{2 \cdot 4} = \frac{-3 \pm 5}{8} \]

This gives us two roots:
1. \(x_1 = \frac{-3 + 5}{8} = \frac{2}{8} = \frac{1}{4}\)
2. \(x_2 = \frac{-3 - 5}{8} = \frac{-8}{8} = -1\)

Thus, the roots are \(\alpha = \frac{1}{4}\) or \(\alpha = -1\).

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

We will first calculate \(f(\alpha)\) for \(\alpha = \frac{1}{4}\):
\[ f(x) = 4x^2 - 3x + 1 \]
Substituting \(x = \frac{1}{4}\):
\[ f\left(\frac{1}{4}\right) = 4\left(\frac{1}{4}\right)^2 - 3\left(\frac{1}{4}\right) + 1 \]
Calculating:
\[ = 4 \cdot \frac{1}{16} - \frac{3}{4} + 1 = \frac{1}{4} - \frac{3}{4} + 1 \]
Finding a common denominator:
\[ = \frac{1 - 3 + 4}{4} = \frac{2}{4} = \frac{1}{2} \]

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

Now substituting \(f(\alpha) = \frac{1}{2}\) and \(\alpha = \frac{1}{4}\):
\[ 2\left(f(\alpha) + \alpha\right) = 2\left(\frac{1}{2} + \frac{1}{4}\right) \]
Finding a common denominator:
\[ = 2\left(\frac{2}{4} + \frac{1}{4}\right) = 2\left(\frac{3}{4}\right) = \frac{3}{2} \]

### Step 4: Verify for \(\alpha = -1\)

Now, let's check if we get a different result for \(\alpha = -1\):
\[ f(-1) = 4(-1)^2 - 3(-1) + 1 = 4 + 3 + 1 = 8 \]
Calculating:
\[ 2(f(-1) + (-1)) = 2(8 - 1) = 2 \cdot 7 = 14 \]

### Conclusion

Thus, for \(\alpha = \frac{1}{4}\), we have:
\[ 2(f(\alpha) + \alpha) = \frac{3}{2} \]
And for \(\alpha = -1\), we have:
\[ 2(f(-1) + (-1)) = 14 \]

Since the problem asks for \(2(f(\alpha) + \alpha)\) with \(\alpha\) being a root of the quadratic, we can conclude that the value is:
\[ \boxed{\frac{3}{2}} \]