If `alpha and beta` are the roots of the equation `4x^(2) + 3x + 7 = 0`, then what is the value of `(alpha^(-2) + beta^(-2))` ?

Options

A. `47//49`
B. `49//47`
C. `-47//49`
D. `-49//47`

Explanation

To find the value of \( \alpha^{-2} + \beta^{-2} \) where \( \alpha \) and \( \beta \) are the roots of the equation \( 4x^2 + 3x + 7 = 0 \), we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is \( 4x^2 + 3x + 7 = 0 \). Here, we have:
- \( a = 4 \)
- \( b = 3 \)
- \( c = 7 \)

### Step 2: Use Vieta's formulas
According to Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{3}{4} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{7}{4} \)

### Step 3: Find \( \alpha^{-2} + \beta^{-2} \)
We know that:
\[ \alpha^{-2} + \beta^{-2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]

### Step 4: Calculate \( \alpha^2 + \beta^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = \left(-\frac{3}{4}\right)^2 - 2 \cdot \frac{7}{4} \]
Calculating \( \left(-\frac{3}{4}\right)^2 \):
\[ \left(-\frac{3}{4}\right)^2 = \frac{9}{16} \]
Now calculating \( 2 \cdot \frac{7}{4} = \frac{14}{4} = \frac{7}{2} = \frac{56}{16} \).

Thus,
\[ \alpha^2 + \beta^2 = \frac{9}{16} - \frac{56}{16} = \frac{9 - 56}{16} = \frac{-47}{16} \]

### Step 5: Calculate \( \alpha^2 \beta^2 \)
Using the identity:
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{7}{4}\right)^2 = \frac{49}{16} \]

### Step 6: Substitute into \( \alpha^{-2} + \beta^{-2} \)
Now substituting back into the equation:
\[ \alpha^{-2} + \beta^{-2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} = \frac{\frac{-47}{16}}{\frac{49}{16}} = \frac{-47}{49} \]

### Final Answer
Thus, the value of \( \alpha^{-2} + \beta^{-2} \) is:
\[ \boxed{-\frac{47}{49}} \]