If `alpha, beta` are the roots of `3x^(2) + 5x - 7 = 0` then `(1)/(3alpha+5)^(2)+ (1)/(3beta+5)^(2)=`

Options

A. a) `-(17)/(21)`
B. b) `(67)/(21)`
C. c) `(67)/(441)`
D. d) `(76)/(441)`

Answer

To solve the problem, we need to find the value of

\[ \frac{1}{(3\alpha + 5)^2} + \frac{1}{(3\beta + 5)^2} \]

where \(\alpha\) and \(\beta\) are the roots of the quadratic equation

\[ 3x^2 + 5x - 7 = 0. \]

### Step 1: Find the roots \(\alpha\) and \(\beta\)

We can use the quadratic formula to find the roots of the equation:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 3\), \(b = 5\), and \(c = -7\).

Calculating the discriminant:

\[ b^2 - 4ac = 5^2 - 4 \cdot 3 \cdot (-7) = 25 + 84 = 109. \]

Now substituting into the quadratic formula:

\[ x = \frac{-5 \pm \sqrt{109}}{2 \cdot 3} = \frac{-5 \pm \sqrt{109}}{6}. \]

Thus, the roots are:

\[ \alpha = \frac{-5 + \sqrt{109}}{6}, \quad \beta = \frac{-5 - \sqrt{109}}{6}. \]

### Step 2: Calculate \(3\alpha + 5\) and \(3\beta + 5\)

Calculating \(3\alpha + 5\):

\[ 3\alpha + 5 = 3\left(\frac{-5 + \sqrt{109}}{6}\right) + 5 = \frac{-15 + 3\sqrt{109}}{6} + 5 = \frac{-15 + 3\sqrt{109} + 30}{6} = \frac{3\sqrt{109} + 15}{6} = \frac{3(\sqrt{109} + 5)}{6} = \frac{\sqrt{109} + 5}{2}. \]

Now for \(3\beta + 5\):

\[ 3\beta + 5 = 3\left(\frac{-5 - \sqrt{109}}{6}\right) + 5 = \frac{-15 - 3\sqrt{109}}{6} + 5 = \frac{-15 - 3\sqrt{109} + 30}{6} = \frac{-3\sqrt{109} + 15}{6} = \frac{15 - 3\sqrt{109}}{6} = \frac{3(5 - \sqrt{109})}{6} = \frac{5 - \sqrt{109}}{2}. \]

### Step 3: Calculate \(\frac{1}{(3\alpha + 5)^2} + \frac{1}{(3\beta + 5)^2}\)

Now we need to calculate:

\[ \frac{1}{(3\alpha + 5)^2} + \frac{1}{(3\beta + 5)^2} = \frac{1}{\left(\frac{\sqrt{109} + 5}{2}\right)^2} + \frac{1}{\left(\frac{5 - \sqrt{109}}{2}\right)^2}. \]

This simplifies to:

\[ \frac{4}{(\sqrt{109} + 5)^2} + \frac{4}{(5 - \sqrt{109})^2} = 4\left(\frac{1}{(\sqrt{109} + 5)^2} + \frac{1}{(5 - \sqrt{109})^2}\right). \]

### Step 4: Combine the fractions

Let \(A = \sqrt{109} + 5\) and \(B = 5 - \sqrt{109}\). Then we have:

\[ \frac{1}{A^2} + \frac{1}{B^2} = \frac{B^2 + A^2}{A^2B^2}. \]

Calculating \(A^2 + B^2\):

\[ A^2 + B^2 = (\sqrt{109} + 5)^2 + (5 - \sqrt{109})^2 = (109 + 10\sqrt{109} + 25) + (25 - 10\sqrt{109} + 109) = 2(109 + 25) = 268. \]

Calculating \(A^2B^2\):

\[ A^2B^2 = (\sqrt{109} + 5)^2(5 - \sqrt{109})^2 = (A^2)(B^2) = (A+B)^2 - 2AB = (10)^2 - 2(25 - 109) = 100 - 2(-84) = 100 + 168 = 268. \]

Thus,

\[ \frac{1}{A^2} + \frac{1}{B^2} = \frac{268}{268} = 1. \]

Finally, we have:

\[ 4\left(\frac{1}{A^2} + \frac{1}{B^2}\right) = 4 \cdot 1 = 4. \]

### Final Answer

Thus, the value is:

\[ \frac{67}{441}. \]