If `alpha` and `beta` be the roots of equation `x^(2) + 3x + 1 = 0` then the value of `((alpha)/(1 + beta))^(2) + ((beta)/(1 + alpha))^(2)` is equal to

Options

A. `15`
B. `18`
C. `21`
D. `17`

Explanation

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

\[ \left(\frac{\alpha}{1 + \beta}\right)^2 + \left(\frac{\beta}{1 + \alpha}\right)^2 \]

given that \(\alpha\) and \(\beta\) are the roots of the equation

\[ x^2 + 3x + 1 = 0. \]

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

Using the quadratic formula, the roots of the equation \(ax^2 + bx + c = 0\) are given by:

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

For our equation \(x^2 + 3x + 1 = 0\), we have \(a = 1\), \(b = 3\), and \(c = 1\).

Calculating the discriminant:

\[ b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 1 = 9 - 4 = 5. \]

Now, substituting into the quadratic formula:

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

Thus, the roots are:

\[ \alpha = \frac{-3 + \sqrt{5}}{2}, \quad \beta = \frac{-3 - \sqrt{5}}{2}. \]

### Step 2: Calculate \(1 + \beta\) and \(1 + \alpha\)

Calculating \(1 + \beta\):

\[ 1 + \beta = 1 + \frac{-3 - \sqrt{5}}{2} = \frac{2 - 3 - \sqrt{5}}{2} = \frac{-1 - \sqrt{5}}{2}. \]

Calculating \(1 + \alpha\):

\[ 1 + \alpha = 1 + \frac{-3 + \sqrt{5}}{2} = \frac{2 - 3 + \sqrt{5}}{2} = \frac{-1 + \sqrt{5}}{2}. \]

### Step 3: Substitute into the expression

Now we substitute these values into our expression:

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

This simplifies to:

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

### Step 4: Simplify each term

Calculating the first term:

\[ \frac{-3 + \sqrt{5}}{-1 - \sqrt{5}} = \frac{(-3 + \sqrt{5})(-1 + \sqrt{5})}{(-1 - \sqrt{5})(-1 + \sqrt{5})} = \frac{(3 - 3\sqrt{5} + \sqrt{5} - 5)}{1 - 5} = \frac{-2 - 2\sqrt{5}}{-4} = \frac{1 + \sqrt{5}}{2}. \]

Thus,

\[ \left(\frac{-3 + \sqrt{5}}{-1 - \sqrt{5}}\right)^2 = \left(\frac{1 + \sqrt{5}}{2}\right)^2 = \frac{(1 + \sqrt{5})^2}{4} = \frac{1 + 2\sqrt{5} + 5}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}. \]

Calculating the second term similarly gives:

\[ \left(\frac{-3 - \sqrt{5}}{-1 + \sqrt{5}}\right)^2 = \left(\frac{1 - \sqrt{5}}{2}\right)^2 = \frac{(1 - \sqrt{5})^2}{4} = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}. \]

### Step 5: Combine the results

Now we combine both results:

\[ \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = \frac{(3 + \sqrt{5}) + (3 - \sqrt{5})}{2} = \frac{6}{2} = 3. \]

### Final Answer

Thus, the value of

\[ \left(\frac{\alpha}{1 + \beta}\right)^2 + \left(\frac{\beta}{1 + \alpha}\right)^2 \]

is

\[ \boxed{3}. \]