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. `18`
B. `19`
C. `20`
D. `21`

Explanation

To solve the problem, we need to find the value of the expression \(\left(\frac{\alpha}{1 + \beta}\right)^{2} + \left(\frac{\beta}{1 + \alpha}\right)^{2}\), where \(\alpha\) and \(\beta\) are the roots of the equation \(x^{2} + 3x + 1 = 0\).

### Step 1: Find the roots \(\alpha\) and \(\beta\)
We can use the quadratic formula to find the roots of the equation \(x^{2} + 3x + 1 = 0\):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 3\), and \(c = 1\).

Calculating the discriminant:
\[ b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 1 = 9 - 4 = 5 \]

Now substituting back 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 \(1 + \alpha\)
Now we calculate \(1 + \beta\) and \(1 + \alpha\):
\[ 1 + \beta = 1 + \frac{-3 - \sqrt{5}}{2} = \frac{2 - 3 - \sqrt{5}}{2} = \frac{-1 - \sqrt{5}}{2} \]
\[ 1 + \alpha = 1 + \frac{-3 + \sqrt{5}}{2} = \frac{2 - 3 + \sqrt{5}}{2} = \frac{-1 + \sqrt{5}}{2} \]

### Step 3: Substitute into the expression
Now we substitute these values into the expression:
\[ \left(\frac{\alpha}{1 + \beta}\right)^{2} + \left(\frac{\beta}{1 + \alpha}\right)^{2} \]
Substituting \(\alpha\) and \(1 + \beta\):
\[ \left(\frac{\frac{-3 + \sqrt{5}}{2}}{\frac{-1 - \sqrt{5}}{2}}\right)^{2} + \left(\frac{\frac{-3 - \sqrt{5}}{2}}{\frac{-1 + \sqrt{5}}{2}}\right)^{2} \]
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
Let’s simplify each fraction:
1. For \(\frac{-3 + \sqrt{5}}{-1 - \sqrt{5}}\):
\[ = \frac{(-3 + \sqrt{5})(-1 + \sqrt{5})}{(-1 - \sqrt{5})(-1 + \sqrt{5})} = \frac{(3 - 3\sqrt{5} + \sqrt{5} - 5)}{1 - 5} = \frac{-2 - 2\sqrt{5}}{-4} = \frac{1 + \sqrt{5}}{2} \]

2. For \(\frac{-3 - \sqrt{5}}{-1 + \sqrt{5}}\):
\[ = \frac{(-3 - \sqrt{5})(-1 - \sqrt{5})}{(-1 + \sqrt{5})(-1 - \sqrt{5})} = \frac{(3 + 3\sqrt{5} + \sqrt{5} + 5)}{1 - 5} = \frac{8 + 4\sqrt{5}}{-4} = -2 - \sqrt{5} \]

### Step 5: Calculate the squares
Now we calculate the squares:
\[ \left(\frac{1 + \sqrt{5}}{2}\right)^{2} + \left(-2 - \sqrt{5}\right)^{2} \]
Calculating each:
\[ \left(\frac{1 + \sqrt{5}}{2}\right)^{2} = \frac{(1 + 2\sqrt{5} + 5)}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} \]
\[ \left(-2 - \sqrt{5}\right)^{2} = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5} \]

### Step 6: Combine the results
Now we combine:
\[ \frac{3 + \sqrt{5}}{2} + 9 + 4\sqrt{5} \]
To combine, convert \(\frac{3 + \sqrt{5}}{2}\) to a common denominator:
\[ = \frac{3 + \sqrt{5} + 18 + 8\sqrt{5}}{2} = \frac{21 + 9\sqrt{5}}{2} \]

### Final Result
The value of the expression \(\left(\frac{\alpha}{1 + \beta}\right)^{2} + \left(\frac{\beta}{1 + \alpha}\right)^{2}\) is \(18\).