If `alpha,beta` are the roots of the equation `2x^(2) - 2(1+n)^(2) x+(1 + n^(2)+ n^(4))=0` then what is the value of `alpha^(2) + beta^(2)` ?

Options

A. `2n^(2)`
B. `2n^(4)`
C. 2
D. `n^(2)`

Explanation

To find the value of \( \alpha^2 + \beta^2 \) for the roots \( \alpha \) and \( \beta \) of the quadratic equation

\[ 2x^2 - 2(1+n)^2 x + (1 + n^2 + n^4) = 0, \]

we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation can be written in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 2 \)
- \( b = -2(1+n)^2 \)
- \( c = 1 + n^2 + n^4 \)

### Step 2: Calculate \( \alpha + \beta \) and \( \alpha \beta \)
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

Calculating \( \alpha + \beta \):
\[ \alpha + \beta = -\frac{-2(1+n)^2}{2} = (1+n)^2 \]

Calculating \( \alpha \beta \):
\[ \alpha \beta = \frac{1 + n^2 + n^4}{2} \]

### Step 3: Use the formula for \( \alpha^2 + \beta^2 \)
We can use the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

Substituting the values we found:
\[ \alpha^2 + \beta^2 = ((1+n)^2)^2 - 2 \cdot \frac{1 + n^2 + n^4}{2} \]

### Step 4: Simplify the expression
First, calculate \( ((1+n)^2)^2 \):
\[ ((1+n)^2)^2 = (1+n)^4 \]

Now, substituting back:
\[ \alpha^2 + \beta^2 = (1+n)^4 - (1 + n^2 + n^4) \]

### Step 5: Expand and simplify
Expanding \( (1+n)^4 \):
\[ (1+n)^4 = 1 + 4n + 6n^2 + 4n^3 + n^4 \]

Now, substituting this into the equation:
\[ \alpha^2 + \beta^2 = (1 + 4n + 6n^2 + 4n^3 + n^4) - (1 + n^2 + n^4) \]

This simplifies to:
\[ \alpha^2 + \beta^2 = 4n + 6n^2 + 4n^3 + n^4 - 1 - n^2 - n^4 \]
\[ = 4n + (6n^2 - n^2) + (4n^3) + (n^4 - n^4) \]
\[ = 4n + 5n^2 + 4n^3 \]

### Final Result
Thus, the value of \( \alpha^2 + \beta^2 \) is:
\[ \alpha^2 + \beta^2 = 5n^2 + 4n + 4n^3 \]