If `alpha, beta` are the roots of `x^(2)+2x-1=0`, then the equation whose roots are `alpha^(2), beta^(2)` is

Options

A. `x^(2)-6x+1=0`
B. `x^(2)+6x-1=0`
C. `x^(2)+6x+1=0`
D. `x^(2)+6x-2=0`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the given quadratic equation and its roots
We start with the quadratic equation:
\[ x^2 + 2x - 1 = 0 \]

Let the roots of this equation be \( \alpha \) and \( \beta \).

### Step 2: Use Vieta's formulas to find \( \alpha + \beta \) and \( \alpha \beta \)
From Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{2}{1} = -2 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{-1}{1} = -1 \)

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

Substituting the values we found:
\[ \alpha^2 + \beta^2 = (-2)^2 - 2(-1) \]
\[ \alpha^2 + \beta^2 = 4 + 2 = 6 \]

### Step 4: Find \( \alpha^2 \beta^2 \)
The product of the squares of the roots can be calculated as:
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = (-1)^2 = 1 \]

### Step 5: Form the new quadratic equation
Now we need to form the quadratic equation whose roots are \( \alpha^2 \) and \( \beta^2 \). The equation can be expressed as:
\[ x^2 - (\alpha^2 + \beta^2)x + \alpha^2 \beta^2 = 0 \]

Substituting the values we calculated:
\[ x^2 - 6x + 1 = 0 \]

### Final Answer
The equation whose roots are \( \alpha^2 \) and \( \beta^2 \) is:
\[ x^2 - 6x + 1 = 0 \]