The equation whose roots are squares of the roots of ` x^3 + 2x ^2 -x +3=0` is

Options

A. `x^3 - 6x^2 -11 x -9=0`
B. `x^3 - 6x^2 -11x +9=0`
C. `x^3 + 6x -11x -9=0`
D. `x^3 +6x ^2 +11 x +9=0`

Answer

To find the equation whose roots are the squares of the roots of the given polynomial \( x^3 + 2x^2 - x + 3 = 0 \), we will follow these steps:

### Step 1: Identify the coefficients of the given polynomial
The given polynomial is:
\[ x^3 + 2x^2 - x + 3 = 0 \]
From this, we can identify the coefficients:
- \( a = 1 \)
- \( b = 2 \)
- \( c = -1 \)
- \( d = 3 \)

### Step 2: Use Vieta's formulas to find sums of roots
Let the roots of the polynomial be \( \alpha, \beta, \gamma \). According to Vieta's formulas:
- The sum of the roots:
\[ \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{2}{1} = -2 \]
- The sum of the products of the roots taken two at a time:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{-1}{1} = -1 \]
- The product of the roots:
\[ \alpha\beta\gamma = -\frac{d}{a} = -\frac{3}{1} = -3 \]

### Step 3: Calculate the new roots
The new roots we need to find are \( \alpha^2, \beta^2, \gamma^2 \).

#### Step 3.1: Find \( \alpha^2 + \beta^2 + \gamma^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Substituting the values:
\[ \alpha^2 + \beta^2 + \gamma^2 = (-2)^2 - 2(-1) = 4 + 2 = 6 \]

#### Step 3.2: Find \( \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 \)
Using the identity:
\[ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) \]
Substituting the values:
\[ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (-1)^2 - 2(-3)(-2) = 1 - 12 = -11 \]

#### Step 3.3: Find \( \alpha^2\beta^2\gamma^2 \)
This is simply:
\[ \alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (-3)^2 = 9 \]

### Step 4: Form the new polynomial
The new polynomial whose roots are \( \alpha^2, \beta^2, \gamma^2 \) can be expressed as:
\[ x^3 - (\alpha^2 + \beta^2 + \gamma^2)x^2 + (\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2)x - \alpha^2\beta^2\gamma^2 = 0 \]
Substituting the values we found:
\[ x^3 - 6x^2 - 11x - 9 = 0 \]

### Final Answer
The equation whose roots are the squares of the roots of the original polynomial is:
\[ \boxed{x^3 - 6x^2 - 11x - 9 = 0} \]