If `alpha,beta` are the roots of the equation `x^(2) + Px + P^(3) = 0, P ne 0` such that `alpha =beta^(2)` then the roots of the given equation are

Options

A. a) 4, -2
B. b) 4, 2
C. c) 1, -1
D. d) 1,1

Answer

To solve the problem, we start with the given quadratic equation:

\[ x^2 + Px + P^3 = 0 \]

where \( P \neq 0 \) and the roots \( \alpha \) and \( \beta \) satisfy the condition \( \alpha = \beta^2 \).

### Step 1: Use Vieta's Formulas
From Vieta's formulas, we know:

1. The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -P \) (Equation 1)
2. The product of the roots \( \alpha \beta = \frac{c}{a} = P^3 \) (Equation 2)

### Step 2: Substitute \( \alpha \) with \( \beta^2 \)
Given that \( \alpha = \beta^2 \), we can substitute this into Equation 2:

\[ \alpha \beta = \beta^2 \cdot \beta = \beta^3 = P^3 \]

Thus, we have:

\[ \beta^3 = P^3 \]

### Step 3: Solve for \( \beta \)
Taking the cube root of both sides, we find:

\[ \beta = P \]

### Step 4: Substitute \( \beta \) back to find \( \alpha \)
Now, substitute \( \beta = P \) back into Equation 1 to find \( \alpha \):

\[ \alpha + P = -P \]

This simplifies to:

\[ \alpha = -P - P = -2P \]

### Step 5: Verify the relationship \( \alpha = \beta^2 \)
Now we check if \( \alpha = \beta^2 \):

\[ \alpha = -2P \]
\[ \beta^2 = P^2 \]

Setting these equal gives:

\[ -2P = P^2 \]

### Step 6: Rearranging the equation
Rearranging this equation leads to:

\[ P^2 + 2P = 0 \]

Factoring out \( P \):

\[ P(P + 2) = 0 \]

Since \( P \neq 0 \), we have:

\[ P + 2 = 0 \]
\[ P = -2 \]

### Step 7: Find the roots
Now substituting \( P = -2 \) back into our expressions for \( \alpha \) and \( \beta \):

1. \( \beta = P = -2 \)
2. \( \alpha = -2P = -2(-2) = 4 \)

Thus, the roots of the equation are:

\[ \alpha = 4 \quad \text{and} \quad \beta = -2 \]

### Final Answer
The roots of the given equation are \( 4 \) and \( -2 \).

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