The roots `alpha and beta` of the quadratic equation `px^(2) + qx + r = 0` are real and of opposite signs. The roots of `alpha(x-beta)^(2) + beta(x-alpha)^(2) = 0` are:

Options

A. positive
B. negative
C. ral and of opposite sign
D. imaginary

Answer

To solve the problem, we need to analyze the given quadratic equation and derive the roots of the new equation step by step.

### Step 1: Understand the given equation
We have the quadratic equation:
\[ px^2 + qx + r = 0 \]
with roots \( \alpha \) and \( \beta \) that are real and of opposite signs.

### Step 2: Write the new equation
The new equation we need to solve is:
\[ \alpha(x - \beta)^2 + \beta(x - \alpha)^2 = 0 \]

### Step 3: Expand the equation
Let's expand the equation:
1. Expand \( (x - \beta)^2 \):
\[ (x - \beta)^2 = x^2 - 2\beta x + \beta^2 \]
2. Expand \( (x - \alpha)^2 \):
\[ (x - \alpha)^2 = x^2 - 2\alpha x + \alpha^2 \]

Now substituting these expansions into the equation:
\[ \alpha(x^2 - 2\beta x + \beta^2) + \beta(x^2 - 2\alpha x + \alpha^2) = 0 \]
This simplifies to:
\[ \alpha x^2 - 2\alpha \beta x + \alpha \beta^2 + \beta x^2 - 2\beta \alpha x + \beta \alpha^2 = 0 \]

### Step 4: Combine like terms
Combine the terms:
\[ (\alpha + \beta)x^2 - 2(\alpha + \beta) x + (\alpha \beta^2 + \beta \alpha^2) = 0 \]
This can be rewritten as:
\[ (\alpha + \beta)x^2 - 2(\alpha + \beta)x + \alpha \beta(\alpha + \beta) = 0 \]

### Step 5: Factor out common terms
Factoring out \( (\alpha + \beta) \):
\[ (\alpha + \beta)(x^2 - 2x + \alpha \beta) = 0 \]

### Step 6: Solve for roots
The roots of the equation are found by setting each factor to zero:
1. \( \alpha + \beta = 0 \) (which is not possible since \( \alpha \) and \( \beta \) are of opposite signs)
2. Solve \( x^2 - 2x + \alpha \beta = 0 \)

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -2, c = \alpha \beta \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot \alpha \beta}}{2 \cdot 1} \]
\[ x = \frac{2 \pm \sqrt{4 - 4\alpha \beta}}{2} \]
\[ x = 1 \pm \sqrt{1 - \alpha \beta} \]

### Step 7: Analyze the roots
Since \( \alpha \) and \( \beta \) are of opposite signs, their product \( \alpha \beta < 0 \). Thus, \( 1 - \alpha \beta > 1 \), which means the square root is real and positive.

### Conclusion
The roots of the equation \( \alpha(x - \beta)^2 + \beta(x - \alpha)^2 = 0 \) are:
\[ x = 1 + \sqrt{1 - \alpha \beta} \quad \text{and} \quad x = 1 - \sqrt{1 - \alpha \beta} \]