If `alpha, beta` are the roots of the quadratic equation `ax^(2) + bx + c = 0` then form the quadratic equation whose roots are `palpha, pbeta` where p is a real number.

Answer

To find the quadratic equation whose roots are \( p\alpha \) and \( p\beta \), where \( \alpha \) and \( \beta \) are the roots of the original quadratic equation \( ax^2 + bx + c = 0 \), we can follow these steps:

### Step 1: Identify the sum and product of the original roots
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

### Step 2: Calculate the sum of the new roots
The new roots are \( p\alpha \) and \( p\beta \). Therefore, the sum of the new roots is:
\[ p\alpha + p\beta = p(\alpha + \beta) = p\left(-\frac{b}{a}\right) \]
This simplifies to:
\[ -\frac{bp}{a} \]

### Step 3: Calculate the product of the new roots
The product of the new roots is:
\[ p\alpha \cdot p\beta = p^2(\alpha \beta) = p^2\left(\frac{c}{a}\right) \]
This simplifies to:
\[ \frac{p^2c}{a} \]

### Step 4: Form the new quadratic equation
Using the sum and product of the new roots, we can form the quadratic equation using the standard form:
\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \]
Substituting the values we found:
\[ x^2 - \left(-\frac{bp}{a}\right)x + \frac{p^2c}{a} = 0 \]
This simplifies to:
\[ x^2 + \frac{bp}{a}x + \frac{p^2c}{a} = 0 \]

### Step 5: Multiply through by \( a \) to eliminate the denominator
To write the equation in standard form without fractions, we can multiply through by \( a \):
\[ ax^2 + bpx + p^2c = 0 \]

### Final Result
Thus, the quadratic equation whose roots are \( p\alpha \) and \( p\beta \) is:
\[ ax^2 + bpx + p^2c = 0 \]
where \( p \) is a real number.

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