IF `alpha , beta ` are the roots of the equation `ax^2+ bx +c=0`then the quadratic equation whose roots are `alpha + beta , alpha beta ` is

Options

A. `a^(2)x^(2)+a(b-c)x+bc=0`
B. `a^(2)x^(2)+a(b-c)x-bc=0`
C. `a^(2)x^(2)+(b+c)x+bc=0`
D. `a^(2)x^(2)-(b+c)x+bc=0`

Answer

To find the quadratic equation whose roots are \( \alpha + \beta \) and \( \alpha \beta \), we start with the given quadratic equation:

\[ ax^2 + bx + c = 0 \]

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

### Step 2: Form the new quadratic equation
We need to form a new quadratic equation with roots \( \alpha + \beta \) and \( \alpha \beta \). The general form of a quadratic equation with roots \( p \) and \( q \) is given by:

\[ x^2 - (p + q)x + pq = 0 \]

Here, let:
- \( p = \alpha + \beta = -\frac{b}{a} \)
- \( q = \alpha \beta = \frac{c}{a} \)

### Step 3: Calculate \( p + q \) and \( pq \)
Now we calculate:
1. \( p + q = \alpha + \beta + \alpha \beta = -\frac{b}{a} + \frac{c}{a} = \frac{c - b}{a} \)
2. \( pq = (\alpha + \beta)(\alpha \beta) = \left(-\frac{b}{a}\right)\left(\frac{c}{a}\right) = -\frac{bc}{a^2} \)

### Step 4: Substitute into the quadratic form
Substituting \( p + q \) and \( pq \) into the quadratic equation form gives:

\[ x^2 - \left(\frac{c - b}{a}\right)x - \left(-\frac{bc}{a^2}\right) = 0 \]

This simplifies to:

\[ x^2 - \frac{c - b}{a}x + \frac{bc}{a^2} = 0 \]

### Step 5: Multiply through by \( a^2 \) to eliminate fractions
To eliminate the fractions, multiply the entire equation by \( a^2 \):

\[ a^2x^2 - (c - b)ax + bc = 0 \]

### Step 6: Rearranging the equation
This can be rearranged to:

\[ a^2x^2 + (b - c)ax - bc = 0 \]

### Final Result
Thus, the quadratic equation whose roots are \( \alpha + \beta \) and \( \alpha \beta \) is:

\[ a^2x^2 + (b - c)ax - bc = 0 \]