If `alpha, beta` are the roots of `ax^(2) + bx + c = 0` then the equation with roots `(1)/(aalpha+b), (1)/(abeta+b)` is

Options

A. `acx^(2)+bx+1` =0
B. `acx^(2)+bx-1=0`
C. `acx^(2)-bx-1=0`
D. `acx^(2)-bx+1=0`

Answer

To find the quadratic equation with roots \( \frac{1}{a\alpha + b} \) and \( \frac{1}{a\beta + b} \), where \( \alpha \) and \( \beta \) are the roots of the equation \( ax^2 + bx + c = 0 \), we will follow these steps:

### Step 1: Find the sum of the new roots
The sum of the new roots \( S \) is given by:
\[ S = \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} \]
To combine these fractions, we find a common denominator:
\[ S = \frac{(a\beta + b) + (a\alpha + b)}{(a\alpha + b)(a\beta + b)} \]
This simplifies to:
\[ S = \frac{a(\alpha + \beta) + 2b}{(a\alpha + b)(a\beta + b)} \]

### Step 2: Substitute the values of \( \alpha + \beta \) and \( \alpha \beta \)
Using Vieta's formulas, we know:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a} \]
Substituting these into the sum:
\[ S = \frac{a\left(-\frac{b}{a}\right) + 2b}{(a\alpha + b)(a\beta + b)} = \frac{-b + 2b}{(a\alpha + b)(a\beta + b)} = \frac{b}{(a\alpha + b)(a\beta + b)} \]

### Step 3: Find the product of the new roots
The product of the new roots \( P \) is given by:
\[ P = \frac{1}{(a\alpha + b)(a\beta + b)} \]

### Step 4: Expand the denominator
We can expand \( (a\alpha + b)(a\beta + b) \):
\[ (a\alpha + b)(a\beta + b) = a^2\alpha\beta + ab(\alpha + \beta) + b^2 \]
Substituting the values of \( \alpha + \beta \) and \( \alpha \beta \):
\[ = a^2\left(\frac{c}{a}\right) + ab\left(-\frac{b}{a}\right) + b^2 = ac - b^2 + b^2 = ac \]
Thus, the product becomes:
\[ P = \frac{1}{ac} \]

### Step 5: Form the quadratic equation
Using the sum and product of the roots, we can write the quadratic equation in the standard form:
\[ x^2 - Sx + P = 0 \]
Substituting \( S \) and \( P \):
\[ x^2 - \frac{b}{ac}x + \frac{1}{ac} = 0 \]
To eliminate the fractions, multiply through by \( ac \):
\[ acx^2 - bx + 1 = 0 \]

### Final Answer
The required quadratic equation is:
\[ acx^2 - bx + 1 = 0 \]