If `alpha, beta` are the roots of `ax^(2)+bx+c=0`, the equation whose roots are `2+ alpha, 2+beta` is

Options

A. `ax^(2)+x(4a-b) +4 a -2b+c=0`
B. `ax^(2)+x(4a-b)+4a +2b+c=0`
C. `ax^(2)+x(b-4a) +4a +2b+c=0`
D. `ax^(2)+x(b-4a)+4a-2b+c=0`

Answer

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

### Step 1: Find the Sum of the New Roots
The new roots are \(2 + \alpha\) and \(2 + \beta\). The sum of these roots can be calculated as follows:

\[ (2 + \alpha) + (2 + \beta) = 4 + (\alpha + \beta) \]

From Vieta's formulas, we know that:

\[ \alpha + \beta = -\frac{b}{a} \]

Thus, substituting this into the equation gives:

\[ \text{Sum of new roots} = 4 - \frac{b}{a} \]

### Step 2: Find the Product of the New Roots
Next, we calculate the product of the new roots:

\[ (2 + \alpha)(2 + \beta) = 2^2 + 2(\alpha + \beta) + \alpha\beta = 4 + 2(\alpha + \beta) + \alpha\beta \]

Again using Vieta's formulas, we substitute:

\[ \alpha \beta = \frac{c}{a} \]

Thus, we have:

\[ \text{Product of new roots} = 4 + 2\left(-\frac{b}{a}\right) + \frac{c}{a} = 4 - \frac{2b}{a} + \frac{c}{a} \]

### Step 3: Form the New Quadratic Equation
The new quadratic equation can be formed using the sum and product of the new roots. The standard form of a quadratic equation with roots \(p\) and \(q\) is:

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

Substituting the values we found:

\[ x^2 - \left(4 - \frac{b}{a}\right)x + \left(4 - \frac{2b}{a} + \frac{c}{a}\right) = 0 \]

### Step 4: Multiply through by \(a\) to eliminate the fraction
To eliminate the fractions, we multiply the entire equation by \(a\):

\[ a x^2 - a\left(4 - \frac{b}{a}\right)x + a\left(4 - \frac{2b}{a} + \frac{c}{a}\right) = 0 \]

This simplifies to:

\[ a x^2 - (4a - b)x + (4a - 2b + c) = 0 \]

### Final Equation
Thus, the equation whose roots are \(2 + \alpha\) and \(2 + \beta\) is:

\[ a x^2 - (4a - b)x + (4a - 2b + c) = 0 \]