If ` alpha, beta` are the roots of the equation ` ax^(2) +2bx +c =0 and alpha +h, beta + h` are the roots of the equation ` Ax^(2) +2Bx + C=0` then

Options

A. `h= b/ a-B/A`
B. `(b^(2)-ac)/(B^(2)-AC)=a^(2)/A^(2)`
C. ` h = (Ac +aC)/(A+a)`
D. `(b^(2)- aC)/(B^(2) -AC)= a/A `

Answer

To solve the problem step by step, we start with the given information about the roots of the equations.

### Step 1: Identify the roots of the first equation
The roots of the equation \( ax^2 + 2bx + c = 0 \) are given as \( \alpha \) and \( \beta \). By Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{2b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

### Step 2: Identify the roots of the second equation
The roots of the equation \( Ax^2 + 2Bx + C = 0 \) are given as \( \alpha + h \) and \( \beta + h \). Again, using Vieta's formulas:
- The sum of the roots \( (\alpha + h) + (\beta + h) = -\frac{2B}{A} \)
- The product of the roots \( (\alpha + h)(\beta + h) = \frac{C}{A} \)

### Step 3: Set up the equation for the sum of the roots
From the sum of the roots of the second equation, we have:
\[ \alpha + \beta + 2h = -\frac{2B}{A} \]
Substituting the value of \( \alpha + \beta \) from Step 1:
\[ -\frac{2b}{a} + 2h = -\frac{2B}{A} \]

### Step 4: Solve for \( h \)
Rearranging the equation to isolate \( h \):
\[ 2h = -\frac{2B}{A} + \frac{2b}{a} \]
Dividing everything by 2:
\[ h = -\frac{B}{A} + \frac{b}{a} \]

### Step 5: Final expression for \( h \)
Thus, we can express \( h \) as:
\[ h = \frac{b}{a} - \frac{B}{A} \]

### Conclusion
The final result is:
\[ h = \frac{b}{a} - \frac{B}{A} \]