If the equation `ax^2+2bx+c=0` and `ax^2+2cx+b=0`, `a!=o`, `b!=c`, have a common too then their other roots are the roots of the quadratic equation

Options

A. `a^(2) x (x+1) +4bc =0`
B. `a ^(2) x (x+1) +8bc =0`
C. `a ^(2)x( x+2) +8bc=0`
D. `a ^(2)x (1+2x) +8bc =0`

Answer

To solve the problem step by step, we will analyze the two quadratic equations given and find their common root. We will then derive the quadratic equation whose roots are the other roots of the two equations.

### Step 1: Identify the equations
We have two quadratic equations:
1. \( ax^2 + 2bx + c = 0 \) (Equation 1)
2. \( ax^2 + 2cx + b = 0 \) (Equation 2)

### Step 2: Assume a common root
Let \( \alpha \) be the common root of both equations. Thus, substituting \( \alpha \) into both equations gives:
1. \( a\alpha^2 + 2b\alpha + c = 0 \) (1)
2. \( a\alpha^2 + 2c\alpha + b = 0 \) (2)

### Step 3: Set the equations equal
Since both equations equal zero, we can set them equal to each other:
\[ a\alpha^2 + 2b\alpha + c = a\alpha^2 + 2c\alpha + b \]

### Step 4: Simplify the equation
By canceling \( a\alpha^2 \) from both sides, we get:
\[ 2b\alpha + c = 2c\alpha + b \]
Rearranging gives:
\[ 2b\alpha - 2c\alpha = b - c \]
Factoring out \( \alpha \):
\[ \alpha(2b - 2c) = b - c \]

### Step 5: Solve for \( \alpha \)
Assuming \( b \neq c \) (as given), we can divide both sides by \( b - c \):
\[ \alpha = \frac{1}{2} \]

### Step 6: Find the other roots
Now, we need to find the other roots of both equations. The sum and product of the roots can be used here.

For Equation 1:
- Sum of roots \( \alpha + \beta = -\frac{2b}{a} \)
- Product of roots \( \alpha \beta = \frac{c}{a} \)

For Equation 2:
- Sum of roots \( \alpha + \gamma = -\frac{2c}{a} \)
- Product of roots \( \alpha \gamma = \frac{b}{a} \)

### Step 7: Calculate \( \beta \) and \( \gamma \)
From the sum of roots for Equation 1:
\[ \beta = -\frac{2b}{a} - \alpha = -\frac{2b}{a} - \frac{1}{2} \]

From the sum of roots for Equation 2:
\[ \gamma = -\frac{2c}{a} - \alpha = -\frac{2c}{a} - \frac{1}{2} \]

### Step 8: Form the new quadratic equation
We need to find the quadratic equation whose roots are \( \beta \) and \( \gamma \).

The sum of roots \( \beta + \gamma \):
\[ \beta + \gamma = \left(-\frac{2b}{a} - \frac{1}{2}\right) + \left(-\frac{2c}{a} - \frac{1}{2}\right) = -\frac{2(b+c)}{a} - 1 \]

The product of roots \( \beta \gamma \):
Using the product of roots formula:
\[ \beta \gamma = \left(-\frac{2b}{a} - \frac{1}{2}\right)\left(-\frac{2c}{a} - \frac{1}{2}\right) \]

### Step 9: Write the final quadratic equation
The quadratic equation can be expressed as:
\[ x^2 - \left(-\frac{2(b+c)}{a} - 1\right)x + \beta \gamma = 0 \]