If the equation `x^(2)+ax+b=0` and `x^(2)+bx+a=0` have a common root, then their other roots satisfy the equation

Options

A. `x^(2) + x + ab = 0`
B. `x^(2) - x + ab = 0`
C. `x^(2) - x - ab = 0`
D. `x^(2) - x - ab = 0`

Answer

To solve the problem, we need to find the quadratic equation whose roots are the other roots of the two given equations, which share a common root. Let's denote the common root as \( \alpha \), and the other roots as \( \beta \) and \( \gamma \).

### Step 1: Set Up the Equations
We have two quadratic equations:
1. \( x^2 + ax + b = 0 \)
2. \( x^2 + bx + a = 0 \)

### Step 2: Identify the Common Root
Let \( \alpha \) be the common root. Then, substituting \( \alpha \) into both equations gives us:
1. \( \alpha^2 + a\alpha + b = 0 \) (Equation 1)
2. \( \alpha^2 + b\alpha + a = 0 \) (Equation 2)

### Step 3: Set the Equations Equal
Since both equations equal zero, we can set them equal to each other:
\[ \alpha^2 + a\alpha + b = \alpha^2 + b\alpha + a \]
Cancel \( \alpha^2 \) from both sides:
\[ a\alpha + b = b\alpha + a \]

### Step 4: Rearrange the Equation
Rearranging gives:
\[ a\alpha - b\alpha = a - b \]
Factoring out \( \alpha \):
\[ (a - b)\alpha = a - b \]

### Step 5: Solve for \( \alpha \)
Assuming \( a \neq b \) (if \( a = b \), both equations would be identical), we can divide both sides by \( a - b \):
\[ \alpha = 1 \]

### Step 6: Substitute \( \alpha \) Back
Now, substituting \( \alpha = 1 \) back into either of the original equations to find \( a + b \):
\[ 1^2 + a(1) + b = 0 \implies 1 + a + b = 0 \implies a + b = -1 \quad \text{(Equation 1)} \]

### Step 7: Find the Other Roots
Next, we need to find the other roots \( \beta \) and \( \gamma \):
- For the first equation \( x^2 + ax + b = 0 \):
- The sum of the roots \( \alpha + \beta = -a \)
- Since \( \alpha = 1 \):
\[ 1 + \beta = -a \implies \beta = -a - 1 \quad \text{(Equation 2)} \]

- For the second equation \( x^2 + bx + a = 0 \):
- The sum of the roots \( \alpha + \gamma = -b \)
- Again, since \( \alpha = 1 \):
\[ 1 + \gamma = -b \implies \gamma = -b - 1 \quad \text{(Equation 3)} \]

### Step 8: Sum and Product of the Roots
Now we can find the sum and product of the roots \( \beta \) and \( \gamma \):
- Sum of the roots:
\[ \beta + \gamma = (-a - 1) + (-b - 1) = -a - b - 2 \]
Using \( a + b = -1 \):
\[ \beta + \gamma = -(-1) - 2 = 1 - 2 = -1 \]

- Product of the roots:
\[ \beta \cdot \gamma = (-a - 1)(-b - 1) = (a + 1)(b + 1) = ab + a + b + 1 \]
Using \( a + b = -1 \):
\[ \beta \cdot \gamma = ab - 1 + 1 = ab \]

### Step 9: Form the Quadratic Equation
The quadratic equation with roots \( \beta \) and \( \gamma \) is given by:
\[ x^2 - (\beta + \gamma)x + \beta\gamma = 0 \]
Substituting the values we found:
\[ x^2 - (-1)x + ab = 0 \implies x^2 + x + ab = 0 \]

### Final Answer
Thus, the required quadratic equation is:
\[ x^2 + x + ab = 0 \]