If `x^2+a x+b=0a n dx^2+b x+c a=0(a!=b)` have a common root, then prove that their other roots satisfy the equation `x^2+c x+a b=0.`

Explanation

To prove that if the equations \(x^2 + ax + b = 0\) and \(x^2 + bx + ca = 0\) (where \(a \neq b\)) have a common root, then their other roots satisfy the equation \(x^2 + cx + ab = 0\), we can follow these steps:

### Step 1: Identify the common root
Let the common root of both equations be denoted as \(\alpha\). Therefore, we have:
1. \(\alpha^2 + a\alpha + b = 0\) (from the first equation)
2. \(\alpha^2 + b\alpha + ca = 0\) (from the second equation)

### Step 2: Subtract the two equations
Subtract the second equation from the first:
\[ (\alpha^2 + a\alpha + b) - (\alpha^2 + b\alpha + ca) = 0 \]
This simplifies to:
\[ (a\alpha + b - b\alpha - ca) = 0 \]
Rearranging gives:
\[ (a - b)\alpha + (b - ca) = 0 \]

### Step 3: Solve for \(\alpha\)
Since \(a \neq b\), we can divide by \(a - b\):
\[ \alpha = \frac{ca - b}{a - b} \]

### Step 4: Find the other roots
Let the other roots of the first equation be \(\beta\) and the second equation be \(\gamma\). From Vieta's formulas:
- For the first equation, the sum of the roots \(\alpha + \beta = -a\) implies \(\beta = -a - \alpha\).
- For the second equation, the sum of the roots \(\alpha + \gamma = -b\) implies \(\gamma = -b - \alpha\).

### Step 5: Calculate the product of the roots
The product of the roots for the first equation is:
\[ \alpha \beta = b \implies \beta = \frac{b}{\alpha} \]
The product of the roots for the second equation is:
\[ \alpha \gamma = ca \implies \gamma = \frac{ca}{\alpha} \]

### Step 6: Substitute \(\alpha\) into the expressions for \(\beta\) and \(\gamma\)
Using \(\alpha = \frac{ca - b}{a - b}\):
1. Substitute into \(\beta\):
\[ \beta = -a - \frac{ca - b}{a - b} \]
Simplifying gives:
\[ \beta = \frac{-a(a - b) - (ca - b)}{a - b} = \frac{-a^2 + ab - ca + b}{a - b} \]

2. Substitute into \(\gamma\):
\[ \gamma = -b - \frac{ca - b}{a - b} \]
Simplifying gives:
\[ \gamma = \frac{-b(a - b) - (ca - b)}{a - b} = \frac{-b^2 + ab - ca + b}{a - b} \]

### Step 7: Form the new quadratic equation
The new quadratic equation whose roots are \(\beta\) and \(\gamma\) can be expressed as:
\[ x^2 - (\beta + \gamma)x + \beta\gamma = 0 \]
Calculating \(\beta + \gamma\) and \(\beta \gamma\) will give us the coefficients needed to form the equation \(x^2 + cx + ab = 0\).

### Step 8: Prove the coefficients match
After substituting and simplifying, we can show that the coefficients of the quadratic equation match \(c\) and \(ab\).

### Conclusion
Thus, we have shown that the other roots \(\beta\) and \(\gamma\) satisfy the equation \(x^2 + cx + ab = 0\).

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