Let `a,b in N, a != b` and the two quadratic equations `(a-1)x^2-(a^2+2)x+a^2+2a=0 and (b-1)x^2-(b^2+2)x+(b^2+2b)=0` have a common root. The value of `ab` is

Options

A. `4`
B. `6`
C. `8`
D. `10

Explanation

To solve the problem, we need to find the value of \( ab \) given that the two quadratic equations have a common root. Let's denote the common root as \( \alpha \).

### Step 1: Write down the two quadratic equations
The two quadratic equations are:
1. \((a-1)x^2 - (a^2 + 2)x + (a^2 + 2a) = 0\)
2. \((b-1)x^2 - (b^2 + 2)x + (b^2 + 2b) = 0\)

### Step 2: Substitute the common root into both equations
Since \( \alpha \) is a common root, we can substitute \( \alpha \) into both equations:
1. \((a-1)\alpha^2 - (a^2 + 2)\alpha + (a^2 + 2a) = 0\) (Equation 1)
2. \((b-1)\alpha^2 - (b^2 + 2)\alpha + (b^2 + 2b) = 0\) (Equation 2)

### Step 3: Multiply the equations to eliminate \( \alpha^2 \)
Multiply Equation 1 by \( (b-1) \) and Equation 2 by \( (a-1) \):
- From Equation 1: \((b-1)((a-1)\alpha^2 - (a^2 + 2)\alpha + (a^2 + 2a)) = 0\)
- From Equation 2: \((a-1)((b-1)\alpha^2 - (b^2 + 2)\alpha + (b^2 + 2b)) = 0\)

### Step 4: Subtract the two equations
Now we subtract the two equations:
\[ (b-1)(a-1)\alpha^2 - (b-1)(a^2 + 2)\alpha + (b-1)(a^2 + 2a) - \left[(a-1)(b-1)\alpha^2 - (a-1)(b^2 + 2)\alpha + (a-1)(b^2 + 2b)\right] = 0 \]
This simplifies to:
\[ (a-b)(ab - a - b + 2)\alpha = 0 \]

### Step 5: Analyze the equation
Since \( a \neq b \) (given), we can set:
\[ ab - a - b + 2 = 0 \]
This leads to:
\[ ab = a + b - 2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ ab - a - b + 2 = 0 \implies ab - a - b = -2 \]

### Step 7: Factor the equation
We can rewrite it as:
\[ (a-1)(b-1) = 3 \]
The pairs of natural numbers that satisfy this equation are:
- \( (1, 3) \)
- \( (3, 1) \)
- \( (2, 2) \) (not valid since \( a \neq b \))

### Step 8: Solve for \( a \) and \( b \)
From \( (a-1)(b-1) = 3 \):
1. If \( a-1 = 1 \) and \( b-1 = 3 \), then \( a = 2 \) and \( b = 4 \).
2. If \( a-1 = 3 \) and \( b-1 = 1 \), then \( a = 4 \) and \( b = 2 \).

### Step 9: Calculate \( ab \)
In both cases:
\[ ab = 2 \times 4 = 8 \]

### Final Answer
Thus, the value of \( ab \) is \( \boxed{8} \).