If `alpha, beta` are the roots of (x-a) (x-b)+c=0,` c ne 0`, then roots of `(alpha beta - c) x^(2) + (alpha + beta) x + 1 = 0` are

Options

A. 1/a,1/b
B. `-1/a, - 1/b`
C. `1/a, -1/b`
D. `-1/a, 1/b`

Answer

To solve the problem step by step, we start with the given equation and find the roots of the new quadratic equation based on the roots of the original equation.

### Step 1: Understand the original equation
The original equation is given as:
\[ (x - a)(x - b) + c = 0 \]
We can expand this equation:
\[ x^2 - (a + b)x + (ab + c) = 0 \]
From this, we can identify the roots \( \alpha \) and \( \beta \).

### Step 2: Find the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = a + b \)
- The product of the roots \( \alpha \beta = ab + c \)

### Step 3: Write the new quadratic equation
We need to find the roots of the new quadratic equation:
\[ (\alpha \beta - c)x^2 + (\alpha + \beta)x + 1 = 0 \]
Substituting the values from Step 2:
\[ ((ab + c) - c)x^2 + (a + b)x + 1 = 0 \]
This simplifies to:
\[ ab x^2 + (a + b)x + 1 = 0 \]

### Step 4: Use the quadratic formula
To find the roots of the equation \( ab x^2 + (a + b)x + 1 = 0 \), we use the quadratic formula:
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Here, \( A = ab \), \( B = a + b \), and \( C = 1 \).

### Step 5: Substitute values into the quadratic formula
Substituting the values into the formula:
\[ x = \frac{-(a + b) \pm \sqrt{(a + b)^2 - 4(ab)(1)}}{2(ab)} \]

### Step 6: Simplify the expression under the square root
Calculating the discriminant:
\[ (a + b)^2 - 4ab = a^2 + 2ab + b^2 - 4ab = a^2 - 2ab + b^2 = (a - b)^2 \]
Thus, the expression becomes:
\[ x = \frac{-(a + b) \pm (a - b)}{2ab} \]

### Step 7: Calculate the two roots
Now we can calculate the two roots:
1. For the plus sign:
\[ x_1 = \frac{-(a + b) + (a - b)}{2ab} = \frac{-2b}{2ab} = -\frac{1}{a} \]

2. For the minus sign:
\[ x_2 = \frac{-(a + b) - (a - b)}{2ab} = \frac{-2a}{2ab} = -\frac{1}{b} \]

### Final Result
The roots of the equation \( (\alpha \beta - c)x^2 + (\alpha + \beta)x + 1 = 0 \) are:
\[ x_1 = -\frac{1}{a}, \quad x_2 = -\frac{1}{b} \]