Given the quadratic equation `(k-1)x^(2)-kx+1=0`, (where `kne1)` , find k so that product of the roots is -3.

Answer

To solve the problem, we need to find the value of \( k \) in the quadratic equation \( (k-1)x^2 - kx + 1 = 0 \) such that the product of the roots is \( -3 \).

### Step 1: Identify the coefficients
The given quadratic equation can be expressed in the standard form \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = k - 1 \)
- \( b = -k \)
- \( c = 1 \)

### Step 2: Use the product of roots formula
For any quadratic equation \( ax^2 + bx + c = 0 \), the product of the roots \( \alpha \) and \( \beta \) is given by:
\[ \alpha \beta = \frac{c}{a} \]
In our case, we know that the product of the roots is \( -3 \). Therefore, we can write:
\[ \alpha \beta = \frac{1}{k - 1} \]

### Step 3: Set up the equation
Since we want the product of the roots to equal \( -3 \), we can set up the equation:
\[ \frac{1}{k - 1} = -3 \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ 1 = -3(k - 1) \]

### Step 5: Distribute and rearrange the equation
Distributing the \( -3 \) on the right side:
\[ 1 = -3k + 3 \]
Now, rearranging the equation to isolate \( k \):
\[ -3k = 1 - 3 \]
\[ -3k = -2 \]

### Step 6: Solve for \( k \)
Dividing both sides by \( -3 \):
\[ k = \frac{-2}{-3} = \frac{2}{3} \]

### Final Answer
Thus, the value of \( k \) such that the product of the roots is \( -3 \) is:
\[ \boxed{\frac{2}{3}} \]