What is the value of k, if one of the zeroes of the quadratic polynomial `(k-1)x^(2)+kx+1` is 3?

Options

A. `(4)/(3)`
B. `(2)/(3)`
C. `(1)/(5)`
D. `(5)/(7)`

Answer

To find the value of \( k \) such that one of the zeroes of the quadratic polynomial \( (k-1)x^2 + kx + 1 \) is 3, we can follow these steps:

### Step 1: Substitute the zero into the polynomial
Since 3 is a zero of the polynomial, we can substitute \( x = 3 \) into the polynomial and set it equal to zero:
\[ (k-1)(3^2) + k(3) + 1 = 0 \]

### Step 2: Simplify the equation
Calculating \( 3^2 \) gives us 9, so we can rewrite the equation:
\[ (k-1)(9) + 3k + 1 = 0 \]
This simplifies to:
\[ 9k - 9 + 3k + 1 = 0 \]

### Step 3: Combine like terms
Now, combine the \( k \) terms and the constant terms:
\[ (9k + 3k) + (-9 + 1) = 0 \]
This results in:
\[ 12k - 8 = 0 \]

### Step 4: Solve for \( k \)
Now, isolate \( k \) by adding 8 to both sides:
\[ 12k = 8 \]
Next, divide both sides by 12:
\[ k = \frac{8}{12} \]
This simplifies to:
\[ k = \frac{2}{3} \]

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