If one of the zeroes of the quadratic polynomial `(k-1)x^2+kx+1` is -3, then find the value of k:

Options

A. `3//4`
B. `4//3`
C. `3//2`
D. `2//4`

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 \( 9 \), so we can rewrite the equation as:

\[ (k-1) \cdot 9 - 3k + 1 = 0 \]

This simplifies to:

\[ 9k - 9 - 3k + 1 = 0 \]

### Step 3: Combine like terms
Now, combine the terms involving \( k \) and the constant terms:

\[ (9k - 3k) + (-9 + 1) = 0 \]

This simplifies to:

\[ 6k - 8 = 0 \]

### Step 4: Solve for \( k \)
To isolate \( k \), add \( 8 \) to both sides:

\[ 6k = 8 \]

Now, divide both sides by \( 6 \):

\[ k = \frac{8}{6} = \frac{4}{3} \]

### Final Answer
Thus, the value of \( k \) is:

\[ \boxed{\frac{4}{3}} \]

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