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 root of the polynomial, we can substitute \( x = -3 \) into the polynomial:

\[ (k - 1)(-3)^2 + k(-3) + 1 = 0 \]

### Step 2: Simplify the equation
Calculating \( (-3)^2 \) gives \( 9 \), so we can rewrite the equation:

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

This simplifies to:

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

Expanding this gives:

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

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

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

This simplifies to:

\[ 6k - 8 = 0 \]

### Step 4: Solve for \( k \)
Next, we isolate \( k \):

\[ 6k = 8 \]

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

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

### Step 5: Simplify the fraction
We can simplify \( \frac{8}{6} \) to:

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

Thus, the value of \( k \) is:

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