If one of the zeroes of the quadratic polynomial \((k-1)x^2 + kx + 1\) is \(-3\), then the value of k is
\(\dfrac{4}{3}\)
\(-\dfrac{4}{3}\)
\(\dfrac{2}{3}\)
\(-\dfrac{2}{3}\)
Step 1: Recall the condition.
If \(-3\) is a zero of the polynomial \((k-1)x^2 + kx + 1\), then by definition:
\((k-1)(-3)^2 + k(-3) + 1 = 0\).
Step 2: Simplify the expression.
\((k-1)(9) - 3k + 1 = 0\)
\(9k - 9 - 3k + 1 = 0\)
Step 3: Combine like terms.
\(9k - 3k = 6k\)
\(-9 + 1 = -8\)
So the equation becomes \(6k - 8 = 0\).
Step 4: Solve for \(k\).
\(6k = 8\)
\(k = \dfrac{8}{6}\)
\(k = \dfrac{4}{3}\)
Final Answer: The required value of \(k\) is \(\dfrac{4}{3}\).