Multiple choice

If one of the zeroes of the quadratic polynomial $\left (k-1 \right) x^{2}+kx+1$ is $\left (-3 \right)$, then k equals to:

  1. $\dfrac{4}{3}$
  2. $-\dfrac {4}{3}$
  3. $\dfrac{2}{3}$
  4. $-\dfrac {2}{3}$
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A Correct answer
Explanation

Substitute x = -3 into the polynomial: (k-1)(-3)^2 + k(-3) + 1 = 0. 9(k-1) - 3k + 1 = 0. 9k - 9 - 3k + 1 = 0. 6k = 8. k = 8/6 = 4/3.

AI explanation

Since -3 is a zero of the polynomial, substituting x = -3 into (k-1)x^2 + kx + 1 = 0 gives (k-1)(9) + k(-3) + 1 = 0. Simplifying this yields 9k - 9 - 3k + 1 = 0, so 6k - 8 = 0. Solving for k gives k = 8/6, which reduces to 4/3.

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