Question

if one of the zeros of the quadratic polynomial (k-1)x^2+kx+1 is 3 then the value of k is
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Answer 1

Answer:

$k = \frac{2}{3}$

Step-by-step explanation:

$(k - 1) {x}^{2} + kx + 1 = 0$

$(k - 1)( {3)}^{2} + k(3) + 1 = 0$

$(k - 1)(9) + 3k + 1 = 0$

$9k - 9 + 3k + 1 = 0$

$12k - 8 = 0$

$12k = 8$

$k = \frac{8}{12}$

$k = \frac{2}{3}$

I hope you got your answer.