Answer:
k = 1¹/₃
Explanation:
Comparing (k-1)x² + kx + 1 with ax + bx + c where α, β be the zeroes of the quadratic equation, then
α + β = -b/a = -k/(k - 1) and αβ = c/a = 1/(k - 1)
Since one of the zeros is -3, β = -3
So,
α + β = -k/(k - 1)
α + (-3) = -k/(k - 1)
α - 3 = -k/(k - 1) (1)
and
αβ = 1/(k - 1)
-3α = 1/(k - 1) (2)
From (1), α = 3 - k/(k - 1) (3)
Substituting equation (3) into (2), we have
-3α = 1/(k - 1)
-3[3 - k/(k - 1)] = 1/(k - 1)
-9 + 3k/(k - 1) = 1/(k - 1)
-9 = 1/(k - 1) - 3k/(k - 1)
-9 = (1 - 3k)/(k - 1)
cross-multiplying, we have
-9(k - 1) = 1 - 3k
expanding the brackets, we have
-9k + 9 = 1 - 3k
collecting like terms, we have
-9k + 3k = 1 - 9
-6k = -8
dividing through by -6, we have
k = -8/-6
k = 4/3
k = 1¹/₃