If one of the zeros of the quadratic polynomial $$(k — 1)x^2 + kx + 1$$ is -3 then the value of $$k =$$
Given
one of the zeros of the quadratic polynomial $$(k — 1)x^2 + kx + 1$$ is -3
so put value of x=-3
$$(k — 1)(-3)^2 + k(-3) + 1=0$$
$$(k — 1)9 - 3k + 1=0$$
$$9k-9 - 3k + 1=0$$
$$6k-8=0$$
$$k=\frac{4}{3}$$
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