Question:
If one zero of the quadratic polynomial $(k-1) x^{2}+k x+1$ is $-4$, then the value of $k$ is
(a) $\frac{-5}{4}$
(b) $\frac{5}{4}$
(c) $\frac{-4}{3}$
(d) $\frac{4}{3}$
Solution:
(b) $\frac{5}{4}$
Since $-4$ is a zero of $(k-1) x^{2}+k x+1$, we have:
$(k-1) \times(-4)^{2}+k \times(-4)+1=0$
$=>16 k-16-4 k+1=0$
$=>12 k-15=0$
$=>k=\frac{15^{5}}{12^{4}}$
$=>k=\frac{5}{4}$
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