# If one zero of the quadratic polynomial

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:

$(k-1) \times(-4)^{2}+k \times(-4)+1=0$

$=>16 k-16-4 k+1=0$

$=>12 k-15=0$

$=>k=\frac{15^{5}}{12^{4}}$

$=>k=\frac{5}{4}$