Question:
If one of the zeroes of the quadratic polynomial $(k-1) x^{2}+k x+1$ is $-3$, then the value of $k$ is
(a) $\frac{4}{3}$
(b) $\frac{-4}{3}$
(c) $\frac{2}{3}$
(d) $\frac{-2}{3}$
Solution:
(a) Given that, one of the zeroes of the quadratic polynomial say p(x) = (k- 1)x2 + kx + 1
is $-3$, then $\quad p(-3)=0$
$\Rightarrow \quad(k-1)(-3)^{2}+k(-3)+1=0$
$\Rightarrow \quad 9(k-1)-3 k+1=0$
$\Rightarrow \quad 9 k-9-3 k+1=0$
$\Rightarrow \quad 6 k-8=0$
$\therefore$ $k=4 / 3$
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