Question:
The zeros of the polynomial $x^{2}+\frac{1}{6} x-2$ are
(a) $-3,4$
(b) $\frac{-3}{2}, \frac{4}{3}$
(c) $\frac{-4}{3}, \frac{3}{2}$
(d) none of these
Solution:
(b) $\frac{-3}{2}, \frac{4}{3}$
Let $f(x)=x^{2}+\frac{1}{6} x-2=0$
$=>6 x^{2}+x-12=0$
$=>6 x^{2}+9 x-8 x-12=0$
$=>3 x(2 x+3)-4(2 x+3)=0$
$=>(2 x+3)(3 x-4)=0$
$\therefore x=\frac{-3}{2}$ or $x=\frac{4}{3}$
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