MCQ
The zeros of the polynomial $\text{x}^2+\frac{1}{6}\text{x}-2$ are :
- A$-3,\ 4$
- ✓$\frac{-3}{2},\ \frac{4}{3}$
- C$\frac{-4}{3},\ \frac{3}{2}$
- DNone of these.
✓
Answer
Correct option: B.
$\frac{-3}{2},\ \frac{4}{3}$
$\text{f}(\text{x})=\text{x}^2+\frac{1}{6}\text{x}-2$
Now $, f(x) = 0$
$\Rightarrow\text{x}^2+\frac{1}{6}\text{x}-2=0$
$\Rightarrow 6 x^2+x-12=0$
$\Rightarrow 6 x^2+9 x-8 x-12=0$
$\Rightarrow 3 x(2 x+3)-4(2 x+3)=0$
$\Rightarrow(2 x+3)(3 x-4)=0$
$\Rightarrow 2 x+3=0 $ or $ 3 x-4=0$
$\Rightarrow\text{x}=-\frac{3}{2}$ or $\text{x}=\frac{4}{3}$
So, the zeros of given polynomial are $-\frac{3}{2}$ and $-\frac{4}{3}$
Now $, f(x) = 0$
$\Rightarrow\text{x}^2+\frac{1}{6}\text{x}-2=0$
$\Rightarrow 6 x^2+x-12=0$
$\Rightarrow 6 x^2+9 x-8 x-12=0$
$\Rightarrow 3 x(2 x+3)-4(2 x+3)=0$
$\Rightarrow(2 x+3)(3 x-4)=0$
$\Rightarrow 2 x+3=0 $ or $ 3 x-4=0$
$\Rightarrow\text{x}=-\frac{3}{2}$ or $\text{x}=\frac{4}{3}$
So, the zeros of given polynomial are $-\frac{3}{2}$ and $-\frac{4}{3}$
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