MCQ
The zeros of the polynomial $4\text{x}^2+5\sqrt2\text{x}-3$ are :
- A$-3\sqrt2,\ \sqrt2$
- B$-3\sqrt2,\ \frac{\sqrt2}{2}$
- ✓$\frac{-3\sqrt2}{2},\ \frac{\sqrt2}{4}$
- DNone of these.
✓
Answer
Correct option: C.
$\frac{-3\sqrt2}{2},\ \frac{\sqrt2}{4}$
$\text{f}(\text{x})=4\text{x}^2+5\sqrt2\text{x}-3$
$=4\text{x}^2+6\sqrt2\text{x}-\sqrt2\text{x}-3$
$=2\sqrt2\text{x}\big(\sqrt2\text{x}+3\big)-1\big(\sqrt2\text{x}+3\big)$
$=\big(2\sqrt2\text{x}-1\big)\big(\sqrt2\text{x}+3\big)$
$\therefore\text{f}(\text{x})=0$
$\Rightarrow\big(2\sqrt2\text{x}-1\big)\big(\sqrt2\text{x}+3\big)=0$
$\Rightarrow2\sqrt2\text{x}-1=0$ or $\sqrt2\text{x}+3=0$
$\Rightarrow\text{x}=\frac{1}{2\sqrt2}$ or $\text{x}=-\frac{3}{\sqrt2}$
$\Rightarrow\text{x}=\frac{\sqrt2}{4}$ or $\text{x}=-\frac{3\sqrt2}{2}$
So, the zeros of given polynomial are $\frac{\sqrt2}{4}$ and $-\frac{3\sqrt2}{2}$
$=4\text{x}^2+6\sqrt2\text{x}-\sqrt2\text{x}-3$
$=2\sqrt2\text{x}\big(\sqrt2\text{x}+3\big)-1\big(\sqrt2\text{x}+3\big)$
$=\big(2\sqrt2\text{x}-1\big)\big(\sqrt2\text{x}+3\big)$
$\therefore\text{f}(\text{x})=0$
$\Rightarrow\big(2\sqrt2\text{x}-1\big)\big(\sqrt2\text{x}+3\big)=0$
$\Rightarrow2\sqrt2\text{x}-1=0$ or $\sqrt2\text{x}+3=0$
$\Rightarrow\text{x}=\frac{1}{2\sqrt2}$ or $\text{x}=-\frac{3}{\sqrt2}$
$\Rightarrow\text{x}=\frac{\sqrt2}{4}$ or $\text{x}=-\frac{3\sqrt2}{2}$
So, the zeros of given polynomial are $\frac{\sqrt2}{4}$ and $-\frac{3\sqrt2}{2}$
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