A) \[\frac{2}{\sqrt{3}}.\frac{\sqrt{3}}{4}\]
B) \[-\frac{2}{\sqrt{3}}.\frac{\sqrt{3}}{4}\]
C) \[\frac{2}{\sqrt{3}}.-\frac{\sqrt{3}}{4}\]
D) \[-\frac{2}{\sqrt{3}}.-\frac{\sqrt{3}}{4}\]
Correct Answer: B
Solution :
| Given polynomial \[=4\sqrt{3}{{x}^{2}}+5x-2\sqrt{3}\] |
| \[=4\sqrt{3}\,\,{{x}^{2}}+(8-3)x-2\sqrt{3}\] |
| (by splitting the middle term) |
| [TRICK |
| Product \[4\sqrt{3}\times 2\sqrt{3}=8\times 3=24\] |
| \[24=2\times 12=6\times 4=8\times 3\] |
| \[\therefore \] We take 8 and 3 as a factors of 24. |
| So, middle term \[5=8-3\]] |
| \[=4\sqrt{3}\,\,{{x}^{2}}+8x-3x-2\sqrt{3}\] |
| \[=4\times (\sqrt{3}x+2)-\sqrt{3}(\sqrt{3}x+2)\] |
| \[=(\sqrt{3}x+2)\,(4x-\sqrt{3})\] |
| For the zeroes of the polynomial, |
| \[(\sqrt{3}x+2)\,\,(4x-\sqrt{3})=0\] |
| \[\Rightarrow \,\,\,\,\sqrt{3}x+2=0\] or \[4x-\sqrt{3}=0\] |
| \[\Rightarrow \,\,\,\,\,x=\frac{-2}{\sqrt{3}}\] or \[x=\frac{\sqrt{3}}{4}\,\,\,\Rightarrow \,\,x=\frac{-2}{\sqrt{3}},\,\frac{\sqrt{3}}{4}\] |
| So, option [b] is correct |
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