A) \[-\frac{1}{2},\frac{3}{2}\]
B) \[\frac{1}{2},-\frac{3}{2}\]
C) \[1,-\frac{3}{2}\]
D) \[1,\frac{\sqrt{3}}{2}\]
Correct Answer: B
Solution :
| [b] The zeroes of the polynomial \[f(x)={{x}^{2}}+x-\frac{3}{4}\]are given by \[f(x)=0\]. |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+x-\frac{3}{4}=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,4{{x}^{2}}+4x-3=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,4{{x}^{2}}+6x-2x-3=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,2x(2x+3)-1(2x+3)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,(2x-1)\,\,(2x+3)=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,2x-1=0\] or \[2x+3=0\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,x=\frac{1}{2}\] or \[x=-\frac{3}{2}\] |
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