A) \[-\frac{3}{4}\]
B) \[-\frac{3}{7}\]
C) \[\frac{3}{7}\]
D) \[\frac{4}{7}\]
E) \[\frac{3}{4}\]
Correct Answer: B
Solution :
\[4{{x}^{2}}+3x+7=0\Rightarrow {{x}^{2}}+\frac{3}{4}x+\frac{7}{4}=4\] \[\Rightarrow \] \[\alpha +\beta =-\frac{3}{4},\alpha \beta =\frac{7}{4}\] \[\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=-\frac{3}{7}\]
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