A) \[\frac{ac}{b}\]
B) \[1\]
C) \[\frac{ab}{c}\]
D) \[\frac{bc}{a}\]
E) \[\frac{b}{ac}\]
Correct Answer: E
Solution :
\[\because \]\[\alpha \]and\[\beta \]are the roots of equation \[a{{x}^{2}}+bx+c=0,\] then \[\alpha +\beta =-\frac{b}{a}\]and \[\alpha \,\beta =\frac{c}{a}\] \[\therefore \] \[\frac{1}{a\alpha +b}+\frac{1}{a\beta +b}=\frac{a\alpha +b+a\beta +b}{(a\alpha +b)(a\beta +b)}\] \[=\frac{a(\alpha +\beta )+2b}{{{a}^{2}}\alpha \beta +ab(\alpha +\beta )+{{b}^{2}}}\] \[=\frac{a\left( -\frac{b}{a} \right)+2b}{{{a}^{2}}\left( \frac{c}{a} \right)+ab\left( -\frac{b}{a} \right)+{{b}^{2}}}\] \[=\frac{2b-b}{ac-{{b}^{2}}+{{b}^{2}}}=\frac{b}{ac}\]
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