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