A) \[\frac{c(a-b)}{{{a}^{2}}}\]
B) 0
C) \[\frac{-bc}{{{a}^{2}}}\]
D) abc
Correct Answer: A
Solution :
Given, \[{{\operatorname{ax}}^{2}}+bx+ c = 0\] and a, P are roots of given equation \[\therefore \,\,\,\alpha +\beta =-\frac{b}{a}\,\,and\,\,\alpha \beta =\frac{c}{a}\] ... (i) Now \[\alpha {{\beta }^{2}}+ {{\alpha }^{2}}\beta + \alpha \beta = \alpha \beta \left( \beta + \alpha \right) + \alpha \beta \] \[=\,\,\frac{c}{a}.\left( -\frac{b}{a} \right)+\frac{c}{a}\] [Using equation (i)] \[=\,\,-\frac{cb}{{{a}^{2}}}+\frac{c}{a}\] \[\Rightarrow \,\,\,\frac{-cb+ac}{{{a}^{2}}}=\frac{c(a-b)}{{{a}^{2}}}\]
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