A) \[-\alpha ,-\beta \]
B) \[\alpha ,\frac{1}{\beta }\]
C) \[\frac{1}{\alpha },\frac{1}{\beta }\]
D) None of these
Correct Answer: C
Solution :
\[\alpha ,\beta \]are roots of \[a{{x}^{2}}+bx+c=0\] Þ \[\alpha +\beta =-\frac{b}{a}\]and \[\alpha \beta =\frac{c}{a}\] Let the roots of \[c{{x}^{2}}+bx+a=0\]be \[{\alpha }',{\beta }'\], then \[{\alpha }'+{\beta }'=-\frac{b}{c}\]and \[{\alpha }'{\beta }'=\frac{a}{c}\] but \[\frac{\alpha +\beta }{\alpha \beta }=\frac{-b/a}{c/a}=\frac{-b}{c}\]Þ\[\frac{1}{\alpha }+\frac{1}{\beta }={\alpha }'+{\beta }'\] Hence \[{\alpha }'=\frac{1}{\alpha }\] and \[{\beta }'=\frac{1}{\beta }\].
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