question_answer

If the ratio of the roots of \[\lambda {{x}^{2}}+\mu x+v=0\] is equal to the ratio of the roots of x2 + x + 1 = 0 then \[\lambda ,\mu ,v\]are in

A)  AP                             

B)  GP

C)  HP                             

D)  None of these

Correct Answer: B

Solution :

Since, ratio of roots for \[\lambda {{x}^{2}}+\mu x+v=0\] and\[{{x}^{2}}+x+1=0\]are equal. \[\therefore \]\[\frac{\alpha }{\beta }=\frac{\alpha '}{\beta '},\]where\[\alpha ,\beta \]are roots of \[\lambda {{x}^{2}}+\mu x+v=0\]and \[\alpha ',\beta '\] are roots of x2 + x + 1 = 0, given by \[\omega \] and \[{{\omega }^{2}}\] \[\therefore \]\[\frac{\alpha }{\beta }=\frac{\omega }{{{\omega }^{2}}}=\frac{1}{\omega }\Rightarrow \beta =\omega \alpha \]and\[\alpha +\beta =-\frac{\mu }{\lambda },\alpha \beta =\frac{v}{\lambda }.\] \[\Rightarrow \]\[\alpha (1+\omega )=-\frac{\mu }{\lambda },{{\alpha }^{2}}\omega =\frac{v}{\lambda }\] \[\Rightarrow \]\[-\alpha {{\omega }^{2}}=\frac{\mu }{-\lambda },{{\alpha }^{2}}\omega =\frac{v}{\lambda }(\because 1+\omega =-{{\omega }^{2}})\] \[\Rightarrow \]\[\frac{{{\mu }^{2}}}{{{\lambda }^{2}}}=\frac{v}{\lambda }\Rightarrow {{\mu }^{2}}=\lambda v\] Hence, and v are in GP.