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Question Number 63154 by Rio Michael last updated on 29/Jun/19

$$Giventhat\alpha and\beta aretherootsoctheequation{ax}^2+bx+c=0$$$$.Showthat\lambda \mu b^2=ac{(\lambda +\mu )}^2,where\frac{\alpha }{\beta }=\frac{\lambda }{\mu }.$$

Commented by Prithwish sen last updated on 30/Jun/19

$$\alpha +\beta =-\frac{\mathrm{b}}{\mathrm{a}}\mathrm{and}\alpha \beta =\frac{\mathrm{c}}{\mathrm{a}}$$$$\mathrm{Now}$$$${(\alpha +\beta )}^2=\frac{\alpha \beta {(\alpha +\beta )}^2}{\alpha \beta }=\alpha \beta (\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+2)$$$$\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{\mathrm{c}}{\mathrm{a}}(\frac{\lambda }{\mu }+\frac{\mu }{\lambda }+2)=\frac{\mathrm{c}}{\mathrm{a}}\frac{{(\lambda +\mu )}^2}{\lambda \mu }$$$$\lambda \mu {\mathrm{b}}^2=\mathrm{ac}{(\lambda +\mu )}^2\mathrm{proved}.$$$$$$$$$$

Commented by Rio Michael last updated on 02/Jul/19

$$coolithinkthatscorrect.$$

Commented by Rio Michael last updated on 02/Jul/19

$$pleasecheckmysolution.$$$$$$$$\alpha +\beta =-\frac{b}{a}and\alpha \beta =\frac{c}{a}$$$$-b=a(\alpha +\beta )$$$$and{b}^2=a^2{(\alpha +\beta )}^2..........\left(i\right)$$$$and\alpha \beta =\frac{c}{a}..........\left(ii\right)$$$$takingeqn\left(i\right)\times \left(ii\right)$$$$\Rightarrow \alpha \beta b^2=\frac{c}{a}\times a^2{(\alpha +\beta )}^2$$$$\alpha \beta b^2=ac(\alpha +\beta )$$$$since\frac{\alpha }{\beta }=\frac{\lambda }{\mu }then\alpha =\lambda and\beta =\mu$$$$andsubstitutingweget.$$$$\lambda \mu b^2=ac{(\lambda +\mu )}^{2}proved.$$$$$$$$pleasemakemorecorrections.$$$$$$$$$$

Commented by Prithwish sen last updated on 02/Jul/19

$$\mathrm{I}\mathrm{think}$$$$\frac{\alpha }{\lambda }=\frac{\beta }{\mu }=\mathrm{k}\left(\mathrm{let}\right)$$$$\mathrm{then}\alpha \beta {\mathrm{b}}^2=\mathrm{ac}{(\alpha +\beta )}^2\Rightarrow {\mathrm{k}}^2\lambda \mu {\mathrm{b}}^2={\mathrm{k}}^2\mathrm{ac}{(\lambda +\mu )}^2$$$$$$

Commented by Rio Michael last updated on 02/Jul/19

$$thatsalsoagoodidea$$

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