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Question Number 214414 by ChantalYah last updated on 07/Dec/24

$$\mathrm{Given}\mathrm{that}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}$$$${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0\mathrm{are}\alpha \mathrm{and}\beta ,$$$$\mathrm{show}\mathrm{that};$$$$\lambda \mu {\mathrm{b}}^2=\mathrm{ac}{(\lambda +\mu )}^2\mathrm{where}\frac{\alpha }{\beta }=\frac{\lambda }{\mu }$$$$MrHans$$

Answered by A5T last updated on 07/Dec/24

$$\frac{\alpha }{\beta }=\frac{\lambda }{\mu }\Rightarrow \frac{\alpha +\beta }{\beta }=\frac{\lambda +\mu }{\mu }\Rightarrow \frac{{(\alpha +\beta )}^2}{{\beta }^2}=\frac{{(\lambda +\mu )}^2}{{\mu }^2}$$$$\alpha +\beta =-\frac{b}{a}\Rightarrow {(\alpha +\beta )}^2=\frac{b^2}{a^2}$$$$\Rightarrow \frac{b^2}{a^2{\beta }^2}=\frac{{(\lambda +\mu )}^2}{{\mu }^2}\Rightarrow \lambda \mu b^2=\frac{\lambda }{\mu }{(\lambda +\mu )}^2{a}^2{\beta }^2$$$$\Rightarrow \lambda \mu b^2=\frac{\alpha {(\lambda +\mu )}^2{a}^2{\beta }^2}{\beta }=\alpha \beta a^2{(\lambda +\mu )}^2$$$$\alpha \beta =\frac{c}{a}\Rightarrow \lambda \mu b^2=ac{(\lambda +\mu )}^2◼$$

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