Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

Question Number 211815 by alcohol last updated on 21/Sep/24

$${ax}^2+bx+c=0hasroots\alpha and\beta$$$$and\frac{\alpha }{\beta }=\frac{\lambda }{\mu }.showthat\lambda \mu b^2=ac{(\lambda +\mu )}^2$$

Answered by som(math1967) last updated on 22/Sep/24

$$let\alpha =k\lambda ,\beta =k\mu$$$$k(\lambda +\mu )=-\frac{b}{a}$$$$\Rightarrow k^2{(\lambda +\mu )}^2=\frac{b^2}{a^2}$$$$k^2=\frac{b^2}{a^2{(\lambda +\mu )}^2}$$$$again{k}^2\lambda \mu =\frac{c}{a}$$$$\Rightarrow \frac{b^2\lambda \mu }{a^2{(\lambda +\mu )}^2}=\frac{c}{a}$$$$\therefore \lambda \mu b^2=ac{(\lambda +\mu )}^2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com