Question Number 92275 by otchereabdullai@gmail.com last updated on 05/May/20
$$\mathrm{Make}\:\mathrm{R}\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}: \\ $$$$\:\:\mathrm{P}=\:\frac{\mathrm{RE}^{\mathrm{2}} }{\left(\mathrm{R}+\mathrm{b}\right)^{\mathrm{2}} } \\ $$
Answered by MJS last updated on 06/May/20
$${P}\left({R}+{b}\right)^{\mathrm{2}} ={E}^{\mathrm{2}} {R} \\ $$$$\left({R}+{b}\right)^{\mathrm{2}} =\frac{{E}^{\mathrm{2}} }{{P}}{R} \\ $$$${R}^{\mathrm{2}} +\mathrm{2}{bR}+{b}^{\mathrm{2}} =\frac{{E}^{\mathrm{2}} }{{P}}{R} \\ $$$${R}^{\mathrm{2}} +\left(\mathrm{2}{b}−\frac{{E}^{\mathrm{2}} }{{P}}\right){R}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${R}^{\mathrm{2}} +\frac{\mathrm{2}{bP}−{E}^{\mathrm{2}} }{{P}}{R}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${R}=−\frac{\mathrm{2}{bP}−{E}^{\mathrm{2}} }{\mathrm{2}{P}}\pm\sqrt{\left(\frac{\mathrm{2}{bP}−{E}^{\mathrm{2}} }{\mathrm{2}{P}}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${R}=\frac{{E}^{\mathrm{2}} }{\mathrm{2}{P}}−{b}\pm\frac{{E}}{\mathrm{2}{P}}\sqrt{{E}^{\mathrm{2}} −\mathrm{4}{bP}} \\ $$$${R}=−{b}+\frac{{E}}{\mathrm{2}{P}}\left({E}\pm\sqrt{{E}^{\mathrm{2}} −\mathrm{4}{bP}}\right) \\ $$
Commented by otchereabdullai@gmail.com last updated on 06/May/20
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{prof}\:\mathrm{mjs} \\ $$