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Question Number 92275 by otchereabdullai@gmail.com last updated on 05/May/20

Make R the subject of: P= ((RE^2 )/((R+b)^2 ))

$$\mathrm{Make}\:\mathrm{R}\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}: \\ $$$$\:\:\mathrm{P}=\:\frac{\mathrm{RE}^{\mathrm{2}} }{\left(\mathrm{R}+\mathrm{b}\right)^{\mathrm{2}} } \\ $$

Answered by MJS last updated on 06/May/20

P(R+b)^2 =E^2 R (R+b)^2 =(E^2 /P)R R^2 +2bR+b^2 =(E^2 /P)R R^2 +(2b−(E^2 /P))R+b^2 =0 R^2 +((2bP−E^2 )/P)R+b^2 =0 R=−((2bP−E^2 )/(2P))±(√((((2bP−E^2 )/(2P)))^2 −b^2 )) R=(E^2 /(2P))−b±(E/(2P))(√(E^2 −4bP)) R=−b+(E/(2P))(E±(√(E^2 −4bP)))

$${P}\left({R}+{b}\right)^{\mathrm{2}} ={E}^{\mathrm{2}} {R} \\ $$$$\left({R}+{b}\right)^{\mathrm{2}} =\frac{{E}^{\mathrm{2}} }{{P}}{R} \\ $$$${R}^{\mathrm{2}} +\mathrm{2}{bR}+{b}^{\mathrm{2}} =\frac{{E}^{\mathrm{2}} }{{P}}{R} \\ $$$${R}^{\mathrm{2}} +\left(\mathrm{2}{b}−\frac{{E}^{\mathrm{2}} }{{P}}\right){R}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${R}^{\mathrm{2}} +\frac{\mathrm{2}{bP}−{E}^{\mathrm{2}} }{{P}}{R}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${R}=−\frac{\mathrm{2}{bP}−{E}^{\mathrm{2}} }{\mathrm{2}{P}}\pm\sqrt{\left(\frac{\mathrm{2}{bP}−{E}^{\mathrm{2}} }{\mathrm{2}{P}}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${R}=\frac{{E}^{\mathrm{2}} }{\mathrm{2}{P}}−{b}\pm\frac{{E}}{\mathrm{2}{P}}\sqrt{{E}^{\mathrm{2}} −\mathrm{4}{bP}} \\ $$$${R}=−{b}+\frac{{E}}{\mathrm{2}{P}}\left({E}\pm\sqrt{{E}^{\mathrm{2}} −\mathrm{4}{bP}}\right) \\ $$

Commented by otchereabdullai@gmail.com last updated on 06/May/20

Thank you prof mjs

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{prof}\:\mathrm{mjs} \\ $$

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