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

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

Answered by MJS last updated on 06/May/20

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

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

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

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