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Question Number 94945 by john santu last updated on 22/May/20

$$\mathrm{How}\mathrm{do}\mathrm{you}\mathrm{make}\mathrm{p}\mathrm{the}\mathrm{subject}$$$$\mathrm{of}\mathrm{equation}q=\frac{m}{\sqrt{p}}+\frac{p^2}{m}$$

Commented by MJS last updated on 22/May/20

$$p>0\iff \sqrt{p}>0$$$$\mathrm{let}p=t^2\wedge t>0$$$$q=\frac{m}{t}+\frac{t^4}{m}$$$$t^5-mqt+m^2=0$$$$\mathrm{we}\mathrm{cannot}\mathrm{generally}\mathrm{solve}\mathrm{this}$$

Commented by mr W last updated on 22/May/20

$$agree$$

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