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

How do you make p the subject  of equation q = (m/((√p) )) + (p^2 /m)

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

Commented by MJS last updated on 22/May/20

p>0 ⇔ (√p)>0  let p=t^2 ∧t>0  q=(m/t)+(t^4 /m)  t^5 −mqt+m^2 =0  we cannot generally solve this

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

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

agree

$${agree} \\ $$

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