Question Number 168064 by MathsFan last updated on 02/Apr/22
$$\:{q}=\frac{{m}}{\:\sqrt{{p}}}+\frac{{p}^{\mathrm{2}} }{{m}} \\ $$$$\:{make}\:\:{p}\:\:{the}\:{subject} \\ $$
Commented by MJS_new last updated on 02/Apr/22
$$\mathrm{you}\:\mathrm{can}'\mathrm{t}.\:\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\left(\sqrt{{p}}\right)^{\mathrm{5}} −{mq}\sqrt{{p}}+{m}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{degree} \\ $$
Commented by MathsFan last updated on 02/Apr/22
$${thank}\:{you} \\ $$
Commented by alephzero last updated on 02/Apr/22
$${q}=\frac{{m}}{\:\sqrt{{p}}}+\frac{{p}^{\mathrm{2}} }{{m}} \\ $$$$\Rightarrow\:{q}=\frac{{m}^{\mathrm{2}} +\sqrt{{p}^{\mathrm{5}} }}{{m}\sqrt{{p}}} \\ $$$$\Rightarrow\:\frac{{m}^{\mathrm{2}} +\sqrt{{p}^{\mathrm{5}} }−{mq}\sqrt{{p}}}{{m}\sqrt{{p}}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\sqrt{{p}}\:^{\mathrm{5}} −{mq}\sqrt{{p}}+{m}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:\sqrt{{p}}\left({p}^{\mathrm{2}} −{mq}\right)+{m}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:\sqrt{{p}}\left({p}^{\mathrm{2}} −{mq}\right)\:=\:−{m}^{\mathrm{2}} \\ $$$$\Rightarrow\:\sqrt{{p}}\:=\:−\frac{{m}^{\mathrm{2}} }{{p}^{\mathrm{2}} −{mq}} \\ $$$$\Rightarrow\:{p}\:=\:\left(\frac{{m}^{\mathrm{2}} }{{p}^{\mathrm{2}} −{mq}}\right)^{\mathrm{2}} \\ $$$$... \\ $$
Commented by MJS_new last updated on 02/Apr/22
$$\mathrm{well},\:\mathrm{as}\:\mathrm{I}\:\mathrm{stated}:\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{solve} \\ $$
Commented by alephzero last updated on 02/Apr/22
$${seem}\:{like}\:{yes}.\:{that}\:{can}\:{only}\:{be}\:{as} \\ $$$${the}\:{equation} \\ $$