q-m-p-p-2-m-make-p-the-subject-

Question Number 168064 by MathsFan last updated on 02/Apr/22

q = \frac{m}{\sqrt{p}} + \frac{p^{2}}{m}

m a k e p t h e s u b j e c t

Commented by MJS_new last updated on 02/Apr/22

you {can}^{'} t . this leads to

{(\sqrt{p})}^{5} - m q \sqrt{p} + m^{2} = 0

{there}^{'} s no solution formula for 5^{th} degree

Commented by MathsFan last updated on 02/Apr/22

t h a n k y o u

Commented by alephzero last updated on 02/Apr/22

q = \frac{m}{\sqrt{p}} + \frac{p^{2}}{m}

\Rightarrow q = \frac{m^{2} + \sqrt{p^{5}}}{m \sqrt{p}}

\Rightarrow \frac{m^{2} + \sqrt{p^{5}} - m q \sqrt{p}}{m \sqrt{p}} = 0

\Rightarrow \sqrt{p}^{5} - m q \sqrt{p} + m^{2} = 0

\Rightarrow \sqrt{p} (p^{2} - m q) + m^{2} = 0

\Rightarrow \sqrt{p} (p^{2} - m q) = - m^{2}

\Rightarrow \sqrt{p} = - \frac{m^{2}}{p^{2} - m q}

\Rightarrow p = {(\frac{m^{2}}{p^{2} - m q})}^{2}

\dots

Commented by MJS_new last updated on 02/Apr/22

well, as I stated : impossible to solve

Commented by alephzero last updated on 02/Apr/22

s e e m l i k e y e s . t h a t c a n o n l y b e a s

t h e e q u a t i o n