Menu Close

m-h-2p-p-m-h-k-q-mk-qh-1-3-k-2q-1-ph-Assume-one-find-the-rest-




Question Number 201965 by ajfour last updated on 17/Dec/23
m−h=2p  p(m−h)=k−q  mk−qh=(1/3)  k−2q=1+ph  Assume one find the rest!  ✓
$${m}−{h}=\mathrm{2}{p} \\ $$$${p}\left({m}−{h}\right)={k}−{q} \\ $$$${mk}−{qh}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${k}−\mathrm{2}{q}=\mathrm{1}+{ph} \\ $$$${Assume}\:{one}\:{find}\:{the}\:{rest}! \\ $$$$\checkmark \\ $$
Answered by Frix last updated on 17/Dec/23
Solve (1) for h then (2) for k then (3) for m  ⇒  h=m−2p=((−12p^3 −6pq+1)/(6p^2 ))  k=2p^2 +q  m=((−6pq+1)/(6p^2 ))  (4) p^3 −(p/4)−(1/(24))=0  ⇒ p∈{−((√3)/3)sin ((2π)/9), −((√3)/3)sin (π/9), ((√3)/3)cos (π/(18))}  We are free to choose q
$$\mathrm{Solve}\:\left(\mathrm{1}\right)\:\mathrm{for}\:{h}\:\mathrm{then}\:\left(\mathrm{2}\right)\:\mathrm{for}\:{k}\:\mathrm{then}\:\left(\mathrm{3}\right)\:\mathrm{for}\:{m} \\ $$$$\Rightarrow \\ $$$${h}={m}−\mathrm{2}{p}=\frac{−\mathrm{12}{p}^{\mathrm{3}} −\mathrm{6}{pq}+\mathrm{1}}{\mathrm{6}{p}^{\mathrm{2}} } \\ $$$${k}=\mathrm{2}{p}^{\mathrm{2}} +{q} \\ $$$${m}=\frac{−\mathrm{6}{pq}+\mathrm{1}}{\mathrm{6}{p}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\right)\:{p}^{\mathrm{3}} −\frac{{p}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{24}}=\mathrm{0} \\ $$$$\Rightarrow\:{p}\in\left\{−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{9}},\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{sin}\:\frac{\pi}{\mathrm{9}},\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{18}}\right\} \\ $$$$\mathrm{We}\:\mathrm{are}\:\mathrm{free}\:\mathrm{to}\:\mathrm{choose}\:{q} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *