Question Number 78153 by ajfour last updated on 14/Jan/20
Commented by ajfour last updated on 14/Jan/20
$${Find}\:{h}\:{in}\:{terms}\:{of}\:{c}. \\ $$
Answered by mr W last updated on 14/Jan/20
$$\sqrt{{h}^{\mathrm{2}} +\left(\frac{{hc}}{{h}+\mathrm{1}}\right)^{\mathrm{2}} }=\mathrm{2}{c}−\frac{{hc}}{{h}+\mathrm{1}} \\ $$$${h}^{\mathrm{2}} =\mathrm{4}{c}^{\mathrm{2}} −\frac{\mathrm{4}{hc}^{\mathrm{2}} }{{h}+\mathrm{1}} \\ $$$${h}^{\mathrm{3}} +{h}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:{h}={x}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} −\frac{{x}}{\mathrm{3}}+\left(\frac{\mathrm{2}}{\mathrm{27}}−\mathrm{4}{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{1}}{\mathrm{27}}−\mathrm{2}{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}^{\mathrm{2}} }}+\mathrm{2}{c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}}−\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{1}}{\mathrm{27}}−\mathrm{2}{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}^{\mathrm{2}} }}−\mathrm{2}{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}} \\ $$$$\Rightarrow{h}=\sqrt[{\mathrm{3}}]{\mathrm{2}{c}\sqrt{{c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}}+\mathrm{2}{c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}{c}\sqrt{{c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}}−\mathrm{2}{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by ajfour last updated on 15/Jan/20
$${Thanks}\:{Sir},\:{i}\:{thought}\:{you} \\ $$$${might}\:{construct}\:{and}\:{get}\:{a} \\ $$$${quadratic},\:{rather}. \\ $$