Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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.

$${Find}\:{h}\:{in}\:{terms}\:{of}\:{c}. \\ $$

Answered by mr W last updated on 14/Jan/20

(√(h^2 +(((hc)/(h+1)))^2 ))=2c−((hc)/(h+1)) h^2 =4c^2 −((4hc^2 )/(h+1)) h^3 +h^2 −4c^2 =0 let h=x−(1/3) x^3 −(x/3)+((2/(27))−4c^2 )=0 x=(((√(((1/(27))−2c^2 )^2 −(1/(27^2 ))))+2c^2 −(1/(27))))^(1/3) −(((√(((1/(27))−2c^2 )^2 −(1/(27^2 ))))−2c^2 +(1/(27))))^(1/3) ⇒h=((2c(√(c^2 −(1/(27))))+2c^2 −(1/(27))))^(1/3) −((2c(√(c^2 −(1/(27))))−2c^2 +(1/(27))))^(1/3) −(1/3)

$$\sqrt{{h}^{\mathrm{2}} +\left(\frac{{hc}}{{h}+\mathrm{1}}\right)^{\mathrm{2}} }=\mathrm{2}{c}−\frac{{hc}}{{h}+\mathrm{1}} \\ $$$${h}^{\mathrm{2}} =\mathrm{4}{c}^{\mathrm{2}} −\frac{\mathrm{4}{hc}^{\mathrm{2}} }{{h}+\mathrm{1}} \\ $$$${h}^{\mathrm{3}} +{h}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:{h}={x}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} −\frac{{x}}{\mathrm{3}}+\left(\frac{\mathrm{2}}{\mathrm{27}}−\mathrm{4}{c}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{1}}{\mathrm{27}}−\mathrm{2}{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}^{\mathrm{2}} }}+\mathrm{2}{c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}}−\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{1}}{\mathrm{27}}−\mathrm{2}{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}^{\mathrm{2}} }}−\mathrm{2}{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}} \\ $$$$\Rightarrow{h}=\sqrt[{\mathrm{3}}]{\mathrm{2}{c}\sqrt{{c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}}+\mathrm{2}{c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}{c}\sqrt{{c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{27}}}−\mathrm{2}{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by ajfour last updated on 15/Jan/20

Thanks Sir, i thought you might construct and get a quadratic, rather.

$${Thanks}\:{Sir},\:{i}\:{thought}\:{you} \\ $$$${might}\:{construct}\:{and}\:{get}\:{a} \\ $$$${quadratic},\:{rather}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com