Question Number 81303 by mr W last updated on 11/Feb/20
Commented by mr W last updated on 11/Feb/20
$${Find}\:{x}=? \\ $$
Answered by mr W last updated on 11/Feb/20
Commented by mr W last updated on 11/Feb/20
Commented by mr W last updated on 11/Feb/20
![DE=(√(x^2 −1)) GH=((DE)/2)=((√(x^2 −1))/2) BG=((√3)/2) HE=((AE)/2)=(x/2) ((FH)/(FE))=((BH)/(DE)) (((x/2)+1)/1)=((((√3)/2)+((√(x^2 −1))/2))/( (√(x^2 −1)))) x+2=(√(3/(x^2 −1)))+1 (x+1)^2 =(3/(x^2 −1)) (x+1)^3 (x−1)−3=0 let t=x+1 t^3 (t−2)−3=0 (t+1)(t^3 −3t^2 +3t−3)=0 (t+1)[(t−1)^3 −2]=0 ⇒t−1=(2)^(1/3) =x](https://www.tinkutara.com/question/Q81330.png)
$${DE}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${GH}=\frac{{DE}}{\mathrm{2}}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}} \\ $$$${BG}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${HE}=\frac{{AE}}{\mathrm{2}}=\frac{{x}}{\mathrm{2}} \\ $$$$\frac{{FH}}{{FE}}=\frac{{BH}}{{DE}} \\ $$$$\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{\mathrm{1}}=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${x}+\mathrm{2}=\sqrt{\frac{\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}}+\mathrm{1} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}−\mathrm{1}\right)−\mathrm{3}=\mathrm{0} \\ $$$${let}\:{t}={x}+\mathrm{1} \\ $$$${t}^{\mathrm{3}} \left({t}−\mathrm{2}\right)−\mathrm{3}=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left[\left({t}−\mathrm{1}\right)^{\mathrm{3}} −\mathrm{2}\right]=\mathrm{0} \\ $$$$\Rightarrow{t}−\mathrm{1}=\sqrt[{\mathrm{3}}]{\mathrm{2}}={x} \\ $$