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
$${DE}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${GH}=\frac{{DE}}{\mathrm{2}}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}} \\ $$$${BG}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${HE}=\frac{{AE}}{\mathrm{2}}=\frac{{x}}{\mathrm{2}} \\ $$$$\frac{{FH}}{{FE}}=\frac{{BH}}{{DE}} \\ $$$$\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{\mathrm{1}}=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${x}+\mathrm{2}=\sqrt{\frac{\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}}+\mathrm{1} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}−\mathrm{1}\right)−\mathrm{3}=\mathrm{0} \\ $$$${let}\:{t}={x}+\mathrm{1} \\ $$$${t}^{\mathrm{3}} \left({t}−\mathrm{2}\right)−\mathrm{3}=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left({t}+\mathrm{1}\right)\left[\left({t}−\mathrm{1}\right)^{\mathrm{3}} −\mathrm{2}\right]=\mathrm{0} \\ $$$$\Rightarrow{t}−\mathrm{1}=\sqrt[{\mathrm{3}}]{\mathrm{2}}={x} \\ $$