Question Number 45427 by ajfour last updated on 12/Oct/18
Commented by ajfour last updated on 12/Oct/18
$${Determine}\:{side}\:\boldsymbol{{a}}\:{of}\:{square}, \\ $$$${without}\:{using}\:{calculator}. \\ $$
Commented by ajfour last updated on 12/Oct/18
Commented by ajfour last updated on 12/Oct/18
$$\frac{{AE}}{{ED}}=\frac{{DH}}{{CH}}\:\:\:\Rightarrow\:\:\:\frac{{AE}}{{a}}\:=\:\frac{{a}}{\mathrm{1}} \\ $$$$\Rightarrow\:\:{AE}\:=\:{a}^{\mathrm{2}} \\ $$$${EG}^{\mathrm{2}} =\:\mathrm{1}=\:{x}^{\mathrm{2}} +\left({a}−{x}\right)^{\mathrm{2}} \:\:\:\:\:….\left({i}\right) \\ $$$${And}\:\:\:\frac{{a}−{x}}{{x}}\:=\:\frac{{a}}{{a}+{AE}}\:=\:\frac{{a}}{{a}+{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\left({a}−{x}\right)\left(\mathrm{1}+{a}\right)={x} \\ $$$${or}\:\:\:\:\:\:{a}+{a}^{\mathrm{2}} −{x}−{ax}\:=\:{x} \\ $$$$\Rightarrow\:\:\:\:{x}=\frac{{a}+{a}^{\mathrm{2}} }{\mathrm{2}+{a}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$${using}\:\left({ii}\right)\:{in}\:\left({i}\right): \\ $$$$\:\:\:\:\:\left(\frac{{a}+{a}^{\mathrm{2}} }{\mathrm{2}+{a}}\right)^{\mathrm{2}} +\left({a}−\frac{{a}+{a}^{\mathrm{2}} }{\mathrm{2}+{a}}\right)^{\mathrm{2}} =\:\mathrm{1} \\ $$$$\Rightarrow\:\:\:{a}^{\mathrm{2}} \left(\mathrm{1}+{a}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} =\left(\mathrm{2}+{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} +\mathrm{2}{a}^{\mathrm{2}} \:=\:\mathrm{4}+\mathrm{4}{a}+{a}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{a}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{a}}−\mathrm{4}\:=\:\mathrm{0} \\ $$$$\:\:\:{a}^{\mathrm{3}} \left({a}+\mathrm{1}\right)+{a}^{\mathrm{2}} \left({a}+\mathrm{1}\right)−\mathrm{4}\left({a}+\mathrm{1}\right)=\mathrm{0} \\ $$$${as}\:\:{a}\neq−\mathrm{1} \\ $$$$\Rightarrow\:\:\:{a}^{\mathrm{3}} +{a}^{\mathrm{2}} −\mathrm{4}\:=\:\mathrm{0} \\ $$$${let}\:\:\:{a}=\:{b}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\:{b}^{\mathrm{3}} −{b}^{\mathrm{2}} +\frac{{b}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{27}}+{b}^{\mathrm{2}} −\frac{\mathrm{2}{b}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\:\:{b}^{\mathrm{3}} −\frac{{b}}{\mathrm{3}}−\frac{\mathrm{106}}{\mathrm{27}}\:=\mathrm{0} \\ $$$$\:{b}\:=\:\left(\frac{\mathrm{53}}{\mathrm{27}}+\sqrt{\left(\frac{\mathrm{53}}{\mathrm{27}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{27}}\right)^{\mathrm{2}} }\:\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:+\left(\frac{\mathrm{53}}{\mathrm{27}}−\sqrt{\left(\frac{\mathrm{53}}{\mathrm{27}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{27}}\right)^{\mathrm{2}} }\:\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:=\:\left(\frac{\mathrm{53}}{\mathrm{27}}+\frac{\sqrt{\mathrm{54}×\mathrm{52}}}{\mathrm{27}}\:\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{\mathrm{53}}{\mathrm{27}}−\frac{\sqrt{\mathrm{54}×\mathrm{52}}}{\mathrm{27}}\:\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:{a}=\:−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\left(\mathrm{53}+\sqrt{\mathrm{54}×\mathrm{52}}\:\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\left(\mathrm{53}−\sqrt{\mathrm{54}×\mathrm{52}}\:\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}}\:\: \\ $$$$\:\:\boldsymbol{{a}}\:\approx\:\mathrm{1}.\mathrm{3146}\: \\ $$
Commented by MrW3 last updated on 13/Oct/18
$${another}\:{nice}\:{application}\:{of}\:{cubic}\:{eqn}.! \\ $$