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Question Number 45427 by ajfour last updated on 12/Oct/18

Commented by ajfour last updated on 12/Oct/18

$$Determineside\mathit{a}ofsquare,$$$$withoutusingcalculator.$$

Commented by ajfour last updated on 12/Oct/18

Commented by ajfour last updated on 12/Oct/18

$$\frac{AE}{ED}=\frac{DH}{CH}\Rightarrow \frac{AE}{a}=\frac{a}{1}$$$$\Rightarrow AE=a^2$$$${EG}^2=1=x^2+{(a-x)}^2....\left(i\right)$$$$And\frac{a-x}{x}=\frac{a}{a+AE}=\frac{a}{a+a^2}$$$$\Rightarrow (a-x)(1+a)=x$$$$ora+a^2-x-ax=x$$$$\Rightarrow x=\frac{a+a^2}{2+a}...\left(ii\right)$$$$using\left(ii\right)in\left(i\right):$$$${\left(\frac{a+a^2}{2+a}\right)}^2+{(a-\frac{a+a^2}{2+a})}^2=1$$$$\Rightarrow a^2{(1+a)}^2+a^2={(2+a)}^2$$$$\Rightarrow a^4+2a^3+2a^2=4+4a+a^2$$$$\Rightarrow {\mathit{a}}^4+2{\mathit{a}}^3+{\mathit{a}}^2-4\mathit{a}-4=0$$$$a^3(a+1)+a^2(a+1)-4(a+1)=0$$$$asa\ne -1$$$$\Rightarrow a^3+a^2-4=0$$$$leta=b-\frac{1}{3}$$$$\Rightarrow b^3-b^2+\frac{b}{3}-\frac{1}{27}+b^2-\frac{2b}{3}+\frac{1}{9}-4=0$$$$\Rightarrow b^3-\frac{b}{3}-\frac{106}{27}=0$$$$b={(\frac{53}{27}+\sqrt{{\left(\frac{53}{27}\right)}^2-{\left(\frac{1}{27}\right)}^2})}^{1/3}$$$$+{(\frac{53}{27}-\sqrt{{\left(\frac{53}{27}\right)}^2-{\left(\frac{1}{27}\right)}^2})}^{1/3}$$$$={(\frac{53}{27}+\frac{\sqrt{54\times 52}}{27})}^{1/3}+{(\frac{53}{27}-\frac{\sqrt{54\times 52}}{27})}^{1/3}$$$$a=-\frac{1}{3}+\frac{{(53+\sqrt{54\times 52})}^{1/3}}{3}$$$$+\frac{{(53-\sqrt{54\times 52})}^{1/3}}{3}$$$$\mathit{a}\approx 1.3146$$

Commented by MrW3 last updated on 13/Oct/18

$$anotherniceapplicationofcubiceqn.!$$

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