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Question Number 68601 by Maclaurin Stickker last updated on 14/Sep/19

$$ABCD\mathrm{is}\mathrm{a}\mathrm{side}\mathrm{square}1.$$$$B,F\mathrm{and}E\mathrm{are}\mathrm{collinear}.$$$$FDE\mathrm{is}\mathrm{a}\mathrm{right}\mathrm{triangle}\mathrm{with}\mathrm{hypotenuse}1$$$$\mathrm{and}\mathrm{the}DE\mathrm{cathetus}\mathrm{is}\mathrm{worth}\mathit{x}.$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}\mathit{x}?$$$$\left(\mathrm{Solve}\mathrm{with}\mathrm{algebra}\right)$$

Commented by Maclaurin Stickker last updated on 14/Sep/19

Commented by mr W last updated on 14/Sep/19

$$BE=\sqrt{{(1+x)}^2+1^2}=\sqrt{x^2+2x+2}$$$$\frac{1}{x}=\frac{\sqrt{x^2+2x+2}}{1+x}$$$$\frac{1+2x+x^2}{x^2}=x^2+2x+2$$$$x^4+2x^3+2x^2=1+2x+x^2$$$$x^4+2x^3+x^2-2x-1=0$$$$\Rightarrow x\approx 0.8832$$$$(exactvaluepossible,butcomplicated)$$

Commented by MJS last updated on 14/Sep/19

$$\mathrm{exact}\mathrm{value},\mathrm{just}\mathrm{for}\mathrm{the}\mathrm{fun}\mathrm{of}\mathrm{it}$$$$x^4+2x^3+x^2-2x-1=0$$$$(x^2+(1-\sqrt{2})x+1-\sqrt{2})(x^2+(1+\sqrt{2})x+1+\sqrt{2})=0$$$$x_1=-\frac{1}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{-1+2\sqrt{2}}}{2}<0\Rightarrow \mathrm{not}\mathrm{valid}$$$$x_2=-\frac{1}{2}+\frac{\sqrt{2}}{2}+\frac{\sqrt{-1+2\sqrt{2}}}{2}>0\Rightarrow \mathrm{is}\mathrm{the}\mathrm{answer}$$$$x_{3,4}=-\frac{1}{2}-\frac{\sqrt{2}}{2}\pm \frac{\sqrt{1+2\sqrt{2}}}{2}\mathrm{i}\notin \mathbb{R}$$

Commented by Maclaurin Stickker last updated on 14/Sep/19

$$\mathrm{Ha},\mathrm{ha}!\mathrm{Thank}\mathrm{you},\mathrm{sir}!$$

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