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Question Number 39371 by ajfour last updated on 05/Jul/18

Commented by ajfour last updated on 05/Jul/18

$$Findx.\left(analytically\right)$$

Answered by MJS last updated on 06/Jul/18

$$△1$$$$x^2+1^2=p^2;\alpha +(90°-\alpha )=90°$$$$△2$$$$q^2+x^2={(1-x)}^2;\alpha +(90°-\alpha )=90°$$$$△3$$$$x^2+{(p+q)}^2=2;(45°-\alpha )+(45°+\alpha )=90°$$$$$$$$x=\tan \alpha \Rightarrow \alpha =\arctan x$$$$x=\sqrt{2}(\sin (45°-\alpha )=\cos \alpha -\sin \alpha \Rightarrow$$$$\Rightarrow x=\frac{1-x}{\sqrt{1+x^2}}$$$$x\approx .474627$$$$$$$$x^4+2x-1=0$$

Commented by ajfour last updated on 06/Jul/18

$$ThankyouSir,rightanswer.$$

Commented by MJS last updated on 06/Jul/18

$$\mathrm{again}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{solve}\mathrm{the}\mathrm{equation}\mathrm{by}$$$$\mathrm{setting}$$$$x^4+2x-1=$$$$=(x-\alpha -\beta \mathrm{i})(x-\alpha +\beta \mathrm{i})(x-\gamma -\delta )(x-\gamma +\delta )$$$$\mathrm{and}\mathrm{proceed}\mathrm{the}\mathrm{usual}\mathrm{way}\mathrm{and}\mathrm{again}\mathrm{the}$$$$\mathrm{exact}\mathrm{solutions}\mathrm{are}\mathrm{not}\mathrm{really}\mathrm{useable}$$

Answered by MrW3 last updated on 06/Jul/18

$$\frac{1}{\sqrt{1+x^2}}=\frac{x}{1-x}$$$$\Rightarrow x\sqrt{1+x^2}=1-x$$$$\Rightarrow x^2(1+x^2)=1+x^2-2x$$$$\Rightarrow x^4+2x-1=0$$$$\Rightarrow x\approx 0.4746$$

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