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

Commented by ajfour last updated on 05/Jul/18

Find x. (analytically)

$${Find}\:{x}.\:\left({analytically}\right) \\ $$

Answered by MJS last updated on 06/Jul/18

△1 x^2 +1^2 =p^2 ; α+(90°−α)=90° △2 q^2 +x^2 =(1−x)^2 ; α+(90°−α)=90° △3 x^2 +(p+q)^2 =2; (45°−α)+(45°+α)=90° x=tan α ⇒ α=arctan x x=(√2)(sin(45°−α)=cos α −sin α ⇒ ⇒ x=((1−x)/(√(1+x^2 ))) x≈.474627 x^4 +2x−1=0

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

Commented by ajfour last updated on 06/Jul/18

Thank you Sir, right answer.

$${Thank}\:{you}\:{Sir},\:{right}\:{answer}. \\ $$

Commented by MJS last updated on 06/Jul/18

again it′s possible to solve the equation by setting x^4 +2x−1= =(x−α−βi)(x−α+βi)(x−γ−δ)(x−γ+δ) and proceed the usual way and again the exact solutions are not really useable

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

Answered by MrW3 last updated on 06/Jul/18

(1/(√(1+x^2 )))=(x/(1−x)) ⇒x(√(1+x^2 ))=1−x ⇒x^2 (1+x^2 )=1+x^2 −2x ⇒x^4 +2x−1=0 ⇒x≈0.4746

$$\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}=\frac{{x}}{\mathrm{1}−{x}} \\ $$$$\Rightarrow{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }=\mathrm{1}−{x} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)=\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{x} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}\approx\mathrm{0}.\mathrm{4746} \\ $$

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