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Question Number 180652 by Shrinava last updated on 14/Nov/22

Answered by aleks041103 last updated on 17/Nov/22

$$letdetX=x\Rightarrow a=1$$$$det\left(A^2\right)={\left(detA\right)}^2=1=det\left(BC\right)=detBdetC$$$$\Rightarrow bc=1\Rightarrow b=\frac{1}{c}$$$$B^2=CA\Rightarrow b^2=c\Rightarrow \frac{1}{c^2}=c$$$$C^2=AB\Rightarrow c^2=b\Rightarrow c^2=\frac{1}{c}$$$$\Rightarrow c^3=1=b^3$$$$\Rightarrow c^2=\frac{1}{c}=b,b^3=1$$$$\Rightarrow x^4+6b^3{x}^3-14c^2{x}^2+6x+1=0$$$$\Rightarrow x^4+6x^3-14{bx}^2+6x+1=0$$$$obv.x\ne 0$$$$\Rightarrow (x^2+\frac{1}{x^2})+6(x+\frac{1}{x})-14b=0$$$$y=x+\frac{1}{x}$$$$y^2=x^2+\frac{1}{x^2}+2$$$$\Rightarrow y^2-2+6y-14b=0$$$$\Rightarrow y^2+6y-2(1+7b)=0$$$$\Rightarrow x+\frac{1}{x}=\frac{-6\pm \sqrt{44+56b}}{2}=-3\pm \sqrt{11+14b}$$$$\Rightarrow x^2-(-3\pm \sqrt{11+14b})x+1=0$$$$\Rightarrow x=\frac{1}{2}(-3\pm \sqrt{11+14b}\pm \sqrt{{(-3\pm \sqrt{11+14b})}^2-4})$$$$b^3=1\Rightarrow b=1,e^{\pm \frac{2i\pi }{3}}$$$$forb=1:$$$$x=\frac{1}{2}(-3\pm 5\pm \sqrt{{(-3\pm 5)}^2-4})=$$$$=1,4\pm \sqrt{15}$$$$...$$

Commented by Shrinava last updated on 17/Nov/22

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{dear}\mathrm{professor}$$$$\mathrm{Your}\mathrm{solutions}\mathrm{are}\mathrm{perfect}$$

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