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Question Number 150883 by mathdanisur last updated on 16/Aug/21

$$\mathrm{prove}\mathrm{that}\mathrm{any}\mathrm{real}\mathrm{root}\mathit{\alpha }\mathrm{of}\mathrm{the}$$$$\mathrm{equation}:{\mathrm{x}}^{6\mathbf{n}}={4\mathrm{x}}^{2\mathbf{n}}+4;\mathrm{n}\in \mathbb{N}-\left\{0\right\}$$$$\mathrm{verify}:\mid \mathit{\alpha }\mid >\sqrt[2\mathbf{n}]{2}$$

Answered by mindispower last updated on 16/Aug/21

$$X^{6n}-4X^{4n}-4=0$$$$X^{2n}=y>0$$$$\Rightarrow y^3-4y^2-4=0$$$$f\left(y\right)=y^3-4y^2-4$$$$f^{\prime }\left(y\right)=y(3y^2-8),f^{\prime }\left(y\right)>0,y>2\sqrt{\frac{2}{3}}=\beta$$$$f\left(0\right)=-4,f\left(\beta \right)<f\left(0\right),f\left(x\right)<0,\mathrm{\forall }x\in [0,\beta [$$$$f\left(2\right)=-12$$$$\Rightarrow f\left(y\right)=0,y>2$$$$\Rightarrow \mid y\mid =y>2$$$$\mid X^{2n}\mid >2\Rightarrow \mid X\mid >\sqrt[2n]{2}$$$$$$$$$$$$$$

Commented by mathdanisur last updated on 16/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser},\mathrm{but}{4\mathrm{x}}^{4\mathbf{n}}\mathrm{no}{4\mathrm{x}}^{2\mathbf{n}}$$$$\Rightarrow {\mathrm{y}}^3-{4\mathrm{y}}^2.?\mathrm{or}4\mathrm{y}$$

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