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Question Number 162791 by mnjuly1970 last updated on 01/Jan/22

Answered by amin96 last updated on 01/Jan/22

$${\mathit{x}}^2+2\mathit{x}-1=0\iff Viets\mathit{\alpha }\mathit{\beta }=-1$$$$\mathit{\alpha }=-1+\sqrt{2}\mathit{\beta }=-1-\sqrt{2}$$$${\mathit{\alpha }}^4{\mathit{\beta }}^9+29\mathit{\alpha }+58={\mathit{\beta }}^5{\left(\mathit{\alpha }\mathit{\beta }\right)}^4+29(\mathit{\alpha }+2)=$$$$={\mathit{\beta }}^5+29(1+\sqrt{2})=29+29\sqrt{2}-{(1+\sqrt{2})}^5=$$$$=29+29\sqrt{2}-{(1+\sqrt{2})}^2{(1+\sqrt{2})}^2(1+\sqrt{2})=$$$$=29+29\sqrt{2}-{(3+2\sqrt{2})}^2(1+\sqrt{2})=$$$$=29+29\sqrt{2}-(17+12\sqrt{2})(1+\sqrt{2})=$$$$=29+29\sqrt{2}-(41+29\sqrt{2})=29-41=-12$$

Commented by mnjuly1970 last updated on 01/Jan/22

$$gratefulsiramin$$

Answered by alephzero last updated on 01/Jan/22

$$x^2+2x-1=0$$$$x_{1;2}=\frac{-2\pm \sqrt{4+4}}{2}=\frac{-2\pm \sqrt{8}}{2}=\frac{-2\pm 2\sqrt{2}}{2}$$$$=-1\pm \sqrt{2}$$$$x_1=-1+\sqrt{2}=\sqrt{2}-1$$$$x_2=-1-\sqrt{2}$$$$\mathcal{A}={(\sqrt{2}-1)}^4{(-1-\sqrt{2})}^9+29(\sqrt{2}-1)+$$$$+58=(17-12\sqrt{2})(-1393-985\sqrt{2})+$$$$+29\sqrt{2}+29+58=-41-29\sqrt{2}+29\sqrt{2}+$$$$+29+58=-41+29+58=46$$$$\mathcal{A}=46$$

Answered by mindispower last updated on 01/Jan/22

$$\alpha \beta =-1$$$${\alpha }^4{\beta }^9={\beta }^5{\left(\alpha \beta \right)}^4={\beta }^5$$$$\iff {\beta }^5+29\alpha +58$$$$x^2=1-2x$$$$x^3=x-2(1-2x)=5x-2$$$$x^4=-2x+5(1-2x)=-12x+5$$$$x^5=5x-12(1-2x)=-12+29x$$$${\beta }^5=-12+29\beta$$$$\iff 29(\alpha +\beta )+46$$$$=29.-2+46=-12$$

Commented by mnjuly1970 last updated on 01/Jan/22

$$bravosirpower...thissolution$$$$isverynice...$$

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