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Question Number 171844 by Mikenice last updated on 21/Jun/22

$$solve:$$$$x^5-5x^4+9x^3-9x^2+5x-1=0$$

Answered by floor(10²Eta[1]) last updated on 21/Jun/22

$$1\mathrm{is}\mathrm{root}:$$$$\mathrm{let}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^5-{5\mathrm{x}}^4+{9\mathrm{x}}^3-{9\mathrm{x}}^2+5\mathrm{x}-1$$$$\mathrm{using}{\mathrm{ruffini}}^{\prime }\mathrm{s}\mathrm{rule}:$$$$\overline{)\begin{array}{ccccccc}1& 1& -5& 9& -9& 5& -1\\ & 1& -4& 5& -4& 1& 0\end{array}}$$$$\Rightarrow \mathrm{p}\left(\mathrm{x}\right)=(\mathrm{x}-1)({\mathrm{x}}^4-{4\mathrm{x}}^3+{5\mathrm{x}}^2-4\mathrm{x}+1)$$$${\mathrm{x}}^4-{4\mathrm{x}}^3+{5\mathrm{x}}^2-4\mathrm{x}+1=0$$$$\mathrm{divide}\mathrm{both}\mathrm{sides}\mathrm{by}{\mathrm{x}}^2:$$$${\mathrm{x}}^2-4\mathrm{x}+5-\frac{4}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2}$$$$={(\mathrm{x}+\frac{1}{\mathrm{x}})}^2-4(\mathrm{x}+\frac{1}{\mathrm{x}})+3=0$$$$\mathrm{x}+\frac{1}{\mathrm{x}}=\frac{4\pm \sqrt{16-12}}{2}=3\mathrm{or}1$$$$\mathrm{x}+\frac{1}{\mathrm{x}}=3\Rightarrow {\mathrm{x}}^2-3\mathrm{x}+1=0\Rightarrow \mathrm{x}=\frac{3\pm \sqrt{5}}{2}$$$$\mathrm{x}+\frac{1}{\mathrm{x}}=1\Rightarrow {\mathrm{x}}^2-\mathrm{x}+1=0\Rightarrow \mathrm{x}=\frac{1\pm \mathrm{i}\sqrt{3}}{2}$$$$\Rightarrow \mathrm{solutions}:\mathrm{x}\in \{\frac{3-\sqrt{5}}{2},1,\frac{3+\sqrt{5}}{2},\frac{1+\mathrm{i}\sqrt{3}}{2},\frac{1-\mathrm{i}\sqrt{3}}{2}\}$$

Commented by Mikenice last updated on 21/Jun/22

$$thankssir$$

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