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Question Number 83554 by Power last updated on 03/Mar/20

Answered by MJS last updated on 03/Mar/20

$$x^4+4x^3-8x+\frac{17}{5}=0$$$$\mathrm{let}x=t-1$$$$t^4-6t^2+\frac{42}{5}=0$$$$t^2=3\pm \frac{\sqrt{15}}{5}$$$$t=\pm \sqrt{3\pm \frac{\sqrt{15}}{5}}$$$$x=-1\pm \sqrt{3\pm \frac{\sqrt{15}}{5}}$$

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