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The-number-of-negative-real-roots-of-x-4-4x-1-0-is-




Question Number 5639 by Daily last updated on 23/May/16
The number of negative real roots   of   x^4 −4x−1=0 is
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{negative}\:\mathrm{real}\:\mathrm{roots}\: \\ $$$$\mathrm{of}\:\:\:{x}^{\mathrm{4}} −\mathrm{4}{x}−\mathrm{1}=\mathrm{0}\:\mathrm{is} \\ $$
Commented by Yozzii last updated on 23/May/16
Descartes rule of signs may help.
$${Descartes}\:{rule}\:{of}\:{signs}\:{may}\:{help}. \\ $$
Answered by Rasheed Soomro last updated on 27/May/16
The equation has no sign variation  so it has no positive root.  Replace x by −x     (−x)^4 −4(−x)−1=0  ⇒x^4 +4x−1=0  This has  one variation of signs  So the given equation has one negative  root.
$${The}\:{equation}\:{has}\:{no}\:{sign}\:{variation} \\ $$$${so}\:{it}\:{has}\:{no}\:{positive}\:{root}. \\ $$$${Replace}\:{x}\:{by}\:−{x} \\ $$$$\:\:\:\left(−{x}\right)^{\mathrm{4}} −\mathrm{4}\left(−{x}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +\mathrm{4}{x}−\mathrm{1}=\mathrm{0} \\ $$$${This}\:{has}\:\:{one}\:{variation}\:{of}\:{signs} \\ $$$${So}\:{the}\:{given}\:{equation}\:{has}\:{one}\:{negative} \\ $$$${root}. \\ $$

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