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x-1-4-lt-5x-3-21x-2-17x-61-find-the-root-x-




Question Number 58045 by ANTARES VY last updated on 17/Apr/19
(x+1)^4 <5x^3 +21x^2 +17x+61  find  the  root   x?
$$\left(\boldsymbol{\mathrm{x}}+\mathrm{1}\right)^{\mathrm{4}} <\mathrm{5}\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{21}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{17}\boldsymbol{\mathrm{x}}+\mathrm{61} \\ $$$$\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{root}}\:\:\:\boldsymbol{\mathrm{x}}? \\ $$
Answered by tanmay last updated on 17/Apr/19
x^4 +4c_1 x^3 +4c_2 x^2 +4c_3 x+1<5x^3 +21x^2 +17x+61  x^4 +4x^3 +6x^2 +4x+1<5x^3 +21x^2 +17x+61  x^4 −x^3 −15x^2 −13x−60<0  −x^4 +x^3 +15x^2 +13x+60>0  f(x)=−x^4 +x^3 +15x^2 +13x+60  f(0)=60   f(0)>0  f(1)>0  f(2)>0  f(3)>0  f(4)>0  f(5)=0  so  when x∈ [0,5)     then f(x)>0
$${x}^{\mathrm{4}} +\mathrm{4}{c}_{\mathrm{1}} {x}^{\mathrm{3}} +\mathrm{4}{c}_{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}{c}_{\mathrm{3}} {x}+\mathrm{1}<\mathrm{5}{x}^{\mathrm{3}} +\mathrm{21}{x}^{\mathrm{2}} +\mathrm{17}{x}+\mathrm{61} \\ $$$${x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}<\mathrm{5}{x}^{\mathrm{3}} +\mathrm{21}{x}^{\mathrm{2}} +\mathrm{17}{x}+\mathrm{61} \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{15}{x}^{\mathrm{2}} −\mathrm{13}{x}−\mathrm{60}<\mathrm{0} \\ $$$$−{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} +\mathrm{13}{x}+\mathrm{60}>\mathrm{0} \\ $$$${f}\left({x}\right)=−{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} +\mathrm{13}{x}+\mathrm{60} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{60}\: \\ $$$${f}\left(\mathrm{0}\right)>\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)>\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)>\mathrm{0} \\ $$$${f}\left(\mathrm{3}\right)>\mathrm{0} \\ $$$${f}\left(\mathrm{4}\right)>\mathrm{0} \\ $$$${f}\left(\mathrm{5}\right)=\mathrm{0} \\ $$$${so}\:\:{when}\:{x}\in\:\left[\mathrm{0},\mathrm{5}\right)\:\:\:\:\:{then}\:{f}\left({x}\right)>\mathrm{0} \\ $$

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