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3x-4-4x-3-7x-2-4x-5-0-x-




Question Number 58046 by ANTARES VY last updated on 17/Apr/19
3x^4 −4x^3 −7x^2 −4x+5=0  x=?
$$\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{4}} −\mathrm{4}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{7}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{5}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}=? \\ $$
Answered by tanmay last updated on 17/Apr/19
f(x)=3x^4 −4x^3 −7x^2 −4x+5  f(0)=5         so    f(0)>0  f(1)=−7              f(1)<0  since f(0)>0 and f(1)<0  so one root lies between  (0,1)←look here    f(2)=48−32−28−8+5          =53−68=−15  since f(1)<0  and f(2)<0   so no root between (1,2)←look here    f(3)=243−108−28−8+5            =248−144            =104  f(3)>0  since f(2)<0   and f(3)>0  so one root between  (2,3)←look here  f(4)=389   f(4)>0  since f(3)>0  and f(4)>0  so no root between (3,4)  i have tried to find the location of roots..
$${f}\left({x}\right)=\mathrm{3}{x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} −\mathrm{7}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{5}\:\:\:\:\:\:\:\:\:{so}\:\:\:\:{f}\left(\mathrm{0}\right)>\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=−\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$$${since}\:{f}\left(\mathrm{0}\right)>\mathrm{0}\:{and}\:{f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$$${so}\:{one}\:{root}\:{lies}\:{between}\:\:\left(\mathrm{0},\mathrm{1}\right)\leftarrow{look}\:{here} \\ $$$$ \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{48}−\mathrm{32}−\mathrm{28}−\mathrm{8}+\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{53}−\mathrm{68}=−\mathrm{15} \\ $$$${since}\:{f}\left(\mathrm{1}\right)<\mathrm{0}\:\:{and}\:{f}\left(\mathrm{2}\right)<\mathrm{0}\: \\ $$$${so}\:\boldsymbol{{no}}\:\boldsymbol{{root}}\:{between}\:\left(\mathrm{1},\mathrm{2}\right)\leftarrow{look}\:{here} \\ $$$$ \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{243}−\mathrm{108}−\mathrm{28}−\mathrm{8}+\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{248}−\mathrm{144} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{104} \\ $$$${f}\left(\mathrm{3}\right)>\mathrm{0} \\ $$$${since}\:{f}\left(\mathrm{2}\right)<\mathrm{0}\:\:\:{and}\:{f}\left(\mathrm{3}\right)>\mathrm{0} \\ $$$${so}\:{one}\:{root}\:{between}\:\:\left(\mathrm{2},\mathrm{3}\right)\leftarrow{look}\:{here} \\ $$$$\boldsymbol{{f}}\left(\mathrm{4}\right)=\mathrm{389}\:\:\:\boldsymbol{{f}}\left(\mathrm{4}\right)>\mathrm{0} \\ $$$$\boldsymbol{{since}}\:\boldsymbol{{f}}\left(\mathrm{3}\right)>\mathrm{0}\:\:\boldsymbol{{and}}\:\boldsymbol{{f}}\left(\mathrm{4}\right)>\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{no}}\:\boldsymbol{{root}}\:\boldsymbol{{between}}\:\left(\mathrm{3},\mathrm{4}\right) \\ $$$${i}\:{have}\:{tried}\:{to}\:{find}\:{the}\:{location}\:{of}\:{roots}.. \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 17/Apr/19
x_1 ≈.571581  x_2 ≈2.41140  x_3 ≈−.824822−.727241  x_4 ≈−.824822+.727241
$${x}_{\mathrm{1}} \approx.\mathrm{571581} \\ $$$${x}_{\mathrm{2}} \approx\mathrm{2}.\mathrm{41140} \\ $$$${x}_{\mathrm{3}} \approx−.\mathrm{824822}−.\mathrm{727241} \\ $$$${x}_{\mathrm{4}} \approx−.\mathrm{824822}+.\mathrm{727241} \\ $$

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