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Question Number 58046 by ANTARES VY last updated on 17/Apr/19

$$3{\mathbf{x}}^4-4{\mathbf{x}}^3-7{\mathbf{x}}^2-4\mathbf{x}+5=0$$$$\mathbf{x}=?$$

Answered by tanmay last updated on 17/Apr/19

$$f\left(x\right)=3x^4-4x^3-7x^2-4x+5$$$$f\left(0\right)=5sof\left(0\right)>0$$$$f\left(1\right)=-7f\left(1\right)<0$$$$sincef\left(0\right)>0andf\left(1\right)<0$$$$soonerootliesbetween(0,1)\leftarrow lookhere$$$$$$$$f\left(2\right)=48-32-28-8+5$$$$=53-68=-15$$$$sincef\left(1\right)<0andf\left(2\right)<0$$$$so\mathit{n}\mathit{o}\mathit{r}\mathit{o}\mathit{o}\mathit{t}between(1,2)\leftarrow lookhere$$$$$$$$f\left(3\right)=243-108-28-8+5$$$$=248-144$$$$=104$$$$f\left(3\right)>0$$$$sincef\left(2\right)<0andf\left(3\right)>0$$$$soonerootbetween(2,3)\leftarrow lookhere$$$$\mathit{f}\left(4\right)=389\mathit{f}\left(4\right)>0$$$$\mathit{s}\mathit{i}\mathit{n}\mathit{c}\mathit{e}\mathit{f}\left(3\right)>0\mathit{a}\mathit{n}\mathit{d}\mathit{f}\left(4\right)>0$$$$\mathit{s}\mathit{o}\mathit{n}\mathit{o}\mathit{r}\mathit{o}\mathit{o}\mathit{t}\mathit{b}\mathit{e}\mathit{t}\mathit{w}\mathit{e}\mathit{e}\mathit{n}(3,4)$$$$ihavetriedtofindthelocationofroots..$$$$$$$$$$

Answered by MJS last updated on 17/Apr/19

$$x_1\approx .571581$$$$x_2\approx 2.41140$$$$x_3\approx -.824822-.727241$$$$x_4\approx -.824822+.727241$$

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