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Question Number 31927 by rahul 19 last updated on 16/Mar/18

$$Thenumberofdistinctrealroots$$$$ofequation{\mathit{x}}^4-4{\mathit{x}}^3+12{\mathit{x}}^2+\mathit{x}-1=0.$$

Answered by MJS last updated on 16/Mar/18

$$f\left(x\right)=x^4-4x^3+12x^2+x-1$$$$f^{\prime }\left(x\right)=4x^3-12x^2+24x+1$$$$f^{″}\left(x\right)=12x^2-24x+24$$$$f^{″}\left(x\right)=0\Rightarrow \mathrm{no}\mathrm{real}\mathrm{solution}\Rightarrow$$$$\Rightarrow f\left(x\right)\mathrm{has}\mathrm{no}\mathrm{inflexion}\mathrm{point}\Rightarrow$$$$\Rightarrow \mathrm{it}\mathrm{looks}\mathrm{like}\mathrm{a}\mathrm{hanging}\mathrm{parabola}$$$$(\mathrm{because}{\mathrm{it}}^{\prime }\mathrm{s}+1\times x^4)\Rightarrow$$$$\Rightarrow \mathrm{it}\mathrm{has}0,1\mathrm{or}2\mathrm{zeros}$$$$f^{\prime }\left(x\right)\mathrm{has}\mathrm{only}1\mathrm{zero}(\mathrm{because}$$$$f^{″}\left(x\right)>0\mathrm{for}x\in \mathbb{R}),\mathrm{by}\mathrm{trying}\mathrm{we}$$$$\mathrm{find}-.041<x<-.040\Rightarrow$$$$\Rightarrow f\left(x\right)\mathrm{has}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{minimum}\mathrm{there}$$$$f(-.045)=-1.02...<0\Rightarrow$$$$\Rightarrow f\left(x\right)\mathrm{has}2\mathrm{real}\mathrm{zeros}(x_1\approx -.314$$$$x_2\approx .259)$$

Commented by rahul 19 last updated on 18/Mar/18

$$thankusir!$$

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