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Question Number 62341 by Tawa1 last updated on 20/Jun/19

$$\mathrm{How}\mathrm{many}\mathrm{real}\mathrm{root}\mathrm{does}\mathrm{the}\mathrm{equation}$$$${\mathrm{x}}^8-{\mathrm{x}}^7+{2\mathrm{x}}^6-{2\mathrm{x}}^5+{3\mathrm{x}}^4-{3\mathrm{x}}^3+{4\mathrm{x}}^2-4\mathrm{x}+\frac{5}{2}=0\mathrm{has}$$

Commented by Kunal12588 last updated on 20/Jun/19

Commented by Kunal12588 last updated on 20/Jun/19

$$zerorealroots$$

Commented by Tawa1 last updated on 20/Jun/19

$$\mathrm{Can}\mathrm{we}\mathrm{show}\mathrm{it}\mathrm{without}\mathrm{graph}\mathrm{or}\mathrm{full}\mathrm{workings}\mathrm{sir}$$

Answered by mr W last updated on 20/Jun/19

$$f\left(x\right)={\mathrm{x}}^8-{\mathrm{x}}^7+{2\mathrm{x}}^6-{2\mathrm{x}}^5+{3\mathrm{x}}^4-{3\mathrm{x}}^3+{4\mathrm{x}}^2-4\mathrm{x}+\frac{5}{2}$$$$=(x-1){\mathrm{x}}^7+2(x-1){\mathrm{x}}^5+3(x-1){\mathrm{x}}^3+4(x-1)\mathrm{x}+\frac{5}{2}$$$$=x(x-1)[{\mathrm{x}}^6+{2\mathrm{x}}^4+{3\mathrm{x}}^2+4]+\frac{5}{2}$$$$=x(x-1)\{[{({\mathrm{x}}^2+1)}^2+2]{\mathrm{x}}^2+4\}+\frac{5}{2}$$$$⩾4x(x-1)+\frac{5}{2}$$$$=4[{(x-\frac{1}{2})}^3-\frac{1}{4}]+\frac{5}{2}$$$$⩾4\times (-\frac{1}{4})+\frac{5}{2}$$$$=\frac{3}{2}$$$$sincef\left(x\right)⩾\frac{3}{2},$$$$\Rightarrow f\left(x\right)=0hasnorealroots.$$

Commented by Tawa1 last updated on 21/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by MJS last updated on 20/Jun/19

$$x^7(x-1)+2x^5(x-1)+3x^3(x-1)+4x(x-1)+\frac{5}{2}=0$$$$x(x-1)(x^6+2x^4+3x^2+4)+\frac{5}{2}=0$$$$$$$$x(x-1)(x^6+2x^4+3x^2+4)=0$$$$x_1=0$$$$x_2=1$$$$g\left(x\right)=x^6+2x^4+3x^2+4\mathrm{obviously}\mathrm{has}\mathrm{got}\mathrm{no}\mathrm{real}$$$$\mathrm{zeros};\mathrm{its}\mathrm{minimum}\mathrm{is}\mathrm{at}\left(\begin{array}{c}0\\ 4\end{array}\right).h\left(x\right)=x(x-1)\mathrm{is}$$$$\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{minimum}\mathrm{at}\left(\begin{array}{c}\frac{1}{2}\\ -\frac{1}{4}\end{array}\right)$$$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{seek}\mathrm{the}\mathrm{minimum}\mathrm{of}$$$$f\left(x\right)=x(x-1)(x^6+2x^4+3x^2+4)\mathrm{with}0<x<1$$$$f(.4)=-1.08847104$$$$f(.5)=-1.22265625$$$$f(.6)=-1.29260544$$$$f(.7)=-1.27424829$$$$f(.65)=-1.29673...$$$$f(.64)=-1.29785...$$$$f(.63)=-1.29796...$$$$f(.62)=-1.29710...$$$$\Rightarrow \mathrm{the}\mathrm{minimum}\mathrm{is}\mathrm{close}\mathrm{to}\left(\begin{array}{c}.63\\ -1.298\end{array}\right)$$$$\Rightarrow f\left(x\right)+\frac{5}{2}\mathrm{has}\mathrm{its}\mathrm{minimum}\mathrm{close}\mathrm{to}$$$$\left(\begin{array}{c}.63\\ 1.202\end{array}\right)\Rightarrow \mathrm{it}\mathrm{has}\mathrm{no}\mathrm{real}\mathrm{zeros}$$

Commented by Tawa1 last updated on 21/Jun/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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