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Question Number 60910 by Tawa1 last updated on 27/May/19

Commented by Rasheed.Sindhi last updated on 27/May/19

$$S\mathrm{olve}\mathrm{the}\mathrm{equation}{({\mathrm{x}}^2-\mathrm{x}+1)}^2-4\mathrm{x}{(\mathrm{x}-1)}^2=0$$$$$$$$x^4+x^2+1-2x^3-2x+2x^2-4x(x^2-2x+1)=0$$$$x^4+x^2+1-2x^3-2x+2x^2-4x^3+8x^2-4x=0$$$$x^4-6x^3+11x^2-6x+1=0$$$$x^2(x^2+\frac{1}{x^2}-6x-\frac{6}{x}+11)=0$$$$x^2=0\mid x^2+\frac{1}{x^2}-6(x+\frac{1}{x})+11=0$$$$x=0\left(\mathcal{E}xtraneousroot\right)$$$$x^2+2+\frac{1}{x^2}-6(x+\frac{1}{x})+11-2=0$$$${(x+\frac{1}{x})}^2-6(x+\frac{1}{x})+9=0$$$${(x+\frac{1}{x}-3)}^2=0$$$$x+\frac{1}{x}-3=0$$$$x+\frac{1}{x}=3$$$$x^2-3x+1=0$$$$x=\frac{3\pm \sqrt{5}}{2}$$$$x=\frac{3\pm \sqrt{5}}{2}$$

Commented by Tawa1 last updated on 27/May/19

$$\mathrm{Prove}\mathrm{that}\mathrm{if}\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{y}+1,\mathrm{then}\frac{{({\mathrm{x}}^2-\mathrm{x}+1)}^2}{\mathrm{x}{(\mathrm{x}-1)}^2}=\frac{{\mathrm{y}}^2}{\mathrm{y}-1}.\mathrm{Hence}\mathrm{or}$$$$\mathrm{otherwise}\mathrm{solve}\mathrm{the}\mathrm{equation}{({\mathrm{x}}^2-\mathrm{x}+1)}^2-4\mathrm{x}{(\mathrm{x}-1)}^2=0$$

Commented by Tawa1 last updated on 27/May/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir},\mathrm{i}\mathrm{appreciate}:$$

Commented by MJS last updated on 27/May/19

$$\mathrm{after}\mathrm{the}\mathrm{proof}:$$$$x+\frac{1}{x}=y+1\wedge \frac{{(x^2-x+1)}^2}{x{(x-1)}^2}=\frac{y^2}{y-1}\Rightarrow$$$$\Rightarrow {(x^2-x+1)}^2-4x{(x-1)}^2=0\mathrm{can}\mathrm{be}\mathrm{written}$$$$\mathrm{as}$$$$\frac{y^2}{y-1}-4=0\Rightarrow y^2-4y+4=0\Rightarrow {(y-2)}^2=0\Rightarrow y=2$$$$x+\frac{1}{x}=y+1\Rightarrow x+\frac{1}{x}-3=0\Rightarrow x^2-3x+1=0\Rightarrow$$$$\Rightarrow x=\frac{3}{2}\pm \frac{\sqrt{5}}{2}$$

Commented by Rasheed.Sindhi last updated on 27/May/19

$$Verysimplesir!$$$$Ialsohadtriedtoreplacexbyybut$$$$I{couldn}^{\prime }tdoitperfectly.$$

Commented by Tawa1 last updated on 27/May/19

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

Commented by maxmathsup by imad last updated on 28/May/19

$$1)letprovethat(y-1){(x^2-x+1)}^2=y^{2}x{(x-1)}^{2}wehave$$$$(y-1){(x^2-x+1)}^2=(x+\frac{1}{x}-2){(x^2-x+1)}^2$$$$=\frac{x^2-2x+1}{x}{(x^2-x+1)}^2=\frac{{(x-1)}^2{(x^2-x+1)}^2}{x}$$$$fromatherside$$$$y^{2}x{(x-1)}^2={(x+\frac{1}{x}-1)}^{2}x{(x-1)}^2=\frac{{(x^2-x+1)}^2{(x-1)}^2}{x}$$$$sotheequalityisproved.$$$$2)wehave{(x^2-x+1)}^2-4x{(x-1)}^2=0\iff {(x^2-x+1)}^2=4x{(x-1)}^2\Rightarrow$$$$\frac{{(x^2-x+1)}^2}{x{(x-1)}^2}=4\Rightarrow \frac{y^2}{y-1}=4withy=x+\frac{1}{x}-1$$$$\Rightarrow y^2=4y-4\Rightarrow y^2-4y+4=0\Rightarrow {(y-2)}^2=0\Rightarrow y=2\Rightarrow$$$$x+\frac{1}{x}-1=2\Rightarrow x+\frac{1}{x}=3\Rightarrow x^2+1=3x\Rightarrow x^2-3x+1=0$$$$\mathrm{\Delta }=9-4=5\Rightarrow x_1=\frac{3+\sqrt{5}}{2}and{x}_2=\frac{3-\sqrt{5}}{2}.$$$$$$

Answered by Rasheed.Sindhi last updated on 27/May/19

$$x+\frac{1}{x}=y+1\stackrel{?}{\Rightarrow }\frac{{(x^2-x+1)}^2}{x{(x-1)}^2}=\frac{y^2}{y-1}$$$$x+\frac{1}{x}=y+1\Rightarrow y=\frac{x^2-x+1}{x}$$$$So,$$$$\frac{{(x^2-x+1)}^2}{x{(x-1)}^2}=\frac{y^2}{y-1}$$$$=\frac{{\left(\frac{x^2-x+1}{x}\right)}^2}{\left(\frac{x^2-x+1}{x}\right)-1}$$$$=\frac{\frac{{(x^2-x+1)}^2}{x^2}}{\frac{x^2-2x+1}{x}}$$$$=\frac{{(x^2-x+1)}^2}{x^2}\times \frac{x}{{(x-1)}^2}$$$$=\frac{{(x^2-x+1)}^2}{x{(x-1)}^2}\left(Proved\right)$$$$Forsolutionoftheequationseemy$$$$commentbelowthequestion.$$

Commented by Tawa1 last updated on 27/May/19

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

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