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Question Number 191552 by MATHEMATICSAM last updated on 25/Apr/23

$$\mathrm{If}{x}^2-3x+1=0\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}$$$${(x^2+x+\frac{1}{x}+\frac{1}{x^2})}^2$$

Answered by mr W last updated on 25/Apr/23

$$x^2-3x+1=0$$$$\Rightarrow x+\frac{1}{x}=3$$$${(x^2+x+\frac{1}{x}+\frac{1}{x^2})}^2$$$$={(x+\frac{1}{x}+x^2+\frac{1}{x^2})}^2$$$$={[(x+\frac{1}{x})+{(x+\frac{1}{x})}^2-2]}^2$$$$={(3+3^2-2)}^2$$$$=100✓$$

Answered by Rasheed.Sindhi last updated on 26/Apr/23

$$\mathrm{AnOther}\mathrm{W}ay...$$$$x^2-3x+1=0\Rightarrow 1=3x-x^2=x(3-x)$$$${(x^2+x+\frac{1}{x}+\frac{1}{x^2})}^2$$$$={(x^2+x+\frac{x(3-x)}{x}+\frac{x(3-x)}{x^2})}^2$$$$={(x^2+x+3-x+\frac{3-x}{x})}^2$$$$={(x^2+3+\frac{3-x}{x})}^2$$$$={\left(\frac{x^3+3x+3-x}{x}\right)}^2$$$$={\left(\frac{x^3+2x+3\times 1}{x}\right)}^2$$$$={\left(\frac{x^3+2x+3x(3-x)}{x}\right)}^2$$$$={\left(\frac{x^3+2x+9x-3x^2}{x}\right)}^2$$$$={(x^2-3x+1+10)}^2$$$$={(0+10)}^2=100✓$$

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