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Question Number 212448 by hardmath last updated on 13/Oct/24

$$\mathrm{If}{\mathrm{x}}^2-\mathrm{x}+3=0$$$$\mathrm{Find}\mathrm{x}+\frac{9}{{\mathrm{x}}^2}=?$$

Answered by A5T last updated on 13/Oct/24

$$x^2-x+3=0\Rightarrow x^3=x^2-3x=x-3-3x=-2x-3$$$$x+\frac{9}{x^2}=\frac{x^3+9}{x^2}=\frac{-2x-3+9}{x-3}=\frac{-2(x-3)}{x-3}=-2$$

Answered by A5T last updated on 13/Oct/24

$$x^2-x+3=0\stackrel{/x^2}{\Rightarrow }1-\frac{1}{x}+\frac{3}{x^2}=0\Rightarrow \frac{9}{x^2}=-3+\frac{3}{x}...\left(i\right)$$$$x^2-x+3=0\stackrel{/x}{\Rightarrow }x-1+\frac{3}{x}=0\Rightarrow x=1-\frac{3}{x}...\left(ii\right)$$$$\left(i\right)+\left(ii\right)\Rightarrow \frac{9}{x^2}+x=-2$$

Answered by Ghisom last updated on 14/Oct/24

$$3=x(1-x)$$$$9=x^2{(1-x)}^2$$$$x+\frac{x^2{(1-x)}^2}{x^2}=x^2-x+1=-2$$

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