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Question Number 27983 by JI Siam last updated on 18/Jan/18

$$1)\mathrm{find}\mathrm{two}\mathrm{factors}\mathrm{of}1000001\mathrm{other}\mathrm{than}1\mathrm{and}1000001$$$$2){({\mathrm{x}}^2-5\mathrm{x}+5)}^{({\mathrm{x}}^2+2\mathrm{x}-24)}=1\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{solutions}?$$

Answered by mrW2 last updated on 18/Jan/18

$$\left(1\right)$$$$1000001=101\times 9901$$$$\left(2\right)$$$$x^2-5x+5=1\Rightarrow x^2-5x+4=0\Rightarrow \mathrm{\Pi }x=4$$$$x^2+2x-24=0\Rightarrow \mathrm{\Pi }x=-24$$$$\Rightarrow \mathrm{\Pi }x=-24\times 4=-96$$

Commented by Rasheed.Sindhi last updated on 18/Jan/18

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Answered by abwayh last updated on 19/Jan/18

$$\mathrm{sol}.\left(2\right)$$$${({\mathrm{x}}^2-5\mathrm{x}+5)}^{({\mathrm{x}}^2+2\mathrm{x}-24)}=1$$$${({\mathrm{x}}^2-5\mathrm{x}+5)}^{({\mathrm{x}}^2+2\mathrm{x}-24)}={({\mathrm{x}}^2-5\mathrm{x}+5)}^0$$$$\therefore {\mathrm{x}}^2+2\mathrm{x}-24=0$$$$(\mathrm{x}+6)(\mathrm{x}-4)=0\Rightarrow \mathrm{x}=-6$$$$\mathrm{x}=4$$

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