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1-find-two-factors-of-1000001-other-than-1-and-1000001-2-x-2-5x-5-x-2-2x-24-1-what-is-the-value-of-the-product-of-the-solutions-




Question Number 27983 by JI Siam last updated on 18/Jan/18
1) find two  factors of 1000001 other than 1 and 1000001  2)(x^2 −5x+5)^((x^2 +2x−24)) =1 what is the value  of the product of the solutions?
$$\left.\mathrm{1}\right)\:\mathrm{find}\:\mathrm{two}\:\:\mathrm{factors}\:\mathrm{of}\:\mathrm{1000001}\:\mathrm{other}\:\mathrm{than}\:\mathrm{1}\:\mathrm{and}\:\mathrm{1000001} \\ $$$$\left.\mathrm{2}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{5}\right)^{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{24}\right)} =\mathrm{1}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solutions}? \\ $$
Answered by mrW2 last updated on 18/Jan/18
(1)  1000001=101×9901  (2)  x^2 −5x+5=1⇒x^2 −5x+4=0⇒Πx=4  x^2 +2x−24=0⇒Πx=−24  ⇒Πx=−24×4=−96
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{1000001}=\mathrm{101}×\mathrm{9901} \\ $$$$\left(\mathrm{2}\right) \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{5}=\mathrm{1}\Rightarrow{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}=\mathrm{0}\Rightarrow\Pi{x}=\mathrm{4} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{24}=\mathrm{0}\Rightarrow\Pi{x}=−\mathrm{24} \\ $$$$\Rightarrow\Pi{x}=−\mathrm{24}×\mathrm{4}=−\mathrm{96} \\ $$
Commented by Rasheed.Sindhi last updated on 18/Jan/18
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Answered by abwayh last updated on 19/Jan/18
sol.(2)  (x^2 −5x+5)^((x^2 +2x−24)) =1  (x^2 −5x+5)^((x^2 +2x−24)) =(x^2 −5x+5)^0   ∴ x^2 +2x−24=0  (x+6)(x−4)=0   ⇒ x=−6                                             x=4
$$\mathrm{sol}.\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{5}\right)^{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{24}\right)} =\mathrm{1} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{5}\right)^{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{24}\right)} =\left(\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{5}\right)^{\mathrm{0}} \\ $$$$\therefore\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{24}=\mathrm{0} \\ $$$$\left(\mathrm{x}+\mathrm{6}\right)\left(\mathrm{x}−\mathrm{4}\right)=\mathrm{0}\:\:\:\Rightarrow\:\mathrm{x}=−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{4} \\ $$

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