Question-225098

Question Number 225098 by Ark55 last updated on 17/Oct/25

Answered by Rasheed.Sindhi last updated on 17/Oct/25

5x+(1/x)=6 (5x+(1/x)=6)^2 25x^2 +10+(1/x^2 )=36 25x^2 +(1/x^2 )=26

$$\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6} \\ $$$$\left(\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6}\right)^{\mathrm{2}} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} +\mathrm{10}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{36} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{26} \\ $$

Commented by fantastic last updated on 17/Oct/25

$${I}\:{think}\:{you}\:{have}\:{to}\:{reverse}\:{your} \\ $$$${assumption} \\ $$

Commented by Rasheed.Sindhi last updated on 17/Oct/25

If the question is meant as: Show that 5x+(1/x)=6 if 25x^2 +(1/x^2 )=26 25x^2 +(1/x^2 )=26 25x^2 +2(5x)((1/x))+(1/x^2 )=26+2(5x)((1/x)) (5x+(1/x))^2 =36 5x+(1/x)=6 ✓ or 5x+(1/x)=−6

$${If}\:{the}\:{question}\:{is}\:{meant}\:{as}: \\ $$$${Show}\:{that}\:\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6}\:{if}\:\mathrm{25}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{26} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{26} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{5}{x}\right)\left(\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{26}+\mathrm{2}\left(\mathrm{5}{x}\right)\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$\left(\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{36} \\ $$$$\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6}\:\checkmark\:{or}\:\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=−\mathrm{6} \\ $$

Commented by fantastic last updated on 17/Oct/25

$${yes} \\ $$

Answered by Rasheed.Sindhi last updated on 17/Oct/25

5x+(1/x)=6 5x^2 =6x−1 x^2 =((6x−1)/5) (1/x^2 )=(5/(6x−1)) 25x^2 +(1/x^2 )=5(6x−1)+(5/(6x−1)) =((5(6x−1)^2 +5)/(6x−1)) =((5(36x^2 −12x+1)+5)/(6x−1)) =((5(36(((6x−1)/5))−12x+1)+5)/(6x−1)) =((36(6x−1)−60x+5+5)/(6x−1)) =((156x−26)/(6x−1))=((26(6x−1))/(6x−1))=26✓

$$\mathrm{5}{x}+\frac{\mathrm{1}}{{x}}=\mathrm{6} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} =\mathrm{6}{x}−\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{6}{x}−\mathrm{1}}{\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{\mathrm{5}}{\mathrm{6}{x}−\mathrm{1}} \\ $$$$\: \\ $$$$\mathrm{25}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{5}\left(\mathrm{6}{x}−\mathrm{1}\right)+\frac{\mathrm{5}}{\mathrm{6}{x}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{5}\left(\mathrm{6}{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{5}}{\mathrm{6}{x}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{5}\left(\mathrm{36}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{1}\right)+\mathrm{5}}{\mathrm{6}{x}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{5}\left(\mathrm{36}\left(\frac{\mathrm{6}{x}−\mathrm{1}}{\mathrm{5}}\right)−\mathrm{12}{x}+\mathrm{1}\right)+\mathrm{5}}{\mathrm{6}{x}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{36}\left(\mathrm{6}{x}−\mathrm{1}\right)−\mathrm{60}{x}+\mathrm{5}+\mathrm{5}}{\mathrm{6}{x}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{156}{x}−\mathrm{26}}{\mathrm{6}{x}−\mathrm{1}}=\frac{\mathrm{26}\cancel{\left(\mathrm{6}{x}−\mathrm{1}\right)}}{\cancel{\mathrm{6}{x}−\mathrm{1}}}=\mathrm{26}\checkmark \\ $$

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