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Question Number 147941 by mathdanisur last updated on 24/Jul/21

x^2 - 6x + 3 = 0 ⇒ ((2x^2 )/3) + (6/x^2 ) = ?

$${x}^{\mathrm{2}} \:-\:\mathrm{6}{x}\:+\:\mathrm{3}\:=\:\mathrm{0}\:\:\:\Rightarrow\:\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}}\:+\:\frac{\mathrm{6}}{{x}^{\mathrm{2}} }\:=\:? \\ $$

Answered by Rasheed.Sindhi last updated on 24/Jul/21

x^2 - 6x + 3 = 0⇒x^2 =6x−3 ((2x^2 )/3) + (6/x^2 ) =((2(6x−3))/3)+(6/(6x−3)) =4x−2+(2/(2x−1))=2(2x−1+(1/(2x−1))) x^2 - 6x + 3 = 0⇒x=((6±(√(36−12)))/2) x=3±(√6)⇒2x−1=5±2(√6) ((2x^2 )/3) + (6/x^2 ) =2(2x−1+(1/(2x−1))) =2(5±2(√6)+(1/(5±2(√6)))) (1/(5±2(√6)))=(1/(5±2(√6)))×((5∓2(√6))/(5∓2(√6)))=((5∓2(√6))/(25−24))=5∓2(√6) 2(5±2(√6)+5∓2(√6))=20

$${x}^{\mathrm{2}} \:-\:\mathrm{6}{x}\:+\:\mathrm{3}\:=\:\mathrm{0}\Rightarrow{x}^{\mathrm{2}} =\mathrm{6}{x}−\mathrm{3} \\ $$ $$\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}}\:+\:\frac{\mathrm{6}}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{2}\left(\mathrm{6}{x}−\mathrm{3}\right)}{\mathrm{3}}+\frac{\mathrm{6}}{\mathrm{6}{x}−\mathrm{3}} \\ $$ $$=\mathrm{4}{x}−\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}{x}−\mathrm{1}}=\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\right) \\ $$ $${x}^{\mathrm{2}} \:-\:\mathrm{6}{x}\:+\:\mathrm{3}\:=\:\mathrm{0}\Rightarrow{x}=\frac{\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{12}}}{\mathrm{2}} \\ $$ $$\:\:\:{x}=\mathrm{3}\pm\sqrt{\mathrm{6}}\Rightarrow\mathrm{2}{x}−\mathrm{1}=\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{6}} \\ $$ $$\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}}\:+\:\frac{\mathrm{6}}{{x}^{\mathrm{2}} }\:=\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\right) \\ $$ $$=\mathrm{2}\left(\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{6}}}\right) \\ $$ $$\:\frac{\mathrm{1}}{\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{6}}}=\frac{\mathrm{1}}{\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{6}}}×\frac{\mathrm{5}\mp\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{5}\mp\mathrm{2}\sqrt{\mathrm{6}}}=\frac{\mathrm{5}\mp\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{25}−\mathrm{24}}=\mathrm{5}\mp\mathrm{2}\sqrt{\mathrm{6}} \\ $$ $$\mathrm{2}\left(\mathrm{5}\pm\cancel{\mathrm{2}\sqrt{\mathrm{6}}}+\mathrm{5}\mp\cancel{\mathrm{2}\sqrt{\mathrm{6}}}\right)=\mathrm{20} \\ $$

Commented bymathdanisur last updated on 24/Jul/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

Answered by lyubita last updated on 25/Jul/21

x^2 + 3 = 6x x + (3/x) = 6 x^2 + (9/x^2 ) = 30 ((2x^2 )/3) + (6/x^(2 ) ) = 20

$${x}^{\mathrm{2}} \:+\:\mathrm{3}\:=\:\mathrm{6}{x} \\ $$ $${x}\:+\:\frac{\mathrm{3}}{{x}}\:=\:\mathrm{6} \\ $$ $${x}^{\mathrm{2}} \:+\:\frac{\mathrm{9}}{{x}^{\mathrm{2}} }\:=\:\mathrm{30} \\ $$ $$\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}}\:+\:\frac{\mathrm{6}}{{x}^{\mathrm{2}\:} }\:=\:\mathrm{20} \\ $$

Commented byRasheed.Sindhi last updated on 25/Jul/21

Nice!

$$\mathcal{N}{ice}! \\ $$

Commented bylyubita last updated on 25/Jul/21

thank you mr rasheed

$${thank}\:{you}\:{mr}\:{rasheed} \\ $$

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