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Question Number 161035 by cortano last updated on 11/Dec/21
Solve for x ε R (√((√3)−x)) = x(√((√3)+x))
SolveforxϵR3−x=x3+x
Answered by Rasheed.Sindhi last updated on 11/Dec/21
(√((√3)−x)) = x(√((√3)+x)) (√3) −x=x^2 ((√3) +x) x^2 =(((√3) −x)/( (√3) +x)).(((√3) −x)/( (√3) −x))=((3+x^2 −2(√3) x)/(3−x^2 )) x^2 (3−x^2 )=x^2 −2(√3) x+3 x^2 −2(√3) x+3−3x^2 +x^4 =0 x^4 −2x^2 −2(√3) x+3=0
3−x=x3+x3−x=x2(3+x)x2=3−x3+x.3−x3−x=3+x2−23x3−x2x2(3−x2)=x2−23x+3x2−23x+3−3x2+x4=0x4−2x2−23x+3=0
Commented by blackmamba last updated on 11/Dec/21
only one is x = −((√3)/3)+(((30(√3)))^(1/(3 )) /3)
onlyoneisx=−33+30333
Commented by cortano last updated on 11/Dec/21
typo
typo
Answered by mr W last updated on 11/Dec/21
(√3)−x=x^2 ((√3)+x) x^3 +(√3)x^2 +x−(√3)=0 to avoid (√3), let x=(√3)t ⇒t^3 +t^2 +(t/3)−(1/3)=0 to avoid t^2 term, let t=s−(1/3) ⇒s^3 =((10)/(27)) ⇒s=(((10))^(1/3) /3) ⇒t=((((10))^(1/3) −1)/3) ⇒x=((((10))^(1/3) −1)/( (√3)))
3−x=x2(3+x)x3+3x2+x−3=0toavoid3,letx=3t⇒t3+t2+t3−13=0toavoidt2term,lett=s−13⇒s3=1027⇒s=1033⇒t=103−13⇒x=103−13
Commented by cortano last updated on 11/Dec/21
no. i want prove that the answer is same.
no.iwantprovethattheanswerissame.
Commented by cortano last updated on 11/Dec/21
⇒x=−((√3)/3) +(((10))^(1/3) /( ((3(√3)))^(1/3) ))=−((√3)/3)+((((10)/(3(√3)))×((3(√3))/(3(√3)))))^(1/3) ⇒x=−((√3)/3)+(((30(√3)))^(1/3) /3)
⇒x=−33+103333=−33+1033×33333⇒x=−33+30333
Commented by mr W last updated on 11/Dec/21
what do you want to say? is x=−((√3)/3)+(((30(√3)))^(1/3) /3) more correct than x=((((10))^(1/3) −1)/( (√3))) ? i prefer ((((10))^(1/3) −1)/( (√3))) to −((√3)/3)+(((30(√3)))^(1/3) /3).
whatdoyouwanttosay?isx=−33+30333morecorrectthanx=103−13?iprefer103−13to−33+30333.
Commented by mr W last updated on 11/Dec/21
yes. they are the same.
yes.theyarethesame.
Commented by cortano last updated on 11/Dec/21
thank you
thankyou

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