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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))

$${Solve}\:{for}\:{x}\:\epsilon\:\mathbb{R}\: \\ $$$$\:\sqrt{\sqrt{\mathrm{3}}−{x}}\:=\:{x}\sqrt{\sqrt{\mathrm{3}}+{x}}\: \\ $$

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

$$\:\sqrt{\sqrt{\mathrm{3}}−{x}}\:=\:{x}\sqrt{\sqrt{\mathrm{3}}+{x}}\: \\ $$$$\sqrt{\mathrm{3}}\:−{x}={x}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}\:+{x}\right) \\ $$$${x}^{\mathrm{2}} =\frac{\sqrt{\mathrm{3}}\:−{x}}{\:\sqrt{\mathrm{3}}\:+{x}}.\frac{\sqrt{\mathrm{3}}\:−{x}}{\:\sqrt{\mathrm{3}}\:−{x}}=\frac{\mathrm{3}+{x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\:{x}}{\mathrm{3}−{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} \left(\mathrm{3}−{x}^{\mathrm{2}} \right)={x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\:{x}+\mathrm{3} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\:{x}+\mathrm{3}−\mathrm{3}{x}^{\mathrm{2}} +{x}^{\mathrm{4}} =\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\:{x}+\mathrm{3}=\mathrm{0} \\ $$

Commented by blackmamba last updated on 11/Dec/21

only one is x = −((√3)/3)+(((30(√3)))^(1/(3 )) /3)

$${only}\:{one}\:{is}\:{x}\:=\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{30}\sqrt{\mathrm{3}}}}{\mathrm{3}} \\ $$

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)))

$$\sqrt{\mathrm{3}}−{x}={x}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}+{x}\right) \\ $$$${x}^{\mathrm{3}} +\sqrt{\mathrm{3}}{x}^{\mathrm{2}} +{x}−\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$${to}\:{avoid}\:\sqrt{\mathrm{3}},\:{let}\:{x}=\sqrt{\mathrm{3}}{t} \\ $$$$\Rightarrow{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +\frac{{t}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$${to}\:{avoid}\:{t}^{\mathrm{2}} \:{term},\:{let}\:{t}={s}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{s}^{\mathrm{3}} =\frac{\mathrm{10}}{\mathrm{27}} \\ $$$$\Rightarrow{s}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}}{\mathrm{3}} \\ $$$$\Rightarrow{t}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}−\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$

Commented by cortano last updated on 11/Dec/21

no. i want prove that the answer is same.

$${no}.\:{i}\:{want}\:{prove}\:{that}\:{the}\:{answer}\:{is}\:{same}.\: \\ $$

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)

$$\Rightarrow{x}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}}{\:\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt{\mathrm{3}}}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{10}}{\mathrm{3}\sqrt{\mathrm{3}}}×\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{3}\sqrt{\mathrm{3}}}} \\ $$$$\Rightarrow{x}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{30}\sqrt{\mathrm{3}}}}{\mathrm{3}} \\ $$

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).

$${what}\:{do}\:{you}\:{want}\:{to}\:{say}? \\ $$$${is}\:{x}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{30}\sqrt{\mathrm{3}}}}{\mathrm{3}}\:{more}\:{correct}\:{than} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:? \\ $$$${i}\:{prefer}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:{to}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\sqrt[{\mathrm{3}}]{\mathrm{30}\sqrt{\mathrm{3}}}}{\mathrm{3}}. \\ $$

Commented by mr W last updated on 11/Dec/21

yes. they are the same.

$${yes}.\:{they}\:{are}\:{the}\:{same}. \\ $$

Commented by cortano last updated on 11/Dec/21

thank you

$${thank}\:{you} \\ $$

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