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Question Number 93170 by ar247 last updated on 11/May/20

x^2 +(1/x^2 )=27 (√x)+(1/(√x))=....

$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{27} \\ $$$$\sqrt{{x}}+\frac{\mathrm{1}}{\sqrt{{x}}}=.... \\ $$

Commented by mr W last updated on 11/May/20

(x+(1/x))^2 =x^2 +(1/x^2 )+2=27+2=29 x+(1/x)=±(√(29)) (√x)+(1/(√x))=t>0 t^2 =x+(1/x)+2=(√(29))+2 ⇒t=(√((√(29))+2))

$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{2}=\mathrm{27}+\mathrm{2}=\mathrm{29} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\pm\sqrt{\mathrm{29}} \\ $$$$\sqrt{{x}}+\frac{\mathrm{1}}{\sqrt{{x}}}={t}>\mathrm{0} \\ $$$${t}^{\mathrm{2}} ={x}+\frac{\mathrm{1}}{{x}}+\mathrm{2}=\sqrt{\mathrm{29}}+\mathrm{2} \\ $$$$\Rightarrow{t}=\sqrt{\sqrt{\mathrm{29}}+\mathrm{2}} \\ $$

Commented by i jagooll last updated on 11/May/20

let (√x) + (1/((√x) )) = t x+(1/x) = t^2 −2 x^2 +(1/x^2 )+2 = (t^2 −2)^2 29 = (t^2 −2)^2 t^2 = 2+(√(29)) t = (√(2+(√(29))))

$$\mathrm{let}\:\sqrt{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\sqrt{\mathrm{x}}\:}\:=\:\mathrm{t}\: \\ $$$$\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{t}^{\mathrm{2}} −\mathrm{2} \\ $$$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\mathrm{2}\:=\:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{29}\:=\:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{t}^{\mathrm{2}} \:=\:\mathrm{2}+\sqrt{\mathrm{29}}\: \\ $$$$\mathrm{t}\:=\:\sqrt{\mathrm{2}+\sqrt{\mathrm{29}}}\: \\ $$

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