Respuesta :

We have

[tex]\left(x+\dfrac1x\right)^2=x^2+\dfrac{2x}x+\dfrac1{x^2}=x^2+2+\dfrac1{x^2}[/tex]

and given that

[tex]x^2+\dfrac1{x^2}=38[/tex]

it follows that

[tex]\left(x+\dfrac1x\right)^2=40\implies x+\dfrac1x=\pm\sqrt{40}[/tex]

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