Question


Answer

\because x^2-3x+1=0

\therefore x+\dfrac{1}{x} =3
(1)

Square both sides, we get,

x^2+\dfrac{1}{x^2} +2 = 9

\therefore x^2+\dfrac{1}{x^2} = 7
(2)

Multiply equation (1) with (2),


x^3+\dfrac{1}{x}+x+\dfrac{1}{x^3} = 21

Plug in (1), we get

x^3+\dfrac{1}{x^3} = 18
(3)

Square both sides of equation (2), we get

x^4+\dfrac{1}{x^4}+2 = 49

\therefore x^4+\dfrac{1}{x^4} = 47
(4)

Now we have come to the end of our conclusion.

Given the condistion x+\dfrac{1}{x} =3 ,

\begin{cases} x^2+\dfrac{1}{x^2} = 7 & \\ \\ x^3+\dfrac{1}{x^3} = 18 & \\ \\ x^4+\dfrac{1}{x^4} = 47 \end{cases}

Steven Zheng posted 3 months ago

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