1.

If x^2-x-1 =0 then what is the value of x^8 + 1/x^8

Answer» \textbf{Answer}


x^8 + 1/x^8 = ?

\textbf{We are given,}
x^2 - x - 1 = 0 -------(1)
\textbf{Dividing} both \textbf{sides of equation(1) by x,}
x^2/x - x/x - 1/x = 0/x
x^2 - 1/x - 1 = 0

=> \textbf{(x - 1/x) = 1}
\textbf{Squaring both sides,}
(x - 1/x)^2 = (1)^2
=> x^2 + 1/x^2 - 2(x)(1/x) = 1

=> (x^2 + 1/x^2) = 3
\textbf{Squaring both sides,}
(x^2 + 1/x^2)^2 = (3)^2
=> x^4 + 1/x^4 + 2(x^4)(1/x^4) = 9
=> x^4 + 1/x^4 + 2 = 9

=> (x^4 + 1/x^4) = 7
\textbf{Squaring both sides,}
x^8 + 1/x^8 + 2(x^8)(1/x^8) = (7)^2
=> x^8 + 1/x^8 + 2 = 49
=> x^8 + 1/x^8 = 49 - 2

=> x^8 + 1/x^8 = 47 \textbf{(Solution)}


\textbf{Hope My Answer Helped}
\textbf{Thanks}


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