If $x^2-6x+1=0$, then find the value of $x^2+\frac{1}{x^2}$.

Given: Quadratic equation: $x^2-6x+1=0$.

To do: To find the value of $x^2+\frac{1}{x^2}$.

Solution:

As given $x^2-6x+1=0$

$\Rightarrow x^2=6x-1$ ....... $( i)$

Put this value in $x^2+\frac{1}{x^2}$

$=( 6x-1)+\frac{1}{( 6x-1)}$

$=\frac{( 6x-1)^2+1}{( 6x-1)}$

$=\frac{36x^2-12x+1+1}{( 6x-1)}$

$=\frac{36x^2-12x+2}{6x-1}$

$=\frac{36x^2}{( 6x-1)}-\frac{2( 6x-1)}{( 6x-1)}$

$=\frac{36( 6x-1)}{( 6x-1)}-\frac{( 6x-1)}{( 6x-1)}$

$=36-2$

$=34$

Thus, $x^2+\frac{1}{x^2}=34$.

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Updated on: 10-Oct-2022

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