We have :

\(x^2-5x+1=0\)

\(\Rightarrow x^2+1=5x\) ( 1 )

\(x^2+\dfrac{1}{x^2}=\left(x^2+\dfrac{1}{x^2}+2.\dfrac{1}{x}.x\right)-2.\dfrac{1}{x}.x=\left(x+\dfrac{1}{x}\right)^2-2\)

                \(=\left(\dfrac{x^2+1}{x}\right)^2-2\) ( 2 )

( 1 )( 2 )\(\Rightarrow x^2+\dfrac{1}{x^2}=\left(\dfrac{5x}{x}\right)^2-2=25-2=23\)

Ans : 23.

We have :

x2−5x+1=0

⇒x2+1=5x

 ( 1 )

x2+1x2=(x2+1x2+2.1x.x)−2.1x.x=(x+1x)2−2

                =(x2+1x)2−2

 ( 2 )

( 1 )( 2 )⇒x2+1x2=(5xx)2−2=25−2=23

Ans : 23.