Question

If $x+\frac{1}{x}=6$ then find $x^2+\frac{1}{x^2}-11$

Answer 1

$x+\frac{1}{x}=6$ [ Given ]

$\Rightarrow$ ${(x+\frac{1}{x})}^2=(6{)}^2$ [ Squaring both sides ]

$\Rightarrow$ $x^2+2\times x\times \frac{1}{x}+\frac{1}{x^2}=36$

$\Rightarrow$ $x^2+\frac{1}{x^2}+2=36$

$\therefore$ $x^2+\frac{1}{x^2}=34$ ------- ( 1 )

Now,

$\Rightarrow$ $x^2+\frac{1}{x^2}-11$

$\Rightarrow$ $34-11$ [ From ( 1 ) ]

$\Rightarrow$ $23$

$\therefore$ $x^2+\frac{1}{x^2}-11=23$