Question 913834
solve {{{1-x^-1-20x^-2=0}}}
let u=x^-1
u^2=x^-2
1-u-20u^2=0
20u^2+u-1=0
(4u+1)(5u-1)=0
u=-1/4=x^-1
-1/4=(1/x)
x=-4
or
u=1/5=x^-1
1/5=1/x
x=5
Check:
f(-4)=1-x^-1-20x^-2
f(-4)=1-1/x-20/x^2
f(-4)=1+1/4-20/16=5/4-5/4=0
f(5)=1-1/5-20/25=4/5-4/5=0
Both zeros check ok