Let the equivalent resistance between `A` and `B` is `x`. We may consider the given circuit as shown in figure.
in this diagram
`R_(AB)=(2x)/(2+x)+1` or `x+(2x)/(2+x)+1` (as `R_(AB)=x)`
or `x(2+x)=2x+2+x` or `x^2-x-x-2=0`
`x=(1+-sqrt(1+8))/2=-1Omega` and `2Omega`
Ignoring the negative value, we have `R_(AB)=x=2Omega`