\(e(aU,bV) = e(U,V)^{ab} = e(abU, V) = e(U, abV ) = e(bU,aV)\)
In this case we will use pairing crypto to prove that we know the value of \(x\) which solves \(x^2 + ax +b=0\). For example if we have \(x^2-x-42=0\) has the solution of \(x=7\) and \(x=-6\) as \((x-7)(x+6)=0\).