For which values of b do the equations:
x^3 + bx^2 + 2bx - 1 = 0 and x^2 + (b-1)x + b = 0
have a common root?
I believe there's a little trick involved here:
Let
(1) $\displaystyle x^3+bx^2+2bx-1=0$
(2)$\displaystyle x^2+(b-1)x+b=0$
Observe that $\displaystyle x=-1$ is not a root of (2) for any choice of b. Hence this can not be a common root.
So we may replace the problem of finding b such that (3),(4) have a common root
(3) $\displaystyle x^3+bx^2+2bx-1=0$
(4) $\displaystyle (x+1)(x^2+(b-1)x+b)=0 $
Then a common root of (3),(4) must also be a root of the difference of (3),(4)
$\displaystyle (3)-(4)= x^3+bx^2+2bx-1 - (x+1)(x^2+(b-1)x+b) = x-(1+b)$
This implies that $\displaystyle x=1+b$ is the only possible common root they can have.
substituting $\displaystyle x= 1+b$ in (2) gives 2 possible values for b.