For which real values of b do the 2 equations below have a common root??

Question

1) x^3 + bx^2 + 2bx - 1 =0
2) x^2 + (b-1)x + b =0

Answer 1

Replace equation (1) by the combination (1)-x(2), namely:
(3) x^2 + b x - 1 = 0
Subtract (3) from (2) and you obtain:
(4) -x + b+1 = 0

So, the only common root (if there is any) must be x = b+1
As (4) is so satisfied, (1) and (2) will be satisfied for x=b+1 if and only if (3) is satisfied as well, which means:
(b+1)^2 + b (b+1) - 1 = 0 which means (2b+3) b = 0

Therefore (1) and (2) have a common root if and only if either b=0 or b=-3/2 (the common root being, respectively, x=1 or x=-1/2).

Answer 2

I am not sure if this is what you mean but it does gives values of b and allow the two equations to have common roots.

Solve the second equation for b
Substitute result into equation 1
Now solve equation 1 for x
Using the x values calculate the corresponding b values.

x^2 + b(1 + x) - x = 0 so b = x(1 - x)/(1 + x)

x^3 + (x^3)(1 - x)/(1 + x) + (2x^2)(1 - x)/(1 + x) - 1 = 0
Multiply through by (1 + x)

(1 + x)x^3 + (1 - x)x^3 + 2x^2 - 2x^3 - x - 1 = 0
2x^2 - x - 1 = 0 or x^2 - x/2 - 1/2 = 0
(x - 1)(x + 1/2) = 0
So x = 1 and x = -1/2

Using b = x(1 - x)/(1 + x) gives b = 0 and b = -3/2