A value of b for which the equations x² + bx - 1 = 0 and x² + x + b = 0 have one root in common is

Let α be the common root (solution) of the two equations. So, it will satisfy both the equations, α² + bα - 1 = 0 and α² + α + b = 0.
Solving by cross-multiplication method, we get
α²/b² - (-1) = α/(-1) - b = 1/1 - b
α²/b² + 1 = α/(-1 - b) = 1/1 - b
On equating first and third part, we get α² = b² + 1/1 - b (1)
On equating second and third part, we get α = -(1 + b)/1 - b (2)
Plug in the value of α from equation (2) in equation (1). So,
( -(1 + b)/(1 - b) )² = (b² + 1)/(1 - b)
(1 + b)²/(1 - b)² = (b² + 1)/(1 - b)
(1 + 2b + b²)/(1 - b) = b² + 1
(1 + 2b + b²)(1 - b) = (b² + 1)(1 - b)²
1 + 2b + b² - b - 2b² - b³ = (b² + 1)(1 - 2b + b²)
1 + b - b² - b³ = b² - 2b³ + b⁴ + 1 - 2b + b²
1 + b - b² - b³ = b⁴ - 2b³ + 2b² - 2b + 1
b⁴ - b³ + 3b² - 3b = 0
b³(b - 1) + 3b(b - 1) = 0
(b - 1)(b³ + 3b) = 0
(b - 1)b(b² + 3) = 0
So, b = 0 or b = 1 or b² = -3
So, b = 0 or b = 1 or b = ±i√3
Since √-1 = i (imaginary number). So, b = ±i√3