Official solution
Given,
x2 + bx - 1 = 0 . . . . .(1)
and x2 + x + b = 0 . . . . . (2)
Performing (1) - (2) we get,
bx - 1 - x - b = 0
\[ \Rightarrow \] x(b - 1) = b + 1
\[ \Rightarrow \] x = \[{{b + 1} \over {b - 1}}\]
putting value of x in equation (2),
\[{\left( {{{b + 1} \over {b - 1}}} \right)^2} + \left( {{{b + 1} \over {b - 1}}} \right) + b = 0\]
\[ \Rightarrow \] (b + 1)2 + (b + 1) (b - 1) + b (b - 1)2 = 0
\[ \Rightarrow \] b2 + 2b + 1 + b2 - 1 + b (b2 - 2b + 1) = 0
\[ \Rightarrow \] 2b3 + 2b + b3 - 2b2 + b = 0
\[ \Rightarrow \] b3 + 3b = 0
\[ \Rightarrow \] b(b2 + 3) = 0
b2 = - 3, b = 0
\[ \therefore \] b = \[ \pm \sqrt 3 i\]
\[ \Rightarrow \] \[\left| b \right|\] = \[\sqrt 3 \]