If the equations ${\mathrm{x}}^2 + \mathrm{bx} – 1 = 0$ and ${\mathrm{x}}^2 + \mathrm{x} + \mathrm{b}=0$ have a common root different from  –1, then |b| is equal to:

$\begin{array}{l}GivenEquation{x}^2+bx-1=0,x^2+x+b=0\\ Let\alpha bethecommonroots\\ \Rightarrow {\alpha }^2+b\alpha -1=0\\ and{\alpha }^2+\alpha +b=0\\ \overline{\alpha (b-1)-1-b=0}\\ \alpha =\frac{b+1}{b-1}\\ subin\left(2\right)\\ {\left(\frac{b+1}{b-1}\right)}^2+\left(\frac{b+1}{b-1}\right)+b=0\\ \Rightarrow {(b+1)}^2+(b+1)(b-1)+b{(b-1)}^2=0\\ \Rightarrow b^2+1+2b+b^2-1+b(b^2+1-2b)=0\\ \Rightarrow 2b^2+2b+b^3+b-2b^2=0\\ \Rightarrow b^3+3b=0\\ \Rightarrow b(b^2+3)=0\\ b=0,b^2=-3\Rightarrow b=\pm \sqrt{3}i\\ \therefore \left|b\right|=\sqrt{3}\end{array}$