A value of $b$ for which the equations $x^2+bx-1=0$ and $x^2+x+b=0$ have one root in common, is

Let $\alpha$ be a common root of the given equations.

Then,

  ${\alpha }^2+b\alpha -1=0$                            …(i)

${\alpha }^2+\alpha +b=0$                               …(ii)

On subtracting, we get

$(b-1)a=b+1>a=\frac{b+1}{b-1}$

Substituting the value of $\alpha$ in (i), we get

$(b+1{)}^2+b(b+1\left)\right(b-1)-(b-1{)}^2=0$

$\begin{array}{l}\Rightarrow b\left(b^2-1\right)+4h=0\\ \Rightarrow b\left(b^2+3\right)=0\Rightarrow b=0,\pm i\sqrt{3}\end{array}$