Quadratics making union

since b is real, $f(x)=x^2+bx-1$ has distinct root.we can say this because $\Delta=b^2+4>4$ for b≠0.we know that for equal roots discriminant must be zero.2 roots found here $$ note that $2x^2-2bx-2=2f(-x)$ . we see that the sum of the roots of f(x) is not zero(b≠0). we conclude that changing signs will yield different roots. 2 more roots. $$ $2+2=\boxed{4}$

Simply Calculate the roots of both the equations

The roots are

$\dfrac{-b\pm\sqrt{b^{2}+4}}{2}, \dfrac{b\pm\sqrt{b^{2}+4}}{2}$

There is no common root. So there are $\boxed{4}$ possible values of b.

Both the quadratics have roots with opposite signs.

Hint: The discriminant of each part is greater than 0