Answer:
Option B
Explanation:
If $a_{1}x^{2}+b_{1}x+c_{1}=0$
and $a_{2}x^{2}+b_{2}x+c_{2}=0$
Have a comon real root , then
$\Rightarrow$ $(a_{1}c_{2}-a_{2}c_{1})^{2}= (b_{1}c_{2}-b_{2}c_{1})(a_{1}b_{2}-a_{2}b_{1})$
$\because$ $x^{2}+bx-1=0$,
$x^{2}+x+b$ , have a common root
$\Rightarrow$$(1+b)^{2}=(b^{2}+1)(1-b)$
$\Rightarrow$ $b^{2}+2b+1=b^{2}-b^{3}+1-b$
$\Rightarrow$ $b^{3}+3b=0\Rightarrow b(b^{2}+3)=0$
$\Rightarrow$ $b=0, \pm \sqrt{3} i$
