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

Solution:

We know that, if ${\mathrm{a}}_1{\mathrm{x}}^2+{\mathrm{b}}_1\mathrm{x}+{\mathrm{c}}_1=0$ and ${\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{b}}_2^2\mathrm{x}+{\mathrm{c}}_2=0$  have a common real root, then ${\left({\mathrm{a}}_1{\mathrm{c}}_2-{\mathrm{a}}_2{\mathrm{c}}_1\right)}^2=\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1\right)\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)$ then ${\left({\mathrm{a}}_1{\mathrm{c}}_2-{\mathrm{a}}_2{\mathrm{c}}_1\right)}^2=\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1\right)\left({\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1\right)$
$\begin{array}{l}\Rightarrow (1+\mathrm{b}{)}^2=\left({\mathrm{b}}^2+1\right)(1-\mathrm{b})\\ \Rightarrow {\mathrm{b}}^2+2\mathrm{b}+1={\mathrm{b}}^2-{\mathrm{b}}^3+1-\mathrm{b}\\ \Rightarrow {\mathrm{b}}^3+3\mathrm{b}=0\Rightarrow \mathrm{b}\left({\mathrm{b}}^2+3\right)=0\\ \Rightarrow \mathrm{b}=0,\pm \mathrm{i}\sqrt{3}\end{array}$