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

A value of b for which the equations x²+bx-1=0 x²+x+b=0 have one root in common is

$x^{2} + bx - 1 = 0$

$x^{2} + x + b = 0$

$\text{To find: a value of } b \text{ for which the two equations share one root}$

$\text{Let } \alpha \text{ be the common root}$

$\text{Plan: subtract to eliminate } \alpha^{2}$

$\alpha^{2} + b\alpha - 1 = 0$

$\alpha^{2} + \alpha + b = 0$

$(b\alpha - 1) - (\alpha + b) = 0$

$(b - 1)\alpha = b + 1$

$b = 1 \Rightarrow 0 = 2\ \text{(impossible)}$

$\alpha = \frac{b+1}{b-1}$

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

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

$= \frac{b^{2}+1}{b-1}$

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

$\times\, (b-1)^{2}$

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

$b^{2} + 2b + 1 + b^{3} - b^{2} + b - 1 = 0$

$b^{3} + 3b = 0$

$b(b^{2} + 3) = 0$

$b = 0 \ \text{ or } \ b^{2} = -3$

$b = \pm i\sqrt{3}$

$b = i\sqrt{3}:\ \alpha = \frac{1+i\sqrt{3}}{-1+i\sqrt{3}} = \frac{1 - i\sqrt{3}}{2}$

$\alpha^{2} + \alpha = -i\sqrt{3}, \quad + b = i\sqrt{3}$

$\Rightarrow 0\ \checkmark$

$b = 0\ (\text{common root } x = -1) \ \text{— not an option}$

$\boxed{b = \pm i\sqrt{3}} \ \Rightarrow\ \text{option (2)}$