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

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$- \sqrt{2}$

$- i \sqrt{3}$

$i \sqrt{5}$

$\sqrt{2}$

answer is B.

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Detailed Solution

Given equation are

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

and $x^{2} + x + b = 0 . . . (i i)$

Subtracting Eq. (ii) from Eq. (i), we get,

$\begin{aligned} (b - 1) x = (b + 1) \\ \Rightarrow x = (\frac{b + 1}{b - 1}) \end{aligned}$

Put the value of $x$ in Eq. (i), we get,

$\begin{array}{l} {(\frac{b + 1}{b - 1})}^{2} + b (\frac{b + 1}{b - 1}) - 1 = 0 \\ \Rightarrow (b + 1)^{2} + b (b + 1) (b - 1) - (b - 1)^{2} = 0 \\ \Rightarrow (b + 1)^{2} + b (b^{2} - 1) - (b - 1)^{2} = 0 \\ \Rightarrow (b^{2} + 2 b + 1 - b^{2} + 2 b - 1) + b (b^{2} - 1) = 0 \end{array}$

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