If the equations $x^{2} + bx - 1 = 0$ and $x^{2} + x + b = 0$ have a common root different from –1, then |b| is equal to:

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$\sqrt{3}$

$\sqrt{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n x^{2} + b x - 1 = 0, x^{2} + x + b = 0 \\ L e t α b e t h e c o m m o n r o o t s \\ \Rightarrow α^{2} + b α - 1 = 0 \\ a n d α^{2} + α + b = 0 \\ \bar{α (b - 1) - 1 - b = 0} \\ α = \frac{b + 1}{b - 1} \\ s u b i n (2) \\ {(\frac{b + 1}{b - 1})}^{2} + (\frac{b + 1}{b - 1}) + b = 0 \\ \Rightarrow {(b + 1)}^{2} + (b + 1) (b - 1) + b {(b - 1)}^{2} = 0 \\ \Rightarrow b^{2} + 1 + 2 b + b^{2} - 1 + b (b^{2} + 1 - 2 b) = 0 \\ \Rightarrow 2 b^{2} + 2 b + b^{3} + b - 2 b^{2} = 0 \\ \Rightarrow b^{3} + 3 b = 0 \\ \Rightarrow b (b^{2} + 3) = 0 \\ b = 0, b^{2} = - 3 \Rightarrow b = \pm \sqrt{3} i \\ ∴ | b | = \sqrt{3} \end{array}$