A value of b for which the equations x^2+bx-1 | vedclass

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(B) Let $\alpha$ be the common root of the equations $x^2+bx-1=0$ and $x^2+x+b=0$.Then,$\alpha^2+b\alpha-1=0$ and $\alpha^2+\alpha+b=0$.Subtracting th

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$-i\sqrt{3}$

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(B) Let $\alpha$ be the common root of the equations $x^2+bx-1=0$ and $x^2+x+b=0$.Then,$\alpha^2+b\alpha-1=0$ and $\alpha^2+\alpha+b=0$.Subtracting the two equations,we get:$(\alpha^2+b\alpha-1) - (\alpha^2+\alpha+b) = 0$$(b-1)\alpha - (1+b) = 0$$(b-1)\alpha = b+1$$\alpha = \frac{b+1}{b-1}$ (assuming $b 
eq 1$).Substituting $\alpha$ into the second equation $\alpha^2+\alpha+b=0$:$\left(\frac{b+1}{b-1}ight)^2 + \frac{b+1}{b-1} + b = 0$Multiply by $(b-1)^2$:$(b+1)^2 + (b+1)(b-1) + b(b-1)^2 = 0$$(b^2+2b+1) + (b^2-1) + b(b^2-2b+1) = 0$$2b^2+2b + b^3-2b^2+b = 0$$b^3+3b = 0$$b(b^2+3) = 0$Since $b 
eq 0$ (otherwise the equations become $x^2-1=0$ and $x^2+x=0$,which have no common root),we have $b^2+3=0$,so $b^2=-3$,which gives $b = \pm i\sqrt{3}$.

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

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