If the equations x^2 + bx - 1 = 0 and x^2 + x | vedclass

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(C) Let $\alpha$ be the common root of the equations $x^2 + bx - 1 = 0$ and $x^2 + x + b = 0$.Since $\alpha$ is a root,we have:$\alpha^2 + b\alpha - 1

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$\sqrt{3}$

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(C) Let $\alpha$ be the common root of the equations $x^2 + bx - 1 = 0$ and $x^2 + x + b = 0$.Since $\alpha$ is a root,we have:$\alpha^2 + b\alpha - 1 = 0$  $(1)$$\alpha^2 + \alpha + b = 0$  $(2)$Subtracting equation $(2)$ from $(1)$:$(b - 1)\alpha - 1 - b = 0$$(b - 1)\alpha = b + 1$Since the root is different from $-1$,we can write $\alpha = \frac{b + 1}{b - 1}$ (provided $b 
eq 1$).Substituting $\alpha$ into equation $(2)$:$\left(\frac{b + 1}{b - 1}ight)^2 + \left(\frac{b + 1}{b - 1}ight) + b = 0$$(b + 1)^2 + (b + 1)(b - 1) + b(b - 1)^2 = 0$$(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 $b$ must be a real number for $|b|$ to be a standard magnitude in this context,we check the roots. If $b^2 + 3 = 0$,then $b^2 = -3$,which gives $b = \pm i\sqrt{3}$. However,if we assume $b$ is real,then $b = 0$. If $b=0$,the equations are $x^2 - 1 = 0$ and $x^2 + x = 0$,which have no common root. Re-evaluating the subtraction: $(b-1)\alpha = 1+b$. If $\alpha = -1$,then $-(b-1) = 1+b \implies -b+1 = 1+b \implies 2b = 0 \implies b=0$. Since the root is not $-1$,$b 
eq 0$. The only remaining possibility is $b^2 = -3$ is not intended,let's re-check the subtraction logic. Actually,for $x^2+bx-1=0$ and $x^2+x+b=0$,the common root $\alpha$ satisfies $\alpha^2 = 1-b\alpha$ and $\alpha^2 = -b-\alpha$. Thus $1-b\alpha = -b-\alpha \implies \alpha(1-b) = -b-1 \implies \alpha = \frac{b+1}{b-1}$. Substituting back leads to $b^3+3b=0$. Given the options,there might be a typo in the question constants. If the equations were $x^2+bx+1=0$ and $x^2+x+b=0$,then $b=1$ or $b=-2$. Given the options,$|b|=\sqrt{3}$ is the derived result.

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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