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

Vedclass · Accepted Answer

$\sqrt{3}$

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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)$$\alpha^2 + \alpha + b = 0$  --- $(2)$Subtracting equation $(2)$ from equation $(1)$:$(\alpha^2 + b\alpha - 1) - (\alpha^2 + \alpha + b) = 0$$\alpha(b - 1) - (1 + b) = 0$$\alpha(b - 1) = b + 1$$\alpha = \frac{b + 1}{b - 1}$ (where $b 
eq 1$)Substitute $\alpha$ 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 the root to be defined in the context of standard quadratic problems,we look for the magnitude. If $b^2 + 3 = 0$,then $b^2 = -3$,which gives $b = \pm i\sqrt{3}$.Thus,$|b| = |\pm i\sqrt{3}| = \sqrt{3}$.

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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