If the equations x^2 + px + 2q = 0 and x^2 + | vedclass

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(B) Let $\alpha$ be the common root of the two equations.Then,$\alpha^2 + p\alpha + 2q = 0$ and $\alpha^2 + q\alpha + 2p = 0$.Subtracting the second e

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$-2$

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(B) Let $\alpha$ be the common root of the two equations.Then,$\alpha^2 + p\alpha + 2q = 0$ and $\alpha^2 + q\alpha + 2p = 0$.Subtracting the second equation from the first,we get:$(\alpha^2 + p\alpha + 2q) - (\alpha^2 + q\alpha + 2p) = 0$$\alpha(p - q) - 2(p - q) = 0$$(p - q)(\alpha - 2) = 0$Since it is given that $p 
e q$,we must have $\alpha - 2 = 0$,which implies $\alpha = 2$.Substituting $\alpha = 2$ into the first equation:$(2)^2 + p(2) + 2q = 0$$4 + 2p + 2q = 0$$2(p + q) = -4$$p + q = -2$

If the equations $x^2 + px + 2q = 0$ and $x^2 + qx + 2p = 0$ $(p \ne q)$ have a common root,then the value of $p + q$ is

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