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 equations $x^2 + px + 2q = 0$ and $x^2 + qx + 2p = 0$.Then,$\alpha^2 + p\alpha + 2q = 0$ and $\alpha^2 + q\

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$-2$

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(B) Let $\alpha$ be the common root of the equations $x^2 + px + 2q = 0$ and $x^2 + qx + 2p = 0$.Then,$\alpha^2 + p\alpha + 2q = 0$ and $\alpha^2 + q\alpha + 2p = 0$.Subtracting the two equations: $(\alpha^2 + p\alpha + 2q) - (\alpha^2 + q\alpha + 2p) = 0$.$\alpha(p - q) - 2(p - q) = 0$.$(p - q)(\alpha - 2) = 0$.Since $p 
eq q$,we must have $\alpha - 2 = 0$,which implies $\alpha = 2$.Substituting $\alpha = 2$ into the first equation: $(2)^2 + p(2) + 2q = 0$.$4 + 2p + 2q = 0$.$2(p + q) = -4$.Therefore,$p + q = -2$.

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

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