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(C) Let $\alpha$ be the common root of $f(x)=0$ and $g(x)=0$. Then,$\alpha^2 + a\alpha + 2 = 0$ and $\alpha^2 + 2\alpha + a = 0$. Subtracting the two

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$\frac{1}{2}$

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(C) Let $\alpha$ be the common root of $f(x)=0$ and $g(x)=0$. Then,$\alpha^2 + a\alpha + 2 = 0$ and $\alpha^2 + 2\alpha + a = 0$. Subtracting the two equations: $(a-2)\alpha + (2-a) = 0$. $(a-2)(\alpha - 1) = 0$. Since the polynomials are distinct,$a 
eq 2$. Thus,$\alpha = 1$. Substituting $\alpha = 1$ into $f(x)=0$: $1^2 + a(1) + 2 = 0 \Rightarrow a = -3$. The equation $f(x)+g(x)=0$ becomes $(x^2-3x+2) + (x^2+2x-3) = 0$. $2x^2 - x - 1 = 0$. The sum of the roots is given by $-\frac{b}{a} = -\frac{-1}{2} = \frac{1}{2}$.

Two distinct polynomials $f(x)$ and $g(x)$ are defined as follows: $f(x)=x^2+ax+2$ and $g(x)=x^2+2x+a$. If the equations $f(x)=0$ and $g(x)=0$ have a common root,then the sum of the roots of the equation $f(x)+g(x)=0$ is:

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