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(C) Let the roots of the equation $x^2+bx+a=0$ be $\alpha$ and $\beta$.From the properties of roots,we have $\alpha+\beta = -b$ and $\alpha\beta = a$.

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$-4$

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(C) Let the roots of the equation $x^2+bx+a=0$ be $\alpha$ and $\beta$.From the properties of roots,we have $\alpha+\beta = -b$ and $\alpha\beta = a$.Given that the roots of the equation $x^2+ax+b=0$ are each $1$ greater than the roots of the first equation,the roots are $(\alpha+1)$ and $(\beta+1)$.From the properties of roots for the second equation,we have:$(\alpha+1)+(\beta+1) = -a \implies \alpha+\beta+2 = -a$.Substituting $\alpha+\beta = -b$,we get $-b+2 = -a$,which simplifies to $a-b = -2$ (Equation $1$).Also,$(\alpha+1)(\beta+1) = b \implies \alpha\beta + \alpha+\beta + 1 = b$.Substituting $\alpha\beta = a$ and $\alpha+\beta = -b$,we get $a-b+1 = b$,which simplifies to $a-2b = -1$ (Equation $2$).Subtracting Equation $2$ from Equation $1$: $(a-b) - (a-2b) = -2 - (-1) \implies b = -1$.Substituting $b = -1$ into Equation $1$: $a - (-1) = -2 \implies a+1 = -2 \implies a = -3$.Therefore,$a+b = -3 + (-1) = -4$.

$A$ student notices that the roots of the equation $x^2+bx+a=0$ are each $1$ less than the roots of the equation $x^2+ax+b=0$. Then,$a+b$ is

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