For a neq b,if the equations x^2+ax+b=0 and x | vedclass

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(A) Let $\alpha$ be the common root of the equations $x^2+ax+b=0$ and $x^2+bx+a=0$. Then,$\alpha^2+a\alpha+b=0$ and $\alpha^2+b\alpha+a=0$. Subtractin

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$-1$

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(A) Let $\alpha$ be the common root of the equations $x^2+ax+b=0$ and $x^2+bx+a=0$. Then,$\alpha^2+a\alpha+b=0$ and $\alpha^2+b\alpha+a=0$. Subtracting the two equations: $(\alpha^2+a\alpha+b) - (\alpha^2+b\alpha+a) = 0$ $a\alpha - b\alpha + b - a = 0$ $\alpha(a-b) - (a-b) = 0$ $(a-b)(\alpha-1) = 0$. Since $a 
eq b$,we must have $\alpha-1=0$,which implies $\alpha=1$. Substituting $\alpha=1$ into the first equation: $1^2 + a(1) + b = 0$ $1 + a + b = 0$ $a + b = -1$.

For $a \neq b$,if the equations $x^2+ax+b=0$ and $x^2+bx+a=0$ have a common root,then the value of $a+b=$

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