If the equations x^2+ax+b=0 and x^2+bx+a=0 (a | vedclass

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(A) Let $\alpha$ be the common root of the two equations.Then,$\alpha^2 + a\alpha + b = 0$ and $\alpha^2 + b\alpha + a = 0$.Subtracting the two equati

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$-1$

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(A) Let $\alpha$ be the common root of the two equations.Then,$\alpha^2 + a\alpha + b = 0$ and $\alpha^2 + b\alpha + a = 0$.Subtracting the two equations: $(\alpha^2 + a\alpha + b) - (\alpha^2 + b\alpha + a) = 0$.$(a-b)\alpha + (b-a) = 0$.$(a-b)\alpha = (a-b)$.Since $a 
eq b$,we can divide by $(a-b)$ to get $\alpha = 1$.Substituting $\alpha = 1$ into the first equation: $1^2 + a(1) + b = 0$.$1 + a + b = 0$.Therefore,$a + b = -1$.

If the equations $x^2+ax+b=0$ and $x^2+bx+a=0$ $(a \neq b)$ have a common root,then $a+b$ is equal to

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