If the equations $x^2 + ax + b=0$ and $x^2 + bx + a = 0$ have exactly one common root, then the numerical value of $a + b$ is

Subtracting the two equations, (x^2+ax+b) - (x^2+bx+a) = 0, simplifies to (a-b)x + (b-a) = 0. Factoring this expression gives (a-b)(x-1) = 0. Since the equations have exactly one common root, a cannot equal b, which forces the common root to be x=1. Substituting x=1 into the first equation yields 1^2+a(1)+b=0, so a+b=-1. The numerical value of a+b is therefore -1.

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