For $ a \neq b , $ if the equations $ x ^ { 2 } + a x + b = 0 $ and $ x ^ { 2 } + b x + a = 0 $ have a common root, then the value of ( a + b ) is

Let the common root be r; substituting it into both equations gives r^2 + ar + b = 0 and r^2 + br + a = 0. Subtracting the second equation from the first yields (a - b)r + (b - a) = 0, which factors to (a - b)(r - 1) = 0. Since a is not equal to b, r must equal 1; substituting r = 1 back into the first equation gives 1 + a + b = 0, meaning a + b equals -1.

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