The value of $'a'$ for which the equations $x^{2}-3x+a=0$ and $x^{2}+ax-3=0$ have a common root is

Let the common root be t. Then t^2 - 3t + a = 0 and t^2 + at - 3 = 0. Subtracting the first equation from the second gives at - 3 - (-3t + a) = 0, which simplifies to t(a + 3) - (a + 3) = 0. Factoring this yields (a + 3)(t - 1) = 0, so if a is not -3, the common root must be 1. Substituting t = 1 into the first equation gives 1 - 3 + a = 0, meaning a = 2. Checking a = -3 reveals the two equations are identical and have more than one common root, but the standard unique common root condition gives a = 2.

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