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Algebra Difficulty 4.3 AIME Find the answer Italy

Problem:

Let aa and bb be two distinct real numbers. It is known that the two equations
x2+ax+3b=0x2+bx+3a=0 \begin{aligned} & x^{2}+a x+3 b=0 \\ & x^{2}+b x+3 a=0 \end{aligned}
have a common solution: what are the possible values for the sum a+ba+b?

Pick one

Solution

Solution:

The answer is (D). Every solution of both equations must also be a solution of their difference, which is
(x2+ax+3b)−(x2+bx+3a)=(a−b)x+3(b−a)=(a−b)(x−3)=0. \left(x^{2}+a x+3 b\right)-\left(x^{2}+b x+3 a\right)=(a-b) x+3(b-a)=(a-b)(x-3)=0 .
Since aa and bb are distinct, the only possibility is x=3x=3, which is therefore the only possible common solution of the two original equations. In particular 3 is also a solution of x2+ax+3b=0x^{2}+a x+3 b=0; substituting, we get 9+3a+3b=09+3 a+3 b=0, from which 3(a+b)=−93(a+b)=-9 and hence a+b=−3a+b=-3.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.