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Algebra Difficulty 4.8 AIME Prove it Slovenia

Let $a$ and $b$ be different real numbers such that the equations $x^2 + a x + b = 0$ and $x^2 + b x + a = 0$ have some common solution. How much is $a + b$ ?

Solution

If we equalize the left sides of the quadratic equations, we get $x^2 + a x + b = x^2 + b x + a$ . Hence $a x + b = b x + a$ and $(x - 1)(a - b) = 0$ . Because $a \neq b$ , it follows that $x = 1$ . If we now substitute $x = 1$ in one of the equations, we get $a + b = -1$ .

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