Let $f(x) = ax^{2} + bx + c, a, b\epsilon R, a\neq 0$ satisfying $f(1) + f(2) = 0$ and and $-1$ is a root of the expression Then the equation $f(x) = 0$ has

Since -1 is a root of the quadratic expression $ax^2 + bx + c$, we have $a(-1)^2 + b(-1) + c = 0$, which simplifies to $a - b + c = 0$. Using the condition $f(1) + f(2) = 0$, we get $(a + b + c) + (4a + 2b + c) = 0$, which simplifies to $5a + 3b + 2c = 0$. Substituting $b = a + c$ into the second equation gives $5a + 3(a + c) + 2c = 0$, resulting in $8a + 5c = 0$ and $c = -8a/5$. The discriminant of the quadratic is $b^2 - 4ac = (a + c)^2 - 4ac = (a - c)^2 = (a + 8a/5)^2 = (13a/5)^2$. Since the discriminant is strictly positive, the equation $f(x) = 0$ has two distinct real roots.

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