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[TEX]f(x) = ax^2+bx+c[/TEX]
Ta có [TEX]f(10) = a.10^2 + 10b + c = 100a + 10b + c[/TEX]
[TEX]f(-3) = a.(-3)^2 + (-3)b + c = 9a - 3b + c [/TEX]
\Rightarrow [TEX]f(10) - f(-3) = 100a + 10b + c - 9a + 3b - c [/TEX]
\Rightarrow [TEX]f(10) - f(-3) = 91a + 13b [/TEX]
\Rightarrow [TEX]f(10) - f(-3) = 13 ( 7a+b ) [/TEX]
mà [TEX]7a+ b= 0[/TEX] \Rightarrow [TEX]f(10) - f(-3) = 13.0 = 0 [/TEX]
\Rightarrow [TEX]f(10) = f(-3) [/TEX] \Rightarrow [TEX]f(10) . f(-3) = f(10)^2 \geq 0 [/TEX]
Vậy [TEX]f(10) . f(-3) [/TEX] không thể là số âm