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Algebra Difficulty 5.0 AIME Prove it Bulgaria

The functions f(x)=2x2+2x4f(x) = 2x^2 + 2x - 4 and g(x)=x2x+2g(x) = x^2 - x + 2 are given. Find all real values of xx such that:

a) f(x)g(x)\frac{f(x)}{g(x)} is a positive integer;

b) the inequality f(x)+g(x)2\sqrt{f(x)} + \sqrt{g(x)} \ge \sqrt{2} holds.

Solution

a) Hint. Set f(x)g(x)=k\frac{f(x)}{g(x)} = k, where kk is a positive integer. Then (2k)x2+(2+k)x2(2+k)=0(2-k)x^2 + (2+k)x - 2(2+k) = 0 and use the fact that the discriminant of this quadratic equation is nonnegative.

Answer. x=3+332,3332,2x = \frac{-3+\sqrt{33}}{2}, \frac{-3-\sqrt{33}}{2}, 2.

b) Answer. x(,2][1,+)x \in (-\infty, -2] \cup [1, +\infty).

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