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Algebra Difficulty 5.0 AIME Find the answer

4. 1. 7 ** If the inequality concerning xx
x2+(2a2+2)x−a2+4a−7x2+(a2+4a−5)x−a2+4a−7<0 \frac{x^{2}+\left(2 a^{2}+2\right) x-a^{2}+4 a-7}{x^{2}+\left(a^{2}+4 a-5\right) x-a^{2}+4 a-7}<0

has a solution set that is the union of some intervals, and the sum of the lengths of these intervals is no less than 4, find the range of real numbers aa.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let the roots of the equation x2+(2a2+2)x−a2+4a−7=0x^{2}+\left(2 a^{2}+2\right) x-a^{2}+4 a-7=0 be x1、x2x_{1} 、 x_{2}, and the roots of the equation x2+x^{2}+ theorem give x1x2=x3x1=−a2+4a−70x_{1} x_{2}=x_{3} x_{1}=-a^{2}+4 a-70, so x3+x4>x1+x2x_{3}+x_{4}>x_{1}+x_{2}, therefore x1<x3<x2<x4x_{1}<x_{3}<x_{2}<x_{4}, hence the interval length is (x4−x2)+(x3−x1)=a2−4a+7⩾4\left(x_{4}-x_{2}\right)+\left(x_{3}-x_{1}\right)=a^{2}-4 a+7 \geqslant 4, solving this gives a⩽1a \leqslant 1 or a⩾3a \geqslant 3.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.