Solve absolute value inequation with fraction

Question

I am new to absolute-value inequalities, how can I solve this kind of inequation?

$$ | \frac{x^2-5x+4}{x^2-4} | \le 1 $$

Answer 1

We need to solve the following system $$\frac{x^2-5x+4}{x^2-4}\leq1$$ and

$$\frac{x^2-5x+4}{x^2-4}\geq-1$$ For the first we need to solve $$\frac{x^2-5x+4}{x^2-4}-1\leq0$$ or $$\frac{8-5x}{(x-2)(x+2)}\leq0,$$ which by the intervals method gives $x>2$ or $-2<x\leq1.6$.

The second inequality gives $$\frac{x^2-5x+4}{x^2-4}+1\geq0$$ or $$\frac{2x^2-5x}{x^2-4}\geq0$$ or $$\frac{x(2x-5)}{(x-2)(x+2)}\geq0,$$ which by the intervals method again gives $x\geq2.5$ or $0\leq x<2$ or $x<-2$.

Thus, after solving of this system we'll get the answer: $$[0,1.6]\cup[2.5,+\infty)$$

Answer 2

Your equation is equivalent to $$-1\leq \frac{x^2-5x+4}{x^2-4}\leq 1$$

Answer 3

The best way to tackle such a question is to square it and then solve the inequality.

Thus, you have $\frac{(x^2-5x+4)^2}{(x^2-4)^2} \le 1$

$(x^2-5x+4)^2 = x^4-10x^3+33x^2-40x+16$

$(x^2-4)^2 = x^4 - 8x^2+16$

Now $x^4-10x^3+33x^2-40x+16\le x^4 - 8x^2+16 $

$-10x^3 +41x^2 - 40x \le 0$

$x(10x^2-41x+40) \ge 0$

Either $x\ge 0 \text{ and } 10x^2-41x+40\ge 0$ or $x\le 0 \text{ and } 10x^2-41x+40\le 0$

Solving the first one gives you $x\ge0$ and $x\le 1.6$ and $x\ge2.5$ The second inequality can never be true because it say $x\le0$ and $x\ge 1.6$ and $x\le2.5$

Giving you the below answer.

If you simplify, you get the $x(10x^2-41x+40)\ge 0$

Then the region where this inequality holds would be $[0,\frac{8}{5}]U[2.5,\infty)$