Algebra · real student question

Solve the inequality (|x − 2| − x) / (x² − 4) ≤ 0 and give the solution in interval notation.

Question

Solve the inequality

x2xx240\frac{|x-2|-x}{x^2-4}\le 0

and write the solution set in interval notation.

Step-by-step solution

  1. Exclude the points where the expression is undefined. A fraction is only defined when its denominator is non-zero. Here x24=(x2)(x+2)x^2-4=(x-2)(x+2), so

    x2andx2.x\ne 2\quad\text{and}\quad x\ne -2.

    These two points can never belong to the solution set, no matter what the numerator does — a \le sign never rescues a zero denominator.

  2. Split the absolute value at its breakpoint. x2|x-2| changes formula at x=2x=2: it equals x2x-2 when x2x\ge 2 and 2x2-x when x<2x<2. Because the whole inequality behaves differently on the two sides, solve it separately on each and then take the union. Do it algebraically rather than by testing sample numbers, so nothing is missed between test points.

  3. Case 1: x>2x>2. Here x2=x2|x-2|=x-2, so the numerator collapses to a constant:

    (x2)xx24=2x24.\frac{(x-2)-x}{x^2-4}=\frac{-2}{x^2-4}.

    For x>2x>2 the denominator x24x^2-4 is positive, so the fraction is negative for every such xx. The whole ray (2,+)(2,+\infty) satisfies the inequality (with x=2x=2 itself excluded from step 1).

  4. Case 2: x<2x<2. Here x2=2x|x-2|=2-x, so the numerator is 22x=2(1x)2-2x=2(1-x) and

    2(1x)(x2)(x+2)0.\frac{2(1-x)}{(x-2)(x+2)}\le 0.

    The numerator is zero at x=1x=1 and positive for x<1x<1; the factor x2x-2 is negative throughout this case; the factor x+2x+2 changes sign at x=2x=-2.

  5. Build the sign table for Case 2. Split the line at the critical values 2-2 and 11:

    interval2(1x)2(1-x)x2x-2x+2x+2fraction
    x<2x<-2++--++
    2<x<1-2<x<1++-++-
    1<x<21<x<2--++++

    Only 2<x<1-2<x<1 makes the fraction negative. At x=1x=1 the numerator is 00 while the denominator is 30-3\ne 0, so the fraction equals 00 — and 000\le 0 is true, so x=1x=1 is included.

  6. Take the union of the two cases. Combining (2,1](-2,1] from Case 2 with (2,+)(2,+\infty) from Case 1:

    (2,1]  (2,+).(-2,\,1]\ \cup\ (2,\,+\infty).

    Spot-checking confirms it: at x=0x=0 the value is 0.5-0.5 (true), at x=1.5x=1.5 it is +0.571+0.571 (false), at x=3x=3 it is 0.4-0.4 (true), and at x=2.5x=-2.5 it is +3.11+3.11 (false).

Answer

x(2,1](2,+)x\in(-2,\,1]\cup(2,\,+\infty)

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