Modulus Inequalities (Absolute Value Inequalities): Very Difficult Problems with Solutions

Solution:
[tex]|A|\gt B \Leftrightarrow A\gt B[/tex] or [tex]A\lt -B[/tex] (this holds for any sign of [tex]B[/tex]):
[tex]x^2-x-6\gt x+3 \Leftrightarrow x^2-2x-9\gt 0 \Leftrightarrow x\lt 1-\sqrt{10}[/tex] or [tex]x\gt 1+\sqrt{10}[/tex]
[tex]x^2-x-6\lt -x-3 \Leftrightarrow x^2-3\lt 0 \Leftrightarrow -\sqrt{3}\lt x\lt \sqrt{3}[/tex]
The answer is the union of the two sets ([tex]1-\sqrt{10}\approx -2. 16\lt -\sqrt{3}\approx -1. 73[/tex]):
Answer: [tex]x\in\left(-\infty, 1-\sqrt{10}\right)\cup\left(-\sqrt{3}, \sqrt{3}\right)\cup\left(1+\sqrt{10}, +\infty\right)[/tex].
Check: [tex]x=-2[/tex] is not a solution ([tex]0\gt 1[/tex] is false), while [tex]x=0[/tex] is ([tex]6\gt 3[/tex]).

Solution:
Both sides are non-negative, so we can square them:
[tex]\left(x^2-8x+15\right)^2-\left(15-x^2\right)^2\leq 0[/tex]
[tex]\left(x^2-8x+15-15+x^2\right)\left(x^2-8x+15+15-x^2\right)\leq 0 \Leftrightarrow \left(2x^2-8x\right)(30-8x)\leq 0[/tex]
[tex]4x(x-4)(15-4x)\leq 0 \Leftrightarrow x(x-4)(4x-15)\geq 0[/tex]
Zeros of the factors: [tex]0, \frac{15}{4}, 4[/tex]. Signs from left to right: [tex]-, +, -, +[/tex].
Answer: [tex]x\in\left[0, \frac{15}{4}\right]\cup[4, +\infty)[/tex].

Solution:
Domain: [tex]|x|\neq 1[/tex], i.e. [tex]x\neq\pm 1[/tex].
The numerator is non-negative. The fraction is [tex]\leq 0[/tex] in two cases.
If the numerator is [tex]0[/tex]: [tex]|2-x|=3 \Leftrightarrow x=-1[/tex] or [tex]x=5[/tex]. The value [tex]x=-1[/tex] is not in the domain, while [tex]x=5[/tex] is a solution.
If the numerator is positive, the denominator must be negative: [tex]|x|-1\lt 0 \Leftrightarrow -1\lt x\lt 1[/tex] (in this interval the numerator is indeed positive).
Answer: [tex]x\in(-1, 1)\cup\{5\}[/tex].

Solution:
Domain: [tex]x\neq 3[/tex].
[tex]x\lt 0:\ \frac{x^2+x-12}{x-3}=\frac{(x+4)(x-3)}{x-3}=x+4\geq 2x \Leftrightarrow x\leq 4[/tex] — true for every [tex]x[/tex] in this interval.
[tex]x\geq 0:\ \frac{x^2-x-12}{x-3}-2x=\frac{x^2-x-12-2x^2+6x}{x-3}=\frac{-x^2+5x-12}{x-3}\geq 0[/tex]
[tex]-x^2+5x-12\lt 0[/tex] for every [tex]x[/tex] [tex](D=25-48\lt 0)[/tex], so we need [tex]x-3\lt 0[/tex], i.e. [tex]0\leq x\lt 3[/tex].
Answer: [tex]x\in(-\infty, 3)[/tex].

Solution:
Domain: [tex]x\neq 0[/tex]. Multiply both sides by [tex]3|x|\gt 0[/tex]:
[tex]9\leq |x|(x+6)[/tex]
[tex]x\gt 0:\ x^2+6x-9\geq 0 \Leftrightarrow x\geq -3+3\sqrt{2}[/tex] (the other root [tex]-3-3\sqrt{2}[/tex] is negative)
[tex]x\lt 0:\ 9\leq -x(x+6) \Leftrightarrow x^2+6x+9\leq 0 \Leftrightarrow (x+3)^2\leq 0 \Leftrightarrow x=-3[/tex]
Answer: [tex]x\in\{-3\}\cup\left[3\sqrt{2}-3, +\infty\right)[/tex].

Solution:
No solutions in [tex][-3, 0][/tex] means that for every [tex]x\in[-3, 0][/tex] the opposite inequality holds:
[tex]-6\leq x^2+4x-a\leq 6 \Leftrightarrow a-6\leq x^2+4x\leq a+6[/tex] (for every [tex]x\in[-3, 0][/tex])
So all values of [tex]x^2+4x[/tex] on [tex][-3, 0][/tex] must lie in [tex][a-6, a+6][/tex].
[tex]x^2+4x=(x+2)^2-4[/tex]: the smallest value is [tex]-4[/tex] (at [tex]x=-2[/tex]) and the largest is [tex]0[/tex] (at [tex]x=0[/tex]; at [tex]x=-3[/tex] the value is [tex]-3[/tex]).
Therefore [tex]a-6\leq -4[/tex] and [tex]0\leq a+6[/tex], i.e. [tex]-6\leq a\leq 2[/tex].
Answer: [tex]a\in[-6, 2][/tex].