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

Solution:

	$(-\infin,-3)$	$[-3,4)$	$(4,+\infin)$
$x+3$	-	+	+
$x-4$	-	-	+

Case 1. $x \in (-\infin,-3)$

$-(x+3)-(x-4)\le 11$
$-x-3-x+4\le 11$
$-2x+1\le 11$
$-2x\le 10$
$x\ge -5$ but $x\in(-\infin,-3) \Rightarrow x\in[-5,-3)$


Case 2. $x \in [-3,4)$

$(x+3)-(x-4)\le 11$
$x+3-x+4\le 11$
$0x\le 4\ \ \ \ - \forall x$ but $x \in [-3,4)$
$\Rightarrow x\in [-3,4)$


Case 3. $x \in (4,+\infin)$

$(x+3)+(x-4)\le 11$
$x+3+x-4\le 11$
$2x-1\le 11$
$2x\le 12$
$x\le 6$ but $x\in(4,+\infin) \Rightarrow x\in[4,6]$


Answer: $x\in[-5,-3) \cup [-3,4) \cup [4,6] \Leftrightarrow x\in[-5, 6]$

Solution:
The expressions under the absolute value signs change sign at [tex]-1[/tex] and [tex]0[/tex].
[tex]x\lt -1:\ -3x\leq 2-x-1 \Leftrightarrow x\geq -\frac{1}{2}[/tex] — no solutions in this interval.
[tex]-1\leq x\lt 0:\ -3x\leq 2+x+1 \Leftrightarrow x\geq -\frac{3}{4} \Rightarrow x\in\left[-\frac{3}{4}, 0\right)[/tex]
[tex]x\geq 0:\ 3x\leq x+3 \Leftrightarrow x\leq \frac{3}{2} \Rightarrow x\in\left[0, \frac{3}{2}\right][/tex]
Answer: [tex]x\in\left[-\frac{3}{4}, \frac{3}{2}\right][/tex].

Solution:
Solve $|2x - 5| + |x + 3| > 10$

Critical points: $x = -3, x = 2.5$

Break into intervals:

For $x < -3$:
$|2x - 5| + |x + 3| = -2x + 5 - x - 3 = -3x + 2 > 10 \Rightarrow x < -\frac{8}{3}$
So valid: $x < -3$

For $-3 \leq x < 2.5$:
$|2x - 5| + |x + 3| = -2x + 5 + x + 3 = -x + 8 > 10 \Rightarrow x < -2$
So valid: $-3 \leq x < -2$

For $x \geq 2.5$:
$|2x - 5| + |x + 3| = 2x - 5 + x + 3 = 3x - 2 > 10 \Rightarrow x > 4$

Solution:
$x \in (-\infty, -3) \cup [-3, -2) \cup (4, \infty)$ or $x \in (-\infty, -2) \cup (4, \infty)$

Solution:
The expressions under the absolute value signs change sign at [tex]-\frac{1}{2}[/tex] and [tex]\frac{1}{5}[/tex].
[tex]x\lt -\frac{1}{2}:\ (1-5x)+(4x+2)=3-x\leq x-3 \Leftrightarrow x\geq 3[/tex] — no solutions in this interval.
[tex]-\frac{1}{2}\leq x\lt \frac{1}{5}:\ (1-5x)-(4x+2)=-1-9x\leq x-3 \Leftrightarrow x\geq \frac{1}{5}[/tex] — no solutions in this interval.
[tex]x\geq \frac{1}{5}:\ (5x-1)-(4x+2)=x-3\leq x-3[/tex] — true for every [tex]x[/tex] in this interval.
Answer: [tex]x\in\left[\frac{1}{5}, +\infty\right)[/tex].

Solution:
Domain: [tex]x^2-5x+6=(x-2)(x-3)\neq 0 \Leftrightarrow x\neq 2,\ x\neq 3[/tex].
In the domain [tex]|x-3|\gt 0[/tex], so the fraction is non-negative exactly when the denominator is positive:
[tex](x-2)(x-3)\gt 0 \Leftrightarrow x\lt 2[/tex] or [tex]x\gt 3[/tex]
Answer: [tex]x\in(-\infty, 2)\cup(3, +\infty)[/tex].

Solution:
[tex]-6\leq x^2-5x\leq 6[/tex]
[tex]x^2-5x-6\leq 0 \Leftrightarrow (x+1)(x-6)\leq 0 \Leftrightarrow -1\leq x\leq 6[/tex]
[tex]x^2-5x+6\geq 0 \Leftrightarrow (x-2)(x-3)\geq 0 \Leftrightarrow x\leq 2[/tex] or [tex]x\geq 3[/tex]
Answer: [tex]x\in[-1, 2]\cup[3, 6][/tex].

Solution:
$|x^2+2x-4|>4 \Leftrightarrow $

$x^2+2x-4 > 4\\ x^2+2x-4 < -4$


$x^2+2x-4 > 4$
$x^2+2x-8 > 0$
$D=36$
$x_1= 2,\ \ x_2=-4$
$x^2+2x-4 > 4 \Leftrightarrow (x-2)(x+4)>0 \Leftrightarrow x\in(-\infin, -4)\cup(2, +\infin)$


$x^2+2x-4 < -4$
$x^2+2x < 0$
$x(x+2) < 0 \Leftrightarrow x\in(-2, 0)$


$x\in(-\infin, -4)\cup(2, +\infin)\\x\in(-2, 0) \\ \Leftrightarrow x\in (-\infin, -4)\cup (-2, 0) \cup (2, +\infin)$

Solution:
[tex]x\geq 0:\ x^2-3\lt 2x \Leftrightarrow x^2-2x-3\lt 0 \Leftrightarrow -1\lt x\lt 3 \Rightarrow x\in[0, 3)[/tex]
[tex]x\lt 0:\ x^2-3\lt 0 \Leftrightarrow -\sqrt{3}\lt x\lt \sqrt{3} \Rightarrow x\in\left(-\sqrt{3}, 0\right)[/tex]
Answer: [tex]x\in\left(-\sqrt{3}, 3\right)[/tex].

Solution:
$x^2+|6x-24| \le 16$
$|6x-24| \le 16 - x^2$

$\begin{array}{|l} 6x-24 \le 16 - x^2 \\ 6x-24 \ge -(16 - x^2) \end{array}$

$\begin{array}{|l} 6x-24 - 16 + x^2\le 0 \\ 6x-24 \ge -16 + x^2 \end{array}$

$\begin{array}{|l} x^2 + 6x-40\le 0 \\ x^2 - 6x + 8 \le 0 \end{array}$

$x^2 + 6x-40\le 0$
$D=196$
$x_1=-10, \ \ \ \ x_2=4$
$x^2 + 6x-40\le 0 \Leftrightarrow (x-4)(x+10) \le 0 \Leftrightarrow x\in[-10,4]$


$x^2 - 6x + 8 \le 0$
$D=4$
$x_1=2, \ \ \ \ x_2=4$
$x^2 - 6x + 8 \le 0 \Leftrightarrow (x-2)(x-4) \le 0 \Leftrightarrow x\in[2,4]$


$\begin{array}{|l}x\in[-10,4] \\ x\in[2,4] \end{array} \Leftrightarrow x\in[2,4]$

Solution:
[tex]x^2-6x+9=(x-3)^2\gt 0[/tex] for [tex]x\neq 3[/tex], so the fraction is negative exactly when the numerator is negative:
[tex]x^2-7|x|+10\lt 0 \Leftrightarrow (|x|-2)(|x|-5)\lt 0 \Leftrightarrow 2\lt |x|\lt 5[/tex]
[tex]x\in(-5, -2)\cup(2, 5),\ x\neq 3[/tex]
Answer: [tex]x\in(-5, -2)\cup(2, 3)\cup(3, 5)[/tex].

Solution:
[tex]x^2-|x|+3=\left(|x|-\frac{1}{2}\right)^2+\frac{11}{4}\gt 0[/tex], so the outer absolute value can be dropped:
[tex]x^2-|x|+3\lt 5 \Leftrightarrow |x|^2-|x|-2\lt 0 \Leftrightarrow (|x|-2)(|x|+1)\lt 0[/tex]
Since [tex]|x|+1\gt 0[/tex], we get [tex]|x|\lt 2[/tex].
Answer: [tex]x\in(-2, 2)[/tex].

Solution:
Domain: [tex]x\neq 1[/tex]. Multiply by [tex]|x-1|\gt 0[/tex]:
[tex]|3x+2|\gt 3|x-1|[/tex]
Both sides are non-negative, so we can square them:
[tex](3x+2)^2-9(x-1)^2\gt 0 \Leftrightarrow (3x+2-3x+3)(3x+2+3x-3)\gt 0 \Leftrightarrow 5(6x-1)\gt 0 \Leftrightarrow x\gt \frac{1}{6}[/tex]
Answer: [tex]x\in\left(\frac{1}{6}, 1\right)\cup(1, +\infty)[/tex].

Solution:
Domain: [tex]|x|\neq 3[/tex]. Let [tex]t=|x|\geq 0[/tex]:
[tex]\frac{1}{t-3}-\frac{1}{2}=\frac{2-(t-3)}{2(t-3)}=\frac{5-t}{2(t-3)}\lt 0 \Leftrightarrow t\lt 3[/tex] or [tex]t\gt 5[/tex]
[tex]|x|\lt 3 \Leftrightarrow -3\lt x\lt 3[/tex]; [tex]|x|\gt 5 \Leftrightarrow x\lt -5[/tex] or [tex]x\gt 5[/tex]
Answer: [tex]x\in(-\infty, -5)\cup(-3, 3)\cup(5, +\infty)[/tex].

Solution:
[tex]\left|x^2-6x+5\right|\leq x-1 \Leftrightarrow -(x-1)\leq x^2-6x+5\leq x-1[/tex]
[tex]x^2-6x+5\leq x-1 \Leftrightarrow x^2-7x+6\leq 0 \Leftrightarrow 1\leq x\leq 6[/tex]
[tex]x^2-6x+5\geq 1-x \Leftrightarrow x^2-5x+4\geq 0 \Leftrightarrow x\leq 1[/tex] or [tex]x\geq 4[/tex]
The solution of the system is the intersection (common part) of the two sets:
Answer: [tex]x\in\{1\}\cup[4, 6][/tex].

Solution:
[tex]x\geq 2:\ |x-2|=x-2[/tex], [tex]|2x-x+2|=x+2\lt 3 \Leftrightarrow x\lt 1[/tex] — no solutions in this interval.
[tex]x\lt 2:\ |x-2|=2-x[/tex], [tex]|3x-2|\lt 3 \Leftrightarrow -3\lt 3x-2\lt 3 \Leftrightarrow -\frac{1}{3}\lt x\lt \frac{5}{3}[/tex]
Answer: [tex]x\in\left(-\frac{1}{3}, \frac{5}{3}\right)[/tex].

Solution:
$x^2-1=(x-1)(x+1)$

	$(-\infin,-1)$	$[-1,0)$	$[0, 1)$	$[1, +\infin)$
$(x-1)(x+1)$	+	-	-	+
$x$	-	-	+	+

Case 1. $x\in(-\infin, -1)$
$x^2 - 1 < x^2-(-x)+1$
$-1 < x+1$
$x>-2$, but $x\in(-\infin, -1) \Rightarrow x\in(-2, -1)$


Case 2. $x\in[-1, 0)$
$-(x^2 - 1) < x^2-(-x)+1$
$-x^2 + 1 < x^2+x+1$
$-2x^2-x<0$
$2x^2+x>0$
$x(2x+1)>0 \Leftrightarrow x \in (-\infin, -\frac12)\cup(0,+\infin)$
But $x\in[-1, 0) \Rightarrow x\in[-1, -\frac12)$


Case 3. $x\in[0,1)$
$-(x^2 - 1) < x^2-x+1$
$-x^2 + 1 < x^2-x+1$
$2x^2-x>0$
$x(2x-1)>0 \Leftrightarrow x \in (-\infin, 0)\cup(\frac12,+\infin)$
But $x\in[0, 1) \Rightarrow x\in[\frac12, 1)$


Case 4. $x\in[1,+\infin)$
$x^2 - 1 < x^2-x+1$
$x<2$
But $x\in[1,+\infin) \Rightarrow x\in[1, 2)$

Answer: $x\in(-2, -\frac12) \cup (\frac12, 2)$

Solution:
$x^2-2x-3 = 0$
$D=16, x_1=3, x_2=-1$
$x^2-2x-3 = (x-3)(x+1)$

	$(-\infin,-1)$	$[-1,2)$	$[2, 3)$	$[3, +\infin)$
$(x-3)(x+1)$	+	-	-	+
$x-2$	-	-	+	+

Case 1. $x\in (-\infin,-1)$
$x^2-2x-3+2(-(x-2))<5$
$x^2-2x-3-2(x-2)<5$
$x^2-2x-3-2x+4<5$
$x^2-4x-4<0$
$D=32,\ \ x_1=2-\sqrt{2},\ \ x_2=2+\sqrt{2}$
$\Rightarrow x^2-4x-4<0 \Leftrightarrow x\in (2-2\sqrt{2}, 2+2\sqrt{2})$
But $x\in (-\infin, -1)$
$\Rightarrow x\in \varnothing$


Case 2. $x\in [-1,2)$
$-(x^2-2x-3)+2(-(x-2))<5$
$-x^2+2x+3-2(x-2)<5$
$-x^2+2x+3-2x+4<5$
$-x^2+7<5$
$-x^2+2<0$
$x^2-2>0$
$(x-\sqrt{2})(x+\sqrt{2})>0 \Leftrightarrow x \in (-\infin, -\sqrt{2})\cup(\sqrt{2}, +\infin)$
But $x\in [-1,2) \Rightarrow x \in (\sqrt2, 2)$


Case 3. $x\in [2,3)$
$-(x^2-2x-3)+2(x-2)<5$
$-x^2+2x+3+2x-4<5$
$-x^2+4x-1<5$
$x^2-4x+6>0$
$D=-8 < 0 \Leftrightarrow \forall x$ is solution.
But $x\in [2,3) \Rightarrow x \in [2,3)$


Case 4. $x\in [3, +\infin)$
$(x^2-2x-3)+2(x-2)<5$
$x^2-2x-3+2x-4<5$
$x^2-12<0 \Leftrightarrow (x-2\sqrt3)(x+2\sqrt3) < 0$
$\Rightarrow x \in (-2\sqrt3, 2\sqrt3)$
But $x\in [3, +\infin) \Rightarrow x \in [3, 2\sqrt3)$


Answer: $x\in(\sqrt2, 2\sqrt3)$

Solution:
For [tex]-3\leq x\leq 3[/tex] we have [tex]x-4\lt 0[/tex], so [tex]|x-4|=4-x[/tex]:
[tex]4-x+k-5\lt 0 \Leftrightarrow k\lt x+1[/tex] for every [tex]x\in[-3, 3][/tex].
The smallest value of [tex]x+1[/tex] on [tex][-3, 3][/tex] is [tex]-2[/tex] (at [tex]x=-3[/tex]), so we need [tex]k\lt -2[/tex].
Answer: [tex]k\lt -2[/tex].

Solution:
Let [tex]a=x-z[/tex], [tex]b=z-y[/tex]. Then [tex]x-y=a+b[/tex], and we have to prove that [tex]|a+b|\leq |a|+|b|[/tex].
Both sides are non-negative, so it suffices to compare their squares:
[tex]|a+b|^2=a^2+2ab+b^2\leq a^2+2|ab|+b^2=\left(|a|+|b|\right)^2[/tex], because [tex]ab\leq |ab|[/tex].
Taking square roots of non-negative numbers preserves the inequality, so [tex]|a+b|\leq |a|+|b|[/tex], which proves the statement.
Geometric meaning: the distance from [tex]x[/tex] to [tex]y[/tex] on the number line is at most the distance from [tex]x[/tex] to [tex]z[/tex] plus the distance from [tex]z[/tex] to [tex]y[/tex].