How do you solve this inequality with absolute values?

Question

$$|2|x|-5| < |4-|x-1||$$

I tried squaring everything to get rid of absolute value markings, but couldn't figure it out.

Answer 1

We can analyze your inequality with respect to 3 cases: $$Case \ 1:x\ge1\\Case \ 2: 0\le x \lt1\\Case \ 3:x\lt 0 $$ Now let's start with the $\color{red}{Case \ 1}:$ $$|x|=x \quad \& \quad |x-1|=x-1 $$ Therefore we have: $$|2x-5|\lt|4-x+1|\\(2x-5)^2<(5-x)^2\\3x^2-10x<0\\0<x<\frac{10}{3} \quad (Note\ that \ x \ge1)\\\underline{ 1\le x<\frac{10}{3}}$$ Now for $\color{red}{\ Case \ 2}:$ $$|x|=x \quad \& \quad |x-1|=-x+1 $$ Therefore we have: $$|2x-5|\lt|4+x-1|\\(2x-5)^2<(3+x)^2\\3x^2-26x+16<0\\ \frac{2}{3}<x<8 \quad (Note\ that \ 0\le x \lt1)\\\underline{ \frac{2}{3}<x<1}$$ Lastly for $\color{red}{\ Case \ 3}:$ $$|x|=-x \quad \& \quad |x-1|=-x+1 $$ Therefore we have: $$|-2x-5|\lt|4+x-1|\\(2x+5)^2<(3+x)^2\\3x^2+14x+16<0\\ \underline{-\frac{8}{3}<x<-2}$$ So the solution to your inequality is: $$ \bbox[5px,border:2px solid black] {-\frac{8}{3}<x<-2\lor\frac{2}{3}<x<\frac{10}{3}}$$ $$$$Here is an image of the region bounded by your inequality:

Answer 2

Consider the following cases:

$(1)$ If $x \in [0,1] $, we have $|x-1| =1-x $ and $|x|=x $. Then the inequality becomes $$|2x-5|<|3+x|$$

Similarly proceed for $x \leq 0$ and for $x>1$ analysing also each subcase that arises due to any condition. Hope it helps.

Answer 3

Graphically observing the values for which $f(x)>g(x)$, $f(x)=|2|x|-5|$ and $g(x)=|4-|x-1||$ is a simple non-algebraic way to solve this. You can graph these functions in stages knowing that $|h(x)|$ reflects $h(x)$ into the positive $y$-axis quadrants and multiplication by $-1$ flips the whole function about the $x$-axis.

Otherwise you can solve this using cases and algebra which is more cumbersome and involves considering subsets of simultaneous inequalities. You were on the right track with squaring everything: using $|x| = \sqrt{x^2}$ produces the inequality $$(2|x|-5)^2<(4-|x-1|)^2$$

and seeing as both sides are squares, moving everything to one side and factorising using $a^2-b^2=(a-b)(a+b)$ yields: $$(2|x|-5-4+|x-1|)(2|x|-5+4-|x-1|)<0$$ $$(2|x|+|x-1|-9)(2|x|-|x-1|-1)<0$$

From here, consider the four cases and solve the inequalities given that some of the solution sets will overlap: $|x|>0 \cup |x-1|>0$, $|x|>0 \cup |x-1|<0$, $|x|<0 \cup |x-1|>0$ and $|x|<0 \cup |x-1|<0$.

You could also use the fact that $|x|<a$ implies $-a<x<a$ but this is more complicated since both sides are absolute values.