What's the solution to $ 2|2x - a| < |x + 3a|$ (found critical values)

Question

$$ 2|2x - a| < |x + 3a|$$

I found the critical values to be $x = \frac{5a}{3}$ and $x = \frac{-a}{5}$ by removing the abs val signs and equation the left side to the +/- right side.

Now this is an inequality, and I have an arbitrary constant $a$. I'm not sure how to find the range.

P.S.: I'd prefer to learn how to solve it algebraically in particular. Also, if I knew a, I'd take those critical values and test inputs between and beyond those values. Not sure what to do here.

Answer 1

For $a=0$ our inequality has no solutions.

But for $a\neq0$ it's equivalent to $$4(2x-a)^2<(x+3a)^2$$ or $$15x^2-22ax-5a^2<0$$ or $$\frac{11a-14|a|}{15}<x<\frac{11a+14|a|}{15}.$$

Answer 2

An inequality $|P(x)|<|Q(x)|$ is the same as $$ (P(x))^2<(Q(x))^2 $$ which becomes $$ (Q(x)-P(x))(Q(x)+P(x))>0 $$ In your case you get $$ (x+3a-4x+2a)(x+3a+4x-2a)>0 $$ and, in shorter form, $$ (3x-5a)(5x+a)<0 $$ Now it’s easier: for $a=0$ you have no solution. For $a>0$, the roots are $-a/5<5a/3$, so $$ -\frac{1}{5}a<x<\frac{5}{3}a $$ You can do the $a<0$ case.