Solve the inequality $|x^2-3x-15|<2x^2-x$

Question

I wish to solve the inequality

$$|x^2-3x-15|<2x^2-x$$

I have tried to solve this inequality in this way

Let $|x^2-3x-15|=f(x)$ and $2x^2-x = g(x)$ then solve for $f(x)<g(x)$ and then put the answer in sign chart and check if the intervals satisfy this inequality $f(x)\ge 0$ and solve for $f(x)>-g(x)$ and check if intervals satisfy this inequality $f(x)<0$.

But in second one the answer come up to be $(-\infty,-5/3)\cup (3,+\infty)$ and if we choose for example $(-3)$ it doesn't satisfy the inequality $f(x)<0$.

Is the way that I am solving wrong if it's so can you please show me an accurate and general way of solving such inequalities.

Answer 1

The answer is right.

I like the following way.

It's equivalent to $$-2x^2+x<x^2-3x-15<2x^2-x,$$ which gives $$x^2+2x+15>0$$ and $$3x^2-4x-15>0$$ and since $$x^2+2x+15=(x+1)^2+14>0,$$ which is true for all reals $x$ and $$3x^2-4x-15=3x^2-9x+5x-15=(x-3)(3x+5),$$ we got the answer: $$\left(-\infty,-\frac{5}{3}\right)\cup(3,+\infty).$$

I used the following fact. $$|f(x)|<g(x)\Leftrightarrow-g(x)<f(x)<g(x).$$