Range of $x$ for which the modulus inequality will be valid : $|x^2-2x-8| > 2x$ : Need clarification about solution set

Question

$|x^2-2x-8| > 2x \\ $

For this problem I took two cases where in one case $ x^2-2x-8 \geq 0$ and in second case $ x^2-2x-8 \lt 0$.

$1st$ Case when $ x^2-2x-8 \geq 0$ :-

$ x^2-2x-8 \gt 2x$

For this case I got the solution set as $x \in (-\infty,-2] \cup \biggl(2(1+\sqrt{3}),\infty\biggr)$ upon solviing the above two equations i.e. $ x^2-2x-8 \geq 0$ and $ x^2-2x-8 \gt 2x$ and then taking the common range of values of $x$ from the solutions of the two equations.

$2nd$ Case when $ x^2-2x-8 \lt 0$ :-

$ -(x^2-2x-8) \gt 2x$

For this case I got the solution set as $x \in (-2,2\sqrt{2})$

I am getting confused how to decide the overall solution set for the original problem as a whole because if we plot the solution set on the number line for both the above cases then as per my understanding the common range from both the solution set should be the answer and if that's the case the answer should be $x \in (-2,2)$ as this is the range which is common to the two solution set but unfortunately the answer is different.

Can someone please help me on this and help with some clarification with in these cases how do you decide the solution set?

Answer 1

You should take the union of the two region as you want to collect all the possible $x$ for the conditions to be true.

If you do so, you get $(-\infty, 2\sqrt2) \cup (2(1+\sqrt{3}, \infty)$.

Answer 2

Consider the first part. You wrote that you were dealing with the case in which $x^2-2x-8\geqslant0$. That means that $x\leqslant-2$ or that $x\geqslant4$. Then you solved the inequation $x^2-2x+8>2x$, and you should have got that $x>2+2\sqrt3$ or that $x<2-2\sqrt3$. So, the solution when you are in the first situation is that $x\in(-\infty,-2]\cup\left[2+2\sqrt3,\infty\right)$

Now, the second part. Asserting that $x^2-2x-8<0$ means that $x\in(-2,4)$. And asserting that $-(x^2-2x-8)>2x$ means that $x\in\left(-2\sqrt2,2\sqrt2\right)$. So, the solution when you are in the second situation is that $x\in\left(-2,2\sqrt2\right)$.

So, the global answer is that $x\in\left(-\infty,2\sqrt2\right]\cup\left[4,\infty\right)$.

Answer 3

You seem to have taken the range common to the two solution sets. In other words, you have taken the intersection of the two sets, which is also done incorrectly because the intersection of the two cases you have taken is a null set because those two conditions cannot be true simultaneously. $x^2-2x-8 \geq 0 $ AND $x^2-2x-8 < 0 $ is not possible. $$ \{x^2-2x-8 \geq 0 \} ~ \cap ~ \{x^2-2x-8 < 0 \} = \phi$$ The final solution set that you are looking for is the union of the two sets because either $x^2-2x-8 \geq 0 $ OR $x^2-2x-8 < 0 $. OR signifies that you are supposed to take the union of the sets. $$ \{x^2-2x-8 \geq 0 \} ~ \cup ~ \{x^2-2x-8 < 0 \} = (-\infty, 2\sqrt2) \cup (2+2\sqrt{3}, \infty) $$