<s>[INST] Quadratic inequality with Modulus of a Linear Equation : Solution Verification & Clarification Needed $x^2-|5x-3|-x<2$
For this problem I took two cases where in one case $5x-3 \geq 0$ and in second case $5x-3 \lt 0$.
$1st$ Case  when $5x-3 \geq 0$ :-
For this case I got the solution set as $x \in [0.6,3+2\sqrt{2}]$.
$2nd$ Case  when $5x-3 \lt 0$ :-
For this case I got the solution set as $x \in [-5,0.6]$.
Upon combining both the cases I am getting the solution set as $x \in [-5,3+2\sqrt{2}]$ but in the answer that has been provided for this problem is not including the points and thus the answer given is $x \in (-5,3+2\sqrt{2})$.
What am I doing wrong here that is making my endpoints of the solution inclusive whereas the the solution set provided has got endpoints exclusive. Please help me on this !
Thanks in advance !
 [/INST] Let me forcus on when $5x-3<0$, we have $x < \frac35$ and
$$x^2-|5x-3|-x<2$$
becomes
$$x^2-(3-5x)-x<2$$
$$x^2+4x-5<0$$
$$(x+5)(x-1)<0$$
Hence we have $-5 < x < 1$ and $x < \frac35$, hence taking intersection, we have $-5<x < \frac35$.
To answer the question, why did we exclude $x=-5$, consider $(x+5)(x-1)<0$, if we substute $x=5$ inside, we get $0<0$ which is not true.

If $5x-3 \ge 0$, we have $x \ge \frac35$ and
$$x^2-|5x-3|-x < 2$$
becomes
$$x^2-5x+3-x < 2$$
$$x^2-6x+1 < 0$$
$$\frac{6-\sqrt{36-4}}{2} < x < \frac{6+\sqrt{36-4}}{2} $$
$$3-2\sqrt2< x < 3+2\sqrt2 $$
Taking intersection with $x \ge \frac35$, we have
$$\frac35 \le x < 3+2\sqrt2$$

Taking union, we have $-5 < x < 3+2\sqrt2$.
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