how to solve inequation involving modulus how to solve $$ǀx^2 + 3xǀ + x^2-2 ≥ 0 ?$$
I got stuck in the above problem.
What would be the classic process to solve these type of problems. Also, if u have some fast processes then please explain it. It would be of great help in my upcoming tests. Please Help.
 A: I would split the domain into three pieces: $x\leq-3,-3\leq x\leq0$ and $x\geq0$.  That gives you two different quadratic polynomials:
$$x^2+3x+x^2-2\geq0(x\leq-3;x\geq0)\\
-x^2-3x+x^2-2\geq0(-3<x<0)$$
Find the roots of the quadratics, and combine them with the three pieces $x<-3,-3<x<0,0<x$.
A: Move to the $2$ to the right-hand side and square both sides to get
$$
\left(x^{2}+3x\right)^{2}x^{4}-4\geq0.
$$
You can expand and then factor this to get
$$
\left(x^{4}+3x^{3}-2\right)\left(x+1\right)\left(x^{3}+2x^{2}-2x+2\right)\geq0.
$$
Now it's just enumeration. Check when an even number of terms are
negative. There are exact (radical) forms for this (notice that the
highest degree is a quartic), but you'll probably just want to do
this numerically unless you're a sadist.
To save you the trouble, approximately,
$$
x\geq0.8069
$$
$$
-2.91964\leq x\leq-1
$$
$$
x\leq-3.06918
$$
If you really want exact forms, I would use a symbolic algebra package.
A: we will break up your inequality into two pieces according to $x^2 + 3x =x(x+3)$ is positive or negative.  we have $$(x^2 + 3x) +x^2 - 2 =2x^2 + 3x - 2 = (2x-1)(x+2)\ge 0 \text{ and } x < -3, x > 0\tag 1$$  or  $$-x^2 -3x +x^2 - 2 \ge 0 \text{ and } -3 < x < 0\tag 2$$
$(1)$  has solution $x < -2$ or $x > \frac 12$  and $(2)$ has solution $-3 < x < -\frac 23$
the final solution is $$x < -\frac 23 \text{ or } x > \frac 12 $$
