Consider this absolute value quadratic inequality
$$ |x^2-4| < |x^2+2| $$
Right side is always positive for all real numbers,so the absolute value is not needed.
Now consider the cases for the left absolute value
- $$ x^2-4 \geq 0 $$
We get $$ x \geq \pm 2 $$
Solve for first case $$ x^2 – 4 < x^2+2 $$ the solution $$ 0 < 6$$ This is true for all real numbers; taking in consideration the boundary that x has to be greater equal +2 and -2 the first part of the solution I think should be $$ L_1 = [2, \infty) $$
- $$ x^2-4 < 0$$ $$ x < \pm 2 $$
Solve for seonc case I get $$ x > \pm 1 $$ Solution for the second case considering the boundary of 2. should be $$ L_2 = (-2,-1) \cup (1,2) $$ Final solution;
$$ L = (-2,-1) \cup (2,\infty) $$
According to the solutions,this is wrong; it should be $$ L = (-\infty,-1) \cup (1,\infty) $$ Rechecking their L1 and L2; L2 should be correct but for L1 they have $$ L_1 = R \backslash (-2,2) $$
So all real numbers except -2 and 2? Can anyone explain how this is the solution.First the sign is greater equal than +2 -2 shouldnt those numbers be included? Also we are looking for numbers GREATER than -2 and +2, since 2 is greater than -2 I assumed we only need to take numbers from to infinity,how does negative infinity come in consideration here?
Thanks in advance!
Best Answer
In 1., $x^2-4\ge0$ is not equivalent to "x has to be greater equal +2 and -2" but to $x\in L_1,$ where $$L_1=(-\infty,-2]\cup[2,+\infty).$$
In 2., the notations $x<\pm2$ and $x>\pm1$ are similarly confusing and incorrect, but the conclusion is ok: $$L_2 = (-2,-1) \cup (1,2).$$
Your final solution would anyway be false, since $[2,\infty)\cup(-2,-1)\cup(1,2)$ is not equal to $(-2,-1)\cup(2,\infty)$ but to $(-2,-1)\cup(1,\infty).$
Now, the correct solution is indeed $$L=L_1\cup L_2=(-\infty,-1) \cup (1,\infty).$$
Note that we knew in advance that each of the sets $L_1,L_2,L$ would be symmetric with respect to $0,$ since they are given by inequations on $x^2.$