The solution of the inequality $x^2 > b^2$, where $b < 0$ is: Trying to learn maths as an adult and currently working my way through a high school textbook. So far, the method taught for solving quadratic inequalities has been to graph it and then read the values off the graph.
So far, I have: Let $x^2 = b^2.$ Then
$$x^2 - b^2 = 0 \Rightarrow (x+b)(x-b) = 0,$$
so $x = \pm b$.
The quadratic crosses the $x$ axis at $-b$ and $b$. But I get stuck when trying to graph this. I've spent a long time on it and must be missing something obvious!
 A: $$x^2>b^2\iff (x+b)(x-b)>0$$
$$\iff \Bigl((x+b)<0 \wedge (x-b)<0\Bigr) \vee \Bigl( (x-b)>0\wedge (x+b)>0\Bigr)$$
$$\iff x<b \;\text{ or } \;x>-b$$
$$\iff x\in(-\infty,b)\cup(-b,+\infty)$$
$$\iff x\in(-\infty,-|b|)\cup (|b|,+\infty)$$
A: " So far, the method taught for solving quadratic inequalities has been to graph it and then read the values off the graph."
So ... the values are $-b$ and $b$.  (DOn't get confused and think $\pm b$ as written with one symbol is a single value.)
Also $b < 0$ so $-b > 0$ and so the $b < 0 < -b$ so don't get confused and assume $b$ is positive and $-b$ is negative.  Is is the exact opposite.
(If you like you could replace $b$ with $-|b|$ and $-b$ with $|b|$.
So take an $x < b < 0$ then $x^2 > b^2$ so graph that part.
Then take $b < x < -b$.  You could take $x = 0$ to make it easy.  $0^2 < b^2$ so don't graph this part.  And then if $0 < -b < x$ then $b^2 < x^2$ so graph this part.
.....
Yes,  I admit it.  $-b$ being positive throws me off too.
So let $a = -b$ so $a > 0$.
Then $x^2 = b^2 =(-a)^2 = a^2$ and
$x^2-a^2 = (x-a)(x+a)=0$ and the points are $\pm a$.  If $x < -a < 0$ then$0 < a < -x$ and $a^2 < (-x)^2 =x^2$.
And if $-a < x < a$ then $|x| < a$ and $x^2 < a^2$.
And if $0 < a < x$ then $a^2 < x^2$.
