How do I solve this quadratic inequality with numbers on both sides? I thought I'd teach myself some A-Level Maths at home and I'm stuck on a problem I got from mymaths. Problem is mymaths don't bother providing answers, and the tutorial section didn't show me how to solve it and I kept getting the answer wrong.
$18< x^2 + 3x<40$
I had to solve this. I decided to do them as separate inequalities, 
$18<x^2+3x$
$x^2+3x<40$
And after factorizing and sketching a rough graph I got the respective values:
$-8<x<5$
$-6<x<3$
The problem is that's as far as I got, and when I put any of the above as my answer, I still got it wrong. What is the answer and how do I get it? Have I gone wrong anywhere?
 A: You solved first inequality wrong.
$$
x^2 + 3x - 18 > 0 \implies (x+6)(x-3) > 0 \implies x \in (\infty, -6) \cup (3, +\infty)
$$
Second one is correct, and solution is $x \in (-8, 5)$. So general solution is intersection of those $x \in (-8, -6) \cup (3, 5)$.
A: You are on the right track by treating them as separate inequalities:-
$$x^2+3x<40\Rightarrow (x^2+3x-40)<0\Rightarrow (x+8)(x-5)<0$$
$$x^2+3x>18\Rightarrow (x^2+3x-18)>0\Rightarrow (x+6)(x-3)>0$$
Now, to satisfy the first inequality, $-8<x<5$, which is what you obtained. To satisfy the second inequality, you need $x>3$ or $x<-6$ (which is where you went wrong).
Here is the key point: you want to satisfy both inequalities - you look for the overlap in the range of values $x$ can take. Thus your answer should be $3<x<5$ and $-8<x<-6$.
A: Just another way, without splitting:
$$18< x^2 + 3x<40 \iff 4.5^2< (x + 1.5)^2 <6.5^2 \iff 4.5 < \pm(x + 1.5) < 6.5$$
$$\iff 3 < x < 5 \quad \text{or} \quad -8 < x < -6 $$
A: HINT: Factor the polynomial like it has equal signs and then factor. you do not separate inequalities. That will result in different answers. Then the solve the inequality like you usually do.
