I'm currently working through Chapter 1 of Spivak Calculus. I'm trying to solve the inequalities on problem 4 the "proper" way but I am having trouble on exercise 4 - (vii).
Problem
Find all the numbers $x$ for which $x^2 - x + 10 > 16$
Relevant Information
The chapter introduces and uses the following properties to solve
Additionally, Spivak introduces properties regarding inequalities
The final piece of relevant information is the assertions given here:
Assertion of ab > 0 if a < 0 and b < 0
My attempt
$ \begin{align} x^2 - x + 10 &> 16 & \text{Given}\\ x^2 - x - 6 &> 0 & \text{P3}\\ (x-3)(x+2) &> 0 & \text{P9}\\ & \text{...}\\ \end{align} $
Question
How to solve from here?
More specifically, how to solve using the properties and assertions that Spivak has introduced. Is my attempt from $(x-3)(x+2) > 0$ onwards (shown below) the appropriate way to use these foundations?
If so, how is it completed?
My attempt continued
The form is of $ab > 0$ this leads me to believe that I would use both P12 - Closure under multiplication and the assertion Spivak shows: $(-a)(-b) > 0$ to consider the following
$ \begin{align} \text{1.} && (x-3)(x+2) \text{ is in P when} && (x-3) \text{ and } (x + 2) \text{ are both in P}\\ \text{ - OR - }\\ \text{2.} && (x-3)(x+2) \text{ is in P when} && -(x-3) \text{ and } -(x + 2) \text{ are both in P}\\ \end{align} $
Starting with 1.
$ \begin{align} \text{1.} && (x-3)(x+2) \text{ is in P when} && (x-3) \text{ and } (x + 2) \text{ are both in P}\\ \end{align} $
$ \begin{align} x - 3 &> 0 & \text{definition of P} && x+2 &> 0 && \text{definition of P} \\ x &> 3 & \text{P3} && x &> -2 && \text{P3}\\ \end{align} $
Now with 2.
$ \begin{align} \text{2.} && (x-3)(x+2) \text{ is in P when} && -(x-3) \text{ and } -(x + 2) \text{ are both in P}\\ \end{align} $
$ \begin{align} -(x - 3) &> 0 & \text{definition of P} && -(x+2) &> 0 && \text{definition of P} \\ 3 &> x & \text{P3} && -2 &> x && \text{P3}\\ \end{align} $
Condensing 1. and 2.
$ \begin{align} x &> 3 & \text{and} && x &> -2\\ \text{- OR -}\\ 3 &> x & \text{and} && -2 &> x\\ \end{align} $
This doesn't make sense to me, how is the answer $x > 3$ and $x < -2$
Edit:
I see it now.
First, it's $x > 3$ or $x < -2$ which is obvious because $ab > 0$ OR $(-a)(-b) < 0$ which was introduced.
Second, I see why in case 1, it's $x > 3$ because $x > 3$ and $x > -2$ can be simplified to $x > 3$. Same logic to simplify case 2 and get final result.
Note that $$x>3 \quad \text{and} \quad x>-2 \iff x>3$$ while $$ x<3 \quad \text{and} \quad x<-2 \iff x<-2 $$ Thus the solution is $x>3$ or $x<-2$.