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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

Properties of Numbers P1-P9

Additionally, Spivak introduces properties regarding inequalities

Inequalities, P10-P12

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.

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    $\begingroup$ Instead of using the "P" and "Pn" notations with no explanation, you would be much more likely to get an answer if you explained what you meant in terms understandable by someone who has never seen this book. $\endgroup$ – rogerl Jan 27 '18 at 18:04
  • $\begingroup$ thank you for the feedback! I've used @StudentsTea awesome answer on a similar problem here as a template for reformatting. $\endgroup$ – Jake Kirsch Jan 28 '18 at 17:21
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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 $x<-2$.

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  • $\begingroup$ thank for your answer, it took me a while to just take a step and think about the ands/ors to see it clearly. $\endgroup$ – Jake Kirsch Jan 28 '18 at 17:27

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