Solving Quadratic Inequalities Using Inequality Properties

[albema]

Hi everyone, I need help again… on “quadratic inequality”.

How come the 1/3 < |x| < 1/2 become 1/3 < x < 1/2 or 1/3 < −x < 1/2?

And, this is the original problem of above solutions: 1/9 < x^2 < 1/4.

I wasn’t able to find the proof from my calculus book because the book says:

1) |x| = x if x positive, −x if x negative.
2) |x| is the distance between x and the origin or 0.
3) √x^2 = |x|.
4) |x| < a then −a < x < a.
5) |x| > a then x > a or x < −a.

That is the only five properties of inequality (in my opinion) I can hold and I can’t derived more the above-mentioned question by myself with only that five properties.

This is the solutions of that problem I’m doing by myself:

1/9 < x^2 < 1/4
√1/9 < √x^2 < √1/4  √x^2 = |x|, thus
√1/9 < |x| < √1/4
1/3 < |x| < 1/2

Then I divide the solutions for 2 part, first:

|x| < 1/2
−1/2 < x < 1/2

second:

|x| > 1/3
x < −1/3 or x > 1/3

Please advance and thanks.

Why do most people very like James Stewart’s Calculus if there available better (in my opinion).
I’ve got everything about Calculus from Thomas’ Calculus, 11E which I think (in my opinion) that their explanations, descriptions, solutions of covered materials can much be understand than James Stewart’s Calculus by person like me or people don’t have really much known about calculus, organized very well.
Or, if I may say, why do people interested to a book that bundled many materials into one and makes it become so un-organized.
Actually, I have the following 3 books of Calculus and short them into OK, good, to very good:

Calculus, 5E
James Stewart
Brooks Cole

Calculus, 8E
Dale E. Varberg, Edwin J. Purcell, Steve E. Rigdon
Prentice Hall

Thomas’ Calculus, 11E
Giordano, Weir
Addsion Wesley

In contradiction, I love them very much except the first one.

 

[red_dog]

What you did is correct.
$$\displaystyle |x|<\frac{1}{2}\Rightarrow x\in\left(-\frac{1}{2},\frac{1}{2}\right)$$
$$\displaystyle |x|>\frac{1}{3}\Rightarrow x\in\left(-\infty,-\frac{1}{3}\right)\cup\left(\frac{1}{3},\infty\right)$$
Now you have to intersect the sets and you get
$$\displaystyle x\in\left(-\frac{1}{2},-\frac{1}{3}\right)\cup\left(\frac{1}{3},\frac{1}{2}\right)$$

 

[albema]

But, that not the answer I want. I just want to know from what inequality properties that the author can be directly derived the solutions into “1/3 < x < 1/2 or 1/3 < −x < 1/2” form instead “−1/2 < x < 1/2, x < −1/3 or x > 1/3” of mine. Anybody can help me on this?