# Help me get the solution of a quadratic inequality

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#### albema

##### Newbie level 4
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

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

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

Well, just follow those definite rules that you now and wrote here. They can be easily checked and proved. The corresponding information may be found in any mathematical text-book devoted to unequality solutions.

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.

These are the main expressions required. Follow them and try to realize ones.

With respect,

Dmitrij

|x| < 1/2
−1/2 < x < 1/2

second:

|x| > 1/3
x < −1/3 or x > 1/3

it is true >>> -1/2 < x < 1/2
>>> x < -1 /3 and x > 1/3
these are solutions of the inequality and if you think that these are sets then their intersection will be the solution so the solution is
-1/3 > x > -1/2 and 1/3 < x < 1/2
it is the first thing you wrote
1/3 < x <1/2 and 1/3 < -x < 1/2 (multiply all sides by -1 and chamge the direction of inequalities )

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