How do I solve absolute value inequalities involving polynomials?

AI Thread Summary
To solve absolute value inequalities involving polynomials, it's essential to first identify the critical values of the polynomial, which divide the number line into intervals. Each interval can then be tested to determine if it satisfies the inequality. The process involves separating the absolute value into two cases: when the expression is equal to zero and when it is less than zero. It's important to remember that the absolute value of a polynomial is always non-negative. Understanding these steps is crucial for accurately finding the solution set.
moondaaay
Messages
2
Reaction score
0
1.
2015-10-10 07.18.11.png

Homework Equations


Solving Polynomial Inequalities

The Attempt at a Solution




2015-10-10 07.18.11.png

Then I used the property of absolute value inequality to get rid of it.

But I really don't know if I'm doing the right step. Is this correct? So that I could separate them in two cases and find the solution set.
2015-10-10 07.39.00.png

 
Physics news on Phys.org
Divide it into two alternative cases.
(1) It's equal to zero. This is the easiest one. What can we say about x in that case?
(2) It's less than zero. What does that imply about the signs of the numerator and the denominator. Is there any part of either that we can ignore?
 
moondaaay said:
1. View attachment 89960

Homework Equations


Solving Polynomial Inequalities

The Attempt at a Solution




View attachment 89960
Then I used the property of absolute value inequality to get rid of it.

But I really don't know if I'm doing the right step. Is this correct? So that I could separate them in two cases and find the solution set.
View attachment 89962
Hello moondaaay. Welcome to PF !

Where is the function on the left side of the inequality zero?

What is its domain?

upload_2015-10-9_22-8-54.png
 
The inequality has four critical values for x, splitting the real number line into five intervals. Each interval can be checked to find which are solutions to the inequality and which are not solutions.
 
symbolipoint said:
The inequality has four critical values for x, splitting the real number line into five intervals. Each interval can be checked to find which are solutions to the inequality and which are not solutions.
But what should I do first to remove the absolute value?
 
moondaaay said:
But what should I do first to remove the absolute value?
Not really important here. Whatever sign the 3x+7, its absolute value will still be POSITIVE OR ZERO.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top