Is the Solution to the Absolute Value Inequality x^2<4 then |x|<=2 Correct?

AI Thread Summary
The discussion centers on the validity of the statement "If x^2 < 4 then |x| <= 2." Participants clarify that solving the inequality yields -2 < x < 2, which implies |x| < 2, not |x| <= 2. There is a consensus that the original text likely contains a typo, as logically, if x^2 < 4, then |x| must indeed be less than 2. The confusion arises from the distinction between open and closed intervals. Ultimately, the conclusion is that the statement is true, but the notation used is misleading.
OceanSpring
Messages
11
Reaction score
0
Member warned about not using the homework template
Question:
True or False If x^2<4 then |x|<=2

My solution:
I get -2<x<2 when I solve the problem so it should be false. Yet the text says its true? Is this a mistake? If |x| is equal to 2 then it should be a closed interval, not an open interval which seems to be correct to me.
 
Last edited by a moderator:
Physics news on Phys.org
OceanSpring said:
Question:
True or False If x^2<4 then |x|<=2

My solution:
I get -2<x<2 when I solve the problem so it should be false. Yet the text says its true? Is this a mistake? If |x| is equal to 2 then it should be a closed interval, not an open interval which seems to be correct to me.
x2 < 4 is equivalent to -2 < x < 2 or |x| < 2. What the text has appears to be a typo.
 
OceanSpring said:
Question:
True or False If x^2<4 then |x|<=2

My solution:
I get -2<x<2 when I solve the problem so it should be false. Yet the text says its true? Is this a mistake? If |x| is equal to 2 then it should be a closed interval, not an open interval which seems to be correct to me.

It's probably a typo, although logically it is true:

If ##x^2 < 4## then ##|x| < 2## hence ##|x| \le 2##

If it were false, then there would be ##x## with ##|x| > 2## yet ##x^2 < 4##
 
  • Like
Likes schlynn
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top