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Re: |x-2| > 3
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19 Oct 2021, 04:14
19 Oct 2021, 04:14
Expert Reply
Tejas110 wrote:
IlCreatore wrote:
Solving the disequality in the header, \(|x-2|>3\), we get \(x<-1, x>5\).
Then, the smallest value for \(|x-3.5|\) is when x = 5, i.e. \(|5-3.5| = 1.5\). The smallest value for \(|x-1.5|\) is when x = -1, i.e. \(|-1-1.5| = 2.5\).
Thus, since 2.5>1.5, quantity B is greater!
Then, the smallest value for \(|x-3.5|\) is when x = 5, i.e. \(|5-3.5| = 1.5\). The smallest value for \(|x-1.5|\) is when x = -1, i.e. \(|-1-1.5| = 2.5\).
Thus, since 2.5>1.5, quantity B is greater!
Can someone please elaborate why he has taken the extreme cases to plug in the value of 'X' like 5, -1, whereas x>5 or x<-1, so why not x = 6 or -2. I tried choosing x=6 wherein A<B, and if x=-2 A>B, it should be 'D'. Can someone else resolve my query here
Please Sir read the explanations above. We discussed extensively above
Re: |x-2| > 3
[#permalink]
21 Jul 2022, 22:50
21 Jul 2022, 22:50
IlCreatore wrote:
Solving the disequality in the header, \(|x-2|>3\), we get \(x<-1, x>5\).
Then, the smallest value for \(|x-3.5|\) is when x = 5, i.e. \(|5-3.5| = 1.5\). The smallest value for \(|x-1.5|\) is when x = -1, i.e. \(|-1-1.5| = 2.5\).
Thus, since 2.5>1.5, quantity B is greater!
Then, the smallest value for \(|x-3.5|\) is when x = 5, i.e. \(|5-3.5| = 1.5\). The smallest value for \(|x-1.5|\) is when x = -1, i.e. \(|-1-1.5| = 2.5\).
Thus, since 2.5>1.5, quantity B is greater!
Can you help me how to solve this inequality? Actually, I have forgotten it
Re: |x-2| > 3
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18 May 2024, 02:02
18 May 2024, 02:02
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
