Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Just want to make sure that my line of thinking is right. I got D as well.
sqrt(x-5)^2 = 5-x ==> |x-5| = 5-x
which means that if
x-5<0, -(x-5) = 5-x = no solution, therefore x<5
x-5>0, x-5 = 5-x, which means that x>5 and x = 5
from (1) -x|x| > 0, we know that x has to be negative -(-x)|-x| is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.
from (2) 5-x>0, therefore x<5.
Can someone please point out if there is something wrong with my reasoning.
Is root(x-5)^2 = 5 - x ? [#permalink]
09 Sep 2009, 14:41
5
This post received KUDOS
Expert's post
6
This post was BOOKMARKED
dhushan wrote:
Just want to make sure that my line of thinking is right. I got D as well.
sqrt(x-5)^2 = 5-x ==> |x-5| = 5-x
which means that if
x-5<0, -(x-5) = 5-x = no solution, therefore x<5
x-5>0, x-5 = 5-x, which means that x>5 and x = 5
from (1) -x|x| > 0, we know that x has to be negative -(-x)|-x| is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.
from (2) 5-x>0, therefore x<5.
Can someone please point out if there is something wrong with my reasoning.
Is \(\sqrt{(x-5)^2}=5-x\)?
First of all, recall that \(\sqrt{x^2}=|x|\).
Is \(\sqrt{(x-5)^2}=5-x\)? --> is \(|x-5|=5-x\)? --> is \(x-5\leq{0}\)? --> is \(x\leq{5}\)?
(1) \(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so for \(-x|x|\) to be positive, \(-x\) must be positive \(-x>0\) --> \(x<0\). Sufficient.
Re: Is root(x-5)^2 = 5 - x ? [#permalink]
12 Jul 2015, 05:49
Bunuel wrote:
dhushan wrote:
Just want to make sure that my line of thinking is right. I got D as well.
sqrt(x-5)^2 = 5-x ==> |x-5| = 5-x
which means that if
x-5<0, -(x-5) = 5-x = no solution, therefore x<5
x-5>0, x-5 = 5-x, which means that x>5 and x = 5
from (1) -x|x| > 0, we know that x has to be negative -(-x)|-x| is the only way to get a number greater than 0. Therefore, this means that of the three possible solutions for x, only this x<5 hold true.
from (2) 5-x>0, therefore x<5.
Can someone please point out if there is something wrong with my reasoning.
Is \(\sqrt{(x-5)^2}=5-x\)?
First of all, recall that \(\sqrt{x^2}=|x|\).
Is \(\sqrt{(x-5)^2}=5-x\)? --> is \(|x-5|=5-x\)? --> is \(x-5\leq{0}\)? --> is \(x\leq{5}\)?
(1) \(-x|x| > 0\) --> \(|x|\) is never negative (positive or zero), so for \(-x|x|\) to be positive, \(-x\) must be positive \(-x>0\) --> \(x<0\). Sufficient.
(2) \(5-x>0\) --> \(x<5\). Sufficient.
Answer: D.
Hi Bunuel,
I didn't understand from here : is |x−5|=5−x? --> is x−5≤0? --> is x≤5?.
Modulus always results in an positive value. If x-5>0 then |x−5| should be equal to x-5 However, If x-5<0 then modulus would result the positive value of it i.e. -(x-5)=5-x
Could you please explain this step |x−5|=5−x? --> is x−5≤0? --> is x≤5?. I know for all values less than or equal to 5 holds good for this equation, however, how do we arrive to that final equation using |x−5|=5−x.
What i do understand is that absolute value of anything has two values. When i simplify the equation above i get x=5 (when i take positive value) and x-5=x-5 (when i take negative value).
Could you please explain this step |x−5|=5−x? --> is x−5≤0? --> is x≤5?. I know for all values less than or equal to 5 holds good for this equation, however, how do we arrive to that final equation using |x−5|=5−x.
What i do understand is that absolute value of anything has two values. When i simplify the equation above i get x=5 (when i take positive value) and x-5=x-5 (when i take negative value).
Thanks.
Let me try to answer.
You know that \(\sqrt{x^2} = |x|\) ---> \(\sqrt{x^2} = \pm x\). Thus \(\sqrt{(x-5)^2} = |x-5|\) --->\(\sqrt{(x-5)^2} = \pm (x-5)\)....(1)
Additionally, you know that \(\sqrt {x} \geq 0\) (for GMAT purposes!)....(2)
Thus based on (1) and (2) above,
Coming back to your question, the questions asks whether \(\sqrt{(x-5)^2} = 5-x\). This can only be possible when \(x\leq 5\) as if x>5 then 5-x will become <0 and will go against (2) above. The most x can go is = 5....(3)
Also, from (1), \(\sqrt{x^2} = - x\) only when x<0 ...(4)
Thus from (3) and (4), you get the rephrase of the question asked as is \(x \leq 5\) ?
The “3 golden nuggets” of MBA admission process With ten years of experience helping prospective students with MBA admissions and career progression, I will be writing this blog through...
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...