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:

Statement (2) by itself is insufficient. From Statement (2): \(x^2 - x^3 = 0\) \(x^2 (1 - x) = 0\)

\(x = 0\) or \(x =1\).

Answer: A

Hi Bunuel

I have a problem with your analysis of statement (1). x=1 does not satisfy the given equation, x = (2)^0.5 *x -1.

Squaring the given equation will not give 'x^2 = 2x - 1' although it will give you a unique value of x that is real. Please check.

Thanks

Not sure what exactly are you doing there...

x=1 satisfies x^2 = 2x - 1: 1^1 = 2*1 - 1.

Bunuel,

If x^2 =1 , then x = 1 or -1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x -1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x - 1".

How did you get from x = (2)^0.5 *x -1 to x^2 = 2x - 1?

Square \(x = \sqrt{2x - 1}\) to get \(x^2 = 2x - 1\).

Bunuel

The way question is getting displayed, it seems that square root is with 2 only and not with the entire thing (x-1). Please re format the question so that the root is displayed as per your comment.

If x^2 =1 , then x = 1 or -1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x -1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x - 1".

You are NOT correct.

If you square \(x = \sqrt{2x - 1}\) you get \(x^2 = 2x - 1\). Next, if you substitute x=1 into \(x = \sqrt{2x - 1}\) the equation will hold true.
_________________

If x^2 =1 , then x = 1 or -1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x -1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x - 1".

You are NOT correct.

If you square \(x = \sqrt{2x - 1}\) you get \(x^2 = 2x - 1\). Next, if you substitute x=1 into \(x = \sqrt{2x - 1}\) the equation will hold true.

I think the confusion is between 2^0.5*x-1 versus (2x-1)^(0.5), in the 2nd case the problem and solution as stated is correct...

If x^2 =1 , then x = 1 or -1. Both these values do not satisfy the given equation in statement (1). When you square a linear equation, you need to check whether the solution(s) satisfy the given equation. If I have to treat this as a PS problem with only statement (1) as the question statement, then x^2=1 will not satisfy the original equation x = (2)^0.5 *x -1. Additionally, when you square this (statement 1) equation, you can not get "x^2 = 2x - 1".

You are NOT correct.

If you square \(x = \sqrt{2x - 1}\) you get \(x^2 = 2x - 1\). Next, if you substitute x=1 into \(x = \sqrt{2x - 1}\) the equation will hold true.

I think the confusion is between 2^0.5*x-1 versus (2x-1)^(0.5), in the 2nd case the problem and solution as stated is correct...

Yes. Exactly. I was checking for X=1 with the square root with 2 only and not the entire expression. It is not clear from the way the question is getting displayed.

hi Bunuel can you please solve my query? if x^2= x^3, does not it imply that X is Positive? then that would eliminate 0 as a possibility, making statement 2 sufficient. please tell me where i am going wrong.

hi Bunuel can you please solve my query? if x^2= x^3, does not it imply that X is Positive? then that would eliminate 0 as a possibility, making statement 2 sufficient. please tell me where i am going wrong.

I think this is a high-quality question and I agree with explanation. Why can't I divide both sides of statement 2 by X^2?? This would take me to X=1 and make the statement sufficient.

I think this is a high-quality question and I agree with explanation. Why can't I divide both sides of statement 2 by X^2?? This would take me to X=1 and make the statement sufficient.

You can not divide by x^2 as you don't know whether x=0 or not. Statement 2 is also satisfied by x=0. Be careful with dividing by variables when you either don't know the sign or the values.

Dividing by x^2 implies that you are assuming x can't be =0

hi Bunuel can you please solve my query? if x^2= x^3, does not it imply that X is Positive? then that would eliminate 0 as a possibility, making statement 2 sufficient. please tell me where i am going wrong.

x^2 = x^3;

x^2 - x^3 = 0;

x^2(1 - x) = 0;

x = 0 or x = 1.

Hi Bunuel,

x^2 = x^3.

why can't we just divide by x^2 to get x = 1?

What am I missing here? I see how both 0 and 1 are solutions but what theorem am I missing? I feel like it is basic and I may just be forgetting.