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:

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

We have to analyze the two cases so 1)Range: \(x+10>0, x>-10\) \(x+10=2x+8\) \(x=2\) \(-2\) is in the range we are analyzing (>-10) so -2 is a solution

2)Range: \(x+10<0,x<-10\) \(-x-10=2x+8\) \(x=-6\) \(-6\) is OUT of the range we are considering (<-10) so is NOT a possible solution

Let me know if it's clear
_________________

It is beyond a doubt that all our knowledge that begins with experience.

My way: standard procedure: 1st: \(x+10=2x+8\) \(x=2\) OR 2nd:\(-(x+10)=2x+8\) \(x=-6\) the condition \(x=i\) is met: Insufficient.

This approach is NOT correct.

Always be sure to plug back the solutions for equations with modulus sign the values that you have found are x = 2 and x = -6 ONLY x=2 satisfies the given condition For X = -6

LHS = |x+10| = 4 RHS = 2x + 8 = -4 SO x = -6 is NO GOOD

This is happening because you have found the roots of \((x+10)^2\) = \((2x+8)^2\)
_________________

This could ALSO be solved YOUR way but you have missed the definition of the modulus sign |x|= x WHEN x >or = 0 |x| = -x WHEN x<0

Now lets come back to the problem,

|x+10| = 2x+8

SO x+10 = 2x+8 when x+10> 0 So it gives x=2 Now lets check: is 2+10 > 0 YES. So this is GOOD Next -(x+10) = 2x+8 when x+10<0 it gives x=-6 Lets check: is -6+10 < 0 NO. So this is what you call an extraneous root. Does no good.
_________________

the sums in parentheses must have opposite signs, so:

\(\frac{3}{2}<x<3\)

consider the condition \(x=i\):

\(x=2\) Sufficient.

2) is not clear:

\(|x+10|=2x+8\)

My way: standard procedure: 1st: \(x+10=2x+8\) \(x=2\) OR 2nd:\(-(x+10)=2x+8\) \(x=-6\) the condition \(x=i\) is met: Insufficient.

The original explanation: (2) \(|x+10|=2x+8\). The left hand side (LHS) is an absolute value, which is always non-negative, hence RHS must also be non-negative: \(2x+8\geq0\) giving us \(x\geq-4\). Now, for this range \(x+10\) is positive, hence\(|x+10|=x+10\). So, \(|x+10|=2x+8\) can be written as \(x+10=2x+8\), solving for \(x\) gives \(x=2\). Sufficient.

Noting that \(2x+8\geq0\) excludes the negative value and leaves off only one value. But what the heck is the mechanics behind this problem that makes the the good old way of solving inequalities insufficiently precise here?

Merging similar topics.

The way you call "standard procedure" is not complete.

When expanding \(|x+10|=2x+8\) you should consider x<-10 range and x>=-10 range:

\(x<-10\) --> \(-(x+10)=2x+8\) --> \(x=-6\). Discard this solution since it's not in the range \(x<-10\). \(x\geq{-10}\) --> \(x+10=2x+8\) --> \(x=2\).

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Re: What is the value of integer x ? I- 2x^2 + 9 < 9x II- |x+10 [#permalink]

Show Tags

12 Jun 2014, 10:59

What is the value of integer x ?

I- 2x^2 + 9 < 9x II- |x+10| = 2x+8

I took some time but solved it my way.

IMO D.

Stmt I. we can make equation \(2x^2 -9x + 9 < 0\)

solve for X = 1.5, 3 -- 2 is the only integer in between satisfies the relation.

Sufficient.

II. Square both side. \((|x+10|) ^2=(2x+8)^2\)

\(x^2 + 4x - 12 = 0\)

solve for X = 2, -6 -- substitute values back to original equation, only 2 satisfies the relation.

Sufficient.
_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Re: What is the value of integer x? (M27-14) [#permalink]

Show Tags

29 Mar 2016, 04:04

Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...