Events & Promotions
|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 0308:30 AM PDT
-09:30 AM PDT
In Episode 7 of our GMAT Ninja CR series, we are rounding up the oddballs, the misfits, and the format-benders: EXCEPT, Fill-In-The-Blanks, and other unusual Critical Reasoning question types. When you see a question that ends with a literal blank line
May 2910:00 AM IST
-11:00 PM IST
Start your journey with a fully customized action plan and work with a dedicated mentor to achieve a 735+ score.- Jun 04
08:30 AM PDT
-09:30 AM PDT
For most test takers, Data Insights is the most challenging section on the GMAT, with test takers scoring several points lower on average on DI than on Quant or Verbal and completing the section with less time to spare. - Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
01 Feb 2012, 04:23
Kudos
Bookmarks
Solve the inequality: sqrt(2x+3) > x
link to the source (example 5 in https://openlearn.open.ac.uk/mod/oucontent/view.php?id=398420§ion=2.2)
I have done it as follows:
Leg1:
=> sqrt(2x+3) > x
=> 2x+3 > x^2
=> x^2 -2x -3 -1 2x+3 > 0
=> x > -3/2
Considering both the solution is -1< x < 3
But the answer is: [-3/2,3)
Where did I go wrong?
link to the source (example 5 in https://openlearn.open.ac.uk/mod/oucontent/view.php?id=398420§ion=2.2)
I have done it as follows:
Leg1:
=> sqrt(2x+3) > x
=> 2x+3 > x^2
=> x^2 -2x -3 -1 2x+3 > 0
=> x > -3/2
Considering both the solution is -1< x < 3
But the answer is: [-3/2,3)
Where did I go wrong?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
01 Feb 2012, 04:53
Kudos
Bookmarks
bsaikrishna
If we approach the way you propose:
Expression under the square root can not be negative: \(2x+3\geq{0}\) --> \(x\geq{-\frac{3}{2}}\). Notice that for \(-\frac{3}{2}\leq{x}<0\) LHS>0>RHS, so for this range the given inequality holds true.
Next, when you square both parts (squaring an inequality is a tricky thing) you'l get: \(x^2-2x-3<0\) --> \(-1<x<3\). Combining both ranges: \(-\frac{3}{2}\leq{0}<3\).
I'd personally find the range \(-\frac{3}{2}\leq{x}<0\) first and then consider the cases for \(x\geq{0}\). We can safely square both part of the inequality if we know that both parts of an inequality are non-negative (so our case) --> \(x^2-2x-3<0\) --> \(-1<x<3\), as we consider the range \(x\geq{0}\) then: \(0\leq{x}<{3}\). Combine two ranges: \(-\frac{3}{2}\leq{0}<3\).
GENERAL RULE:
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
\(2<4\) --> we can square both sides and write: \(2^2<4^2\);
\(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);
But if either of side is negative then raising to even power doesn't always work.
For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.
B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
\(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\);
\(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).
Hope it helps.
01 Feb 2012, 05:52
Kudos
Bookmarks
Yes, I got where I went wrong. I shouldn't square on both the sides here because 'x' can take negative values too.
So,let us say I am not going to square it both the sides.
Then for a problem like this sqrt( 2x + 3) > x
all I know is 2x+3 >=0, implies x belongs to [-3/2, infinity)
So, how do I solve it from there on?
So,let us say I am not going to square it both the sides.
Then for a problem like this sqrt( 2x + 3) > x
all I know is 2x+3 >=0, implies x belongs to [-3/2, infinity)
So, how do I solve it from there on?
