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:

Something interesting that i read while searching for material on how to solve inequalities with roots.

Though i would share it and also clarify a few doubts. I have highlighted the portions that were confusing, in blue

Kind of backbone for solving inequalities with roots, √x>y OR √x<y

• √x is undefined if x<0 • both sides can be squared when x≥0 and y≥0 • if √x>y is identically true if √x≥0 and y<0 what does identically true/false mean • But √x<y is identically false if y<0

e.g., √(2x+3) > x

when, √(2x+3) ≥ 0 and x<0

left side, 2x + 3 ≥ 0 => x ≥ -1.5

right side, x < 0

thus, -1.5 ≤ x <0 (partial solution)

2nd Condition, where both left and right side, ≥ 0

Identically true means true for all values of the variable.

For example: sqrt(x)>y if sqrt(x)>=0 and y<0 means that this relation holds for all values of x and y that satisfy sqrt(x)>=0 and y<0.

The graph of the parabola is attached.

Please note: The graph of ax^2+bx+c: (a) Is a parabola always (b) Opens upwards (towards +y axis) if a>0 and downwards (towards -y axis) if a<0 (c) Intersects the x axis at the roots of the equation ax^2 + bx + c = 0 (d) Will not touch the x axis if the roots are not real (if b^2 - 4ac < 0) (e) Has its lowest point at x= -b/2a

I have forgotten how to solve these. Thanks for reminding.

You can visualize very well the solution to this by plotting the graph. Not that you can do it during the actual exam, but for reference and to remember how the typical functions look like you can use //rechneronline.de/function-graphs/ to graph any function.

• √x is undefined if x<0 • both sides can be squared when x≥0 and y≥0 • if √x>y is identically true if √x≥0 and y<0 what does identically true/false mean • But √x<y is identically false if y<0

Some Explanations: 1. √x implies the positive square root of x. So √x is positive. But we do not know whether the right side of the inequality is positive or negative. Hence, we cannot square the inequality. Note: Only when both sides are positive, you can square the inequality and still retain the same relation. e.g. 2 < 3 2^2 < 3^2

But -2 < 1 4 not less than 1

Similarly -4 < -2 16 not less than 4

So before you square both sides of the inequality, always ensure that both sides are positive

2. √x>y We know that √x is positive. If y is negative, then this inequality will always hold since positive > negative.

3. √x<y We know that √x is positive. If y is negative, then this inequality will never hold since positive is never less than negative. _________________

1. √(3x-2) < 2x-3 Left hand side is positive, we know. Also, the term under the root i.e. (3x - 2) should be positive or 0. 3x - 2>= 0 i.e. x >= 2/3.

What about the right hand side? Here we can say that the right hand side will definitely be positive too since left hand side (a positive number (√3x-2)) is less than the right hand side. Hence, 2x - 3 > 0 x > 3/2

Let's square both sides now: 3x - 2 < (2x - 3)^2 4x^2 -15x + 11 > 0 4(x - 1)(x - 11/4) > 0 Since the right most region is positive, we will get: positive ... 1 ... negative ... 11/4 ... positive

So, 1 > x or x > 11/4.

But we saw above that x > 3/2 So x > 11/4 _________________

Identically true means true for all values of the variable.

For example: sqrt(x)>y if sqrt(x)>=0 and y<0 means that this relation holds for all values of x and y that satisfy sqrt(x)>=0 and y<0.

The graph of the parabola is attached.

Please note: The graph of ax^2+bx+c: (a) Is a parabola always (b) Opens upwards (towards +y axis) if a>0 and downwards (towards -y axis) if a<0 (c) Intersects the x axis at the roots of the equation ax^2 + bx + c = 0 (d) Will not touch the x axis if the roots are not real (if b^2 - 4ac < 0) (e) Has its lowest point at x= -b/2a

Thanks GyanOne for the explanation and the graph. I get it now. _________________

• √x is undefined if x<0 • both sides can be squared when x≥0 and y≥0 • if √x>y is identically true if √x≥0 and y<0 what does identically true/false mean • But √x<y is identically false if y<0

Some Explanations: 1. √x implies the positive square root of x. So √x is positive. But we do not know whether the right side of the inequality is positive or negative. Hence, we cannot square the inequality. Note: Only when both sides are positive, you can square the inequality and still retain the same relation. e.g. 2 < 3 2^2 < 3^2

But -2 < 1 4 not less than 1

Similarly -4 < -2 16 not less than 4

So before you square both sides of the inequality, always ensure that both sides are positive

2. √x>y We know that √x is positive. If y is negative, then this inequality will always hold since positive > negative.

3. √x<y We know that √x is positive. If y is negative, then this inequality will never hold since positive is never less than negative.

Thanks karishma for the explanation.

i now realize that when i read the material found on one of the math websites, i kind of just tried to mug the concept without really understanding it. By now the concept it clear. _________________

If x ≥ 5/2 then -4x <= -10 ==> -4x+3 <= -7 ---------------------(2)

Now as per your question - -4x + 3 is less than 0...This is the super set of (2), so there is no other unique solution... other than x ≥ 5/2 _________________

Labor cost for typing this post >= Labor cost for pushing the Kudos Button http://gmatclub.com/forum/kudos-what-are-they-and-why-we-have-them-94812.html

Karishma, here's me trying to solve the second problem. Let me know if i got it right.

2. √(2x - 5) > -4x + 3

Since sq.root is always positive, thus:

√(2x - 5) ≥ 0 2x - 5 ≥ 0 x ≥ 5/2

No problems till here... we know now that x must be > or equal to 5/2.

Next think, we have LHS > RHS. RHS can be negative or positive. So there are two cases possible: Case 1: -4x + 3 < 0 x > 3/4 ----- (1) Since in this case, RHS is negative and LHS will always be positive, LHS will be > than RHS. So the inequality will hold whenever x > 3/4 and x >= 5/2. So if x >= 5/2, the inequality will hold.

Case 2: -4x + 3 > 0 x < 3/4 ------(2) Now think, is it possible that x is < than 3/4 and greater than 5/2? No! So this case doesn't give any solutions.

Hence, the only solution is x >= 5/2

and yes, don't try to learn up Mathematical concepts. Try to understand them. That way, you will never forget them. _________________

and yes, don't try to learn up Mathematical concepts. Try to understand them. That way, you will never forget them.

Thanks for this advice. This makes soo much sense. Earlier i would just learn how to solve one particular problem without really understanding the concept, but then i would always make a mistake on a different problem based on the same concept.

Anyways, my POA now is to brush my basic quant skills. _________________

Use the concepts given above to try to solve these:

\(x+2 < \sqrt{(x+14)}\)

The quantity under the root must be non negative so x >= -14

Now left hand side i.e. x+2 can be positive, 0 or negative. Take two cases:

Case 1: x + 2 >= 0 x >= -2 Now both sides of the inequality are non negative so we can square it: \((x + 2)^2 < (x + 14)\) \(x^2 + 3x - 10 < 0\) \((x + 5) (x - 2) < 0\) \(-5 < x < 2\)

Since x >= -2, we get -2 <= x < 2

Case 2: x + 2 < 0 x < -2 The left hand side i.e. x + 2 is always negative in this case while right hand side is always non negative so this inequality will hold for all values in this range. -14 <= x < -2

So the overall acceptable range is -14<= x < 2 _________________

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...