Find all School-related info fast with the new School-Specific MBA Forum

It is currently 21 Oct 2014, 02:22

Close

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

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Inequalities and Roots

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Manager
Manager
User avatar
Joined: 06 Apr 2011
Posts: 77
Location: India
GMAT 1: Q V
Followers: 0

Kudos [?]: 10 [1] , given: 22

Inequalities and Roots [#permalink] New post 09 Aug 2011, 11:56
1
This post received
KUDOS
2
This post was
BOOKMARKED
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

(√(2x+3))^2 > x^2
=> 2x + 3 > x^2
=> x^2 -2x -3 <0
=> (x+1) (x-3)
=> x = -1 or 3

Since the graph is a parabola, it attains its negative values at -1<x<3
since x ≥ 0, thus 0 ≤ x < 3 (partial solution)

So the inequality becomes, -1.5 ≤ x <3

could some show me how the parabola would look like.

BTW, here are a few more inequalities, if someone wants to try out:

1. √(3x-2) < 2x-3
2. √(2x - 5) > -4x + 3

please share your thoughts on this.
_________________

Regards,
Asher

Director
Director
User avatar
Status:
Joined: 24 Jul 2011
Posts: 557
GMAT 1: 780 Q51 V48
GRE 1: 1540 Q800 V740
Followers: 64

Kudos [?]: 288 [0], given: 11

Re: Inequalities and Roots [#permalink] New post 09 Aug 2011, 12:22
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
Attachments

parabola.jpg
parabola.jpg [ 20.1 KiB | Viewed 3123 times ]


_________________

GyanOne | http://www.GyanOne.com | +91 9899831738

Get a free detailed MBA profile evaluation

Top MBA Rankings and MBA Admissions blog


Image


Last edited by GyanOne on 09 Aug 2011, 20:08, edited 1 time in total.
Intern
Intern
avatar
Joined: 01 Aug 2011
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Inequalities and Roots [#permalink] New post 09 Aug 2011, 12:52
Nice.

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.
Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4876
Location: Pune, India
Followers: 1149

Kudos [?]: 5348 [1] , given: 165

Re: Inequalities and Roots [#permalink] New post 09 Aug 2011, 20:03
1
This post received
KUDOS
Expert's post
Asher wrote:

• √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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Expert Post
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4876
Location: Pune, India
Followers: 1149

Kudos [?]: 5348 [2] , given: 165

Re: Inequalities and Roots [#permalink] New post 09 Aug 2011, 20:05
2
This post received
KUDOS
Expert's post
Let's try one of the questions you have given.

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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager
Manager
User avatar
Joined: 06 Apr 2011
Posts: 77
Location: India
GMAT 1: Q V
Followers: 0

Kudos [?]: 10 [0], given: 22

Re: Inequalities and Roots [#permalink] New post 10 Aug 2011, 01:29
GyanOne wrote:
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.
_________________

Regards,
Asher

Manager
Manager
User avatar
Joined: 06 Apr 2011
Posts: 77
Location: India
GMAT 1: Q V
Followers: 0

Kudos [?]: 10 [0], given: 22

Re: Inequalities and Roots [#permalink] New post 10 Aug 2011, 01:38
VeritasPrepKarishma wrote:
Asher wrote:

• √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.
_________________

Regards,
Asher

Manager
Manager
User avatar
Joined: 06 Apr 2011
Posts: 77
Location: India
GMAT 1: Q V
Followers: 0

Kudos [?]: 10 [0], given: 22

Re: Inequalities and Roots [#permalink] New post 10 Aug 2011, 02:33
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

If, -4x + 3 < 0
x > 3/4 ----- (1)

If, -4x + 3 > 0
x < 3/4 ------(2)

How can the left side be both x> 3/4 and x< 3/4

For, -4x + 3 > 0 we get x < 3/4 (can we get rid of this ans right away since LHS with sq. root is x ≥ 5/2 ),

and only compare RHS x > 3/4 and LHS x ≥ 5/2 to conclude that the ans. is x ≥ 5/2


Anyways, If we square both side when both sides for ≥ 0, we get no real roots.

(√(2x - 5))^2 > (-4x + 3)^2
2x - 5 > 16x^2 - 24x + 9
0> 16x^2 - 26x + 14
0> 8x^2 - 13x + 7
Here b^2 - 4ac = 169 - 224 = - 55 (no roots)
_________________

Regards,
Asher

1 KUDOS received
Manager
Manager
User avatar
Status: On...
Joined: 16 Jan 2011
Posts: 189
Followers: 3

Kudos [?]: 35 [1] , given: 62

Re: Inequalities and Roots [#permalink] New post 10 Aug 2011, 16:45
1
This post received
KUDOS
1
This post was
BOOKMARKED
Asher wrote:
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

If, -4x + 3 < 0
x > 3/4 ----- (1)

If, -4x + 3 > 0
x < 3/4 ------(2)

How can the left side be both x> 3/4 and x< 3/4

For, -4x + 3 > 0 we get x < 3/4 (can we get rid of this ans right away since LHS with sq. root is x ≥ 5/2 ),

and only compare RHS x > 3/4 and LHS x ≥ 5/2 to conclude that the ans. is x ≥ 5/2


Anyways, If we square both side when both sides for ≥ 0, we get no real roots.

(√(2x - 5))^2 > (-4x + 3)^2
2x - 5 > 16x^2 - 24x + 9
0> 16x^2 - 26x + 14
0> 8x^2 - 13x + 7
Here b^2 - 4ac = 169 - 224 = - 55 (no roots)


√(2x - 5) > -4x + 3

Since sq.root is always positive, thus:

√(2x - 5) ≥ 0
2x - 5 ≥ 0
x ≥ 5/2 ------------------------------(1)

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
kudos-what-are-they-and-why-we-have-them-94812.html

Manager
Manager
User avatar
Joined: 06 Apr 2011
Posts: 77
Location: India
GMAT 1: Q V
Followers: 0

Kudos [?]: 10 [0], given: 22

Re: Inequalities and Roots [#permalink] New post 11 Aug 2011, 19:15
Quote:
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


Thanks krishp84.
_________________

Regards,
Asher

Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4876
Location: Pune, India
Followers: 1149

Kudos [?]: 5348 [0], given: 165

Re: Inequalities and Roots [#permalink] New post 11 Aug 2011, 22:22
Expert's post
Asher wrote:
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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager
Manager
User avatar
Joined: 06 Apr 2011
Posts: 77
Location: India
GMAT 1: Q V
Followers: 0

Kudos [?]: 10 [0], given: 22

Re: Inequalities and Roots [#permalink] New post 13 Aug 2011, 02:18
Now i get it. :)

Quote:
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.
_________________

Regards,
Asher

Senior Manager
Senior Manager
User avatar
Joined: 23 Mar 2011
Posts: 474
Location: India
GPA: 2.5
WE: Operations (Hospitality and Tourism)
Followers: 11

Kudos [?]: 110 [0], given: 59

Re: Inequalities and Roots [#permalink] New post 15 Jan 2012, 04:28
great explanations all. i got some good insights in an area i hv been struggling on (apart from sequences :cry: )
_________________

"When the going gets tough, the tough gets going!"

Bring ON SOME KUDOS MATES+++



-----------------------------
Quant Notes consolidated: consolodited-quant-guides-of-forum-most-helpful-in-preps-151067.html#p1217652

My GMAT journey begins: my-gmat-journey-begins-122251.html

All about Richard Ivey: all-about-richard-ivey-148594.html#p1190518

Manager
Manager
avatar
Joined: 27 Dec 2011
Posts: 52
Followers: 0

Kudos [?]: 3 [0], given: 11

Re: Inequalities and Roots [#permalink] New post 30 Apr 2012, 00:19
hi All,

:? Might sound lame but what is the question being asked here:

Let's try one of the questions you have given.

1. √(3x-2) < 2x-3

Whats the question here? what are we trying to solve for? Is a P.S question or D.s question?

thanks,
Intern
Intern
avatar
Joined: 12 Mar 2013
Posts: 12
Followers: 0

Kudos [?]: 0 [0], given: 14

Re: Inequalities and Roots [#permalink] New post 12 Mar 2013, 08:58
Hi Karishma,

Thanks for the great explanation. However, I am stuck in solving the below question:

sqroot(11-5x)>x-1

Could you please help?
Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4876
Location: Pune, India
Followers: 1149

Kudos [?]: 5348 [0], given: 165

Re: Inequalities and Roots [#permalink] New post 12 Mar 2013, 20:08
Expert's post
keenys wrote:
sqroot(11-5x)>x-1



\sqrt{11 - 5x} > x - 1

Terms inside square roots are never negative so,
11 - 5x >= 0
x <= 11/5
x <= 2.2

You need to take two cases because the right hand side can be positive or negative.

If (x - 1) < 0
x < 1
In that case, the inequality will always hold because left hand side will be non-negative.

If (x - 1) >= 0, x >= 1
Square both sides of \sqrt{11 - 5x} > x - 1
11 - 5x > x^2 + 1 - 2x
x^2 + 3x - 10 < 0
(x + 5) (x - 2) < 0
-5 < x < 2
Since x >= 1, 1 <= x < 2

So, either x should be less than 1 or between 1 (inclusive) and 2. Hence, whenever x < 2, the inequality will hold.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Senior Manager
Senior Manager
avatar
Joined: 12 Mar 2010
Posts: 385
Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34
Followers: 2

Kudos [?]: 31 [0], given: 87

Re: Inequalities and Roots [#permalink] New post 23 Jul 2013, 04:54
Could somebody help me with the below problems?

1. x+2 < sqrt (x+14)

2. x-1 < sqrt(7-x)
Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4876
Location: Pune, India
Followers: 1149

Kudos [?]: 5348 [0], given: 165

Re: Inequalities and Roots [#permalink] New post 23 Jul 2013, 21:26
Expert's post
gmatter0913 wrote:
Could somebody help me with the below problems?

1. x+2 < sqrt (x+14)

2. x-1 < sqrt(7-x)


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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Senior Manager
Senior Manager
avatar
Joined: 12 Mar 2010
Posts: 385
Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34
Followers: 2

Kudos [?]: 31 [0], given: 87

Re: Inequalities and Roots [#permalink] New post 24 Jul 2013, 00:53
In one of your earlier posts, you mentioned

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.

Could you help me understand this please?
Senior Manager
Senior Manager
avatar
Joined: 12 Mar 2010
Posts: 385
Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34
Followers: 2

Kudos [?]: 31 [0], given: 87

Re: Inequalities and Roots [#permalink] New post 25 Jul 2013, 10:39
Hi Karishma,

I tried the problem x-1 < sqrt (7-x) as below:

As 7-x is under sqrt, it is +ve. Therefore, 7-x>=0 ; x<=7 ----->(1)

x-1 can be -ve or +ve

When x-1<=0; x<=1 --------> (2)

When x-1>=0; x>=1 --------> (3)

As both sides are +ve, we can square both the sides

(x-1)^2 < 7-x
x^2 -x -6<0
(x-3)(x+2)<0

-2<x<3 ------------>(4)

The answer to this problem is (x<3). I am not sure how to arrive at that from hereon. Could you please help me?
Re: Inequalities and Roots   [#permalink] 25 Jul 2013, 10:39
    Similar topics Author Replies Last post
Similar
Topics:
inequality, Ravshonbek 11 09 Sep 2007, 12:18
Inequalities Hermione 3 09 Dec 2006, 04:28
inequalities ugo_castelo 1 06 Aug 2006, 07:23
inequalities mand-y 3 05 Dec 2005, 11:18
inequalities mandy 2 15 Jun 2005, 06:47
Display posts from previous: Sort by

Inequalities and Roots

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2    Next  [ 24 posts ] 



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.