It is currently 21 Oct 2017, 18:02

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Inequalities and Roots

Author Message
TAGS:

### Hide Tags

Manager
Joined: 06 Apr 2011
Posts: 77

Kudos [?]: 15 [1], given: 22

Location: India

### Show Tags

09 Aug 2011, 12:56
1
KUDOS
5
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

_________________

Regards,
Asher

Kudos [?]: 15 [1], given: 22

VP
Joined: 24 Jul 2011
Posts: 1347

Kudos [?]: 643 [0], given: 20

GMAT 1: 780 Q51 V48
GRE 1: 1540 Q800 V740

### Show Tags

09 Aug 2011, 13:22
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.

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 [ 20.1 KiB | Viewed 6048 times ]

_________________

GyanOne | Top MBA Rankings and MBA Admissions Blog

Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching

Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738

Last edited by GyanOne on 09 Aug 2011, 21:08, edited 1 time in total.

Kudos [?]: 643 [0], given: 20

Intern
Joined: 01 Aug 2011
Posts: 6

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

### Show Tags

09 Aug 2011, 13:52
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.

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

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7676

Kudos [?]: 17381 [2], given: 232

Location: Pune, India

### Show Tags

09 Aug 2011, 21:03
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
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

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17381 [2], given: 232 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7676 Kudos [?]: 17381 [4], given: 232 Location: Pune, India Re: Inequalities and Roots [#permalink] ### Show Tags 09 Aug 2011, 21:05 4 This post received KUDOS Expert's post 2 This post was BOOKMARKED 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 Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17381 [4], given: 232

Manager
Joined: 06 Apr 2011
Posts: 77

Kudos [?]: 15 [0], given: 22

Location: India

### Show Tags

10 Aug 2011, 02:29
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.

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

Kudos [?]: 15 [0], given: 22

Manager
Joined: 06 Apr 2011
Posts: 77

Kudos [?]: 15 [0], given: 22

Location: India

### Show Tags

10 Aug 2011, 02:38
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

Kudos [?]: 15 [0], given: 22

Manager
Joined: 06 Apr 2011
Posts: 77

Kudos [?]: 15 [0], given: 22

Location: India

### Show Tags

10 Aug 2011, 03:33
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

Kudos [?]: 15 [0], given: 22

Manager
Status: On...
Joined: 16 Jan 2011
Posts: 184

Kudos [?]: 70 [1], given: 62

### Show Tags

10 Aug 2011, 17:45
1
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
http://gmatclub.com/forum/kudos-what-are-they-and-why-we-have-them-94812.html

Kudos [?]: 70 [1], given: 62

Manager
Joined: 06 Apr 2011
Posts: 77

Kudos [?]: 15 [0], given: 22

Location: India

### Show Tags

11 Aug 2011, 20:15
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

Kudos [?]: 15 [0], given: 22

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7676

Kudos [?]: 17381 [0], given: 232

Location: Pune, India

### Show Tags

11 Aug 2011, 23:22
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

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17381 [0], given: 232 Manager Joined: 06 Apr 2011 Posts: 77 Kudos [?]: 15 [0], given: 22 Location: India Re: Inequalities and Roots [#permalink] ### Show Tags 13 Aug 2011, 03:18 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 Kudos [?]: 15 [0], given: 22 Senior Manager Joined: 23 Mar 2011 Posts: 461 Kudos [?]: 280 [0], given: 59 Location: India GPA: 2.5 WE: Operations (Hospitality and Tourism) Re: Inequalities and Roots [#permalink] ### Show Tags 15 Jan 2012, 05:28 great explanations all. i got some good insights in an area i hv been struggling on (apart from sequences ) _________________ "When the going gets tough, the tough gets going!" Bring ON SOME KUDOS MATES+++ ----------------------------- Quant Notes consolidated: http://gmatclub.com/forum/consolodited-quant-guides-of-forum-most-helpful-in-preps-151067.html#p1217652 My GMAT journey begins: http://gmatclub.com/forum/my-gmat-journey-begins-122251.html All about Richard Ivey: http://gmatclub.com/forum/all-about-richard-ivey-148594.html#p1190518 Kudos [?]: 280 [0], given: 59 Manager Joined: 27 Dec 2011 Posts: 68 Kudos [?]: 22 [0], given: 12 Re: Inequalities and Roots [#permalink] ### Show Tags 30 Apr 2012, 01:19 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, Kudos [?]: 22 [0], given: 12 Intern Joined: 12 Mar 2013 Posts: 14 Kudos [?]: [0], given: 14 Re: Inequalities and Roots [#permalink] ### Show Tags 12 Mar 2013, 09:58 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? Kudos [?]: [0], given: 14 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7676 Kudos [?]: 17381 [0], given: 232 Location: Pune, India Re: Inequalities and Roots [#permalink] ### Show Tags 12 Mar 2013, 21:08 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 Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17381 [0], given: 232

Senior Manager
Joined: 12 Mar 2010
Posts: 357

Kudos [?]: 270 [0], given: 87

Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34

### Show Tags

23 Jul 2013, 05:54
Could somebody help me with the below problems?

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

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

Kudos [?]: 270 [0], given: 87

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7676

Kudos [?]: 17381 [1], given: 232

Location: Pune, India

### Show Tags

23 Jul 2013, 22:26
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
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

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17381 [1], given: 232

Senior Manager
Joined: 12 Mar 2010
Posts: 357

Kudos [?]: 270 [0], given: 87

Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34

### Show Tags

24 Jul 2013, 01:53
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?

Kudos [?]: 270 [0], given: 87

Senior Manager
Joined: 12 Mar 2010
Posts: 357

Kudos [?]: 270 [0], given: 87

Concentration: Marketing, Entrepreneurship
GMAT 1: 680 Q49 V34

### Show Tags

25 Jul 2013, 11:39
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?

Kudos [?]: 270 [0], given: 87

Re: Inequalities and Roots   [#permalink] 25 Jul 2013, 11:39

Go to page    1   2    Next  [ 26 posts ]

Display posts from previous: Sort by