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

It is currently 27 Aug 2014, 11:04

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

M12 Q17

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Director
Director
avatar
Joined: 27 May 2008
Posts: 552
Followers: 5

Kudos [?]: 171 [0], given: 0

GMAT Tests User
M12 Q17 [#permalink] New post 09 Aug 2008, 06:21
3
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

53% (01:38) correct 47% (00:55) wrong based on 137 sessions
How many roots does this equation have?

sqrt(x^2+1) + sqrt(x^2+2) = 2

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions
[Reveal] Spoiler: OA
Kaplan Promo CodeKnewton GMAT Discount CodesGMAT Pill GMAT Discount Codes
2 KUDOS received
Director
Director
avatar
Joined: 10 Sep 2007
Posts: 950
Followers: 7

Kudos [?]: 185 [2] , given: 0

GMAT Tests User
Re: M12 Q17 [#permalink] New post 09 Aug 2008, 08:09
2
This post received
KUDOS
Square original equation
x^2+1 + x^2+2 + 2sqrt[(x^2+1)(x^2+2)] = 4
=> 2sqrt[(x^2+1)(x^2+2)] = 1 - 2x^2

Square again
4(x^4+3x^2+2) = 1 + 4x^4 - 4x^2
=>8x^2 = -7
=> x^2 = -7/8

As x^2 is -ve, so this equation cannot have solution for any real value of x. Although complex numbers solutions can be obtained, these are not part of regular GMAT.

Answer A.
1 KUDOS received
Director
Director
avatar
Joined: 27 May 2008
Posts: 552
Followers: 5

Kudos [?]: 171 [1] , given: 0

GMAT Tests User
Re: M12 Q17 [#permalink] New post 09 Aug 2008, 18:15
1
This post received
KUDOS
abhijit_sen wrote:
Square original equation
x^2+1 + x^2+2 + 2sqrt[(x^2+1)(x^2+2)] = 4
=> 2sqrt[(x^2+1)(x^2+2)] = 1 - 2x^2

Square again
4(x^4+3x^2+2) = 1 + 4x^4 - 4x^2
=>8x^2 = -7
=> x^2 = -7/8

As x^2 is -ve, so this equation cannot have solution for any real value of x. Although complex numbers solutions can be obtained, these are not part of regular GMAT.

Answer A.


thanks Abhijit,
technically the language of the question is worng, should be changed to
"How many real roots does this equation have"
9 KUDOS received
GMAT Instructor
avatar
Joined: 24 Jun 2008
Posts: 967
Location: Toronto
Followers: 253

Kudos [?]: 658 [9] , given: 3

GMAT Tests User
Re: M12 Q17 [#permalink] New post 12 Aug 2008, 08:43
9
This post received
KUDOS
durgesh79 wrote:
How many roots does this equation have?

sqrt(x^2+1) + sqrt(x^2+2) = 2

0
1
2
3
4


I don't think we need to do any algebra here: x^2 \geq 0, so \sqrt{x^2+1} + \sqrt{x^2+2} \geq 1 + \sqrt{2}, which is larger than 2. Hence no (real) solutions for x.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

1 KUDOS received
SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1829
Location: New York
Followers: 26

Kudos [?]: 441 [1] , given: 5

GMAT Tests User
Re: M12 Q17 [#permalink] New post 14 Aug 2008, 12:09
1
This post received
KUDOS
IanStewart wrote:
durgesh79 wrote:
How many roots does this equation have?

sqrt(x^2+1) + sqrt(x^2+2) = 2

0
1
2
3
4


I don't think we need to do any algebra here: x^2 \geq 0, so \sqrt{x^2+1} + \sqrt{x^2+2} \geq 1 + \sqrt{2}, which is larger than 2. Hence no (real) solutions for x.


Agreed .. good point.
+1 for you.

Durgesh,
Question wording need to be changed to real roots to make it clear.
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

CIO
CIO
avatar
Joined: 02 Oct 2007
Posts: 1218
Followers: 87

Kudos [?]: 656 [0], given: 334

GMAT ToolKit User GMAT Tests User
Re: M12 Q17 [#permalink] New post 21 Aug 2008, 01:42
Thanks guys. We'll change the wording of the question.

+1 for everyone.
_________________

Welcome to GMAT Club! :)
Facebook TwitterGoogle+LinkedIn
Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Please read this before posting in GMAT Club Tests forum
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
User avatar
Joined: 16 Feb 2010
Posts: 3
Followers: 0

Kudos [?]: 8 [0], given: 1

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 06:39
Quote:
Square original equation
x^2+1 + x^2+2 + 2sqrt[(x^2+1)(x^2+2)] = 4
=> 2sqrt[(x^2+1)(x^2+2)] = 1 - 2x^2

Square again
4(x^4+3x^2+2) = 1 + 4x^4 - 4x^2
=>8x^2 = -7 should be 16x^2=-7
=> x^2 = -7/8 then x^2=-7/16

As x^2 is -ve, so this equation cannot have solution for any real value of x. Although complex numbers solutions can be obtained, these are not part of regular GMAT.

Answer A.


But even answer is right, A :)
Manager
Manager
avatar
Joined: 18 Jul 2009
Posts: 54
Followers: 2

Kudos [?]: 38 [0], given: 7

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 09:42
IanStewart wrote:

I don't think we need to do any algebra here: x^2 \geq 0, so \sqrt{x^2+1} + \sqrt{x^2+2} \geq 1 + \sqrt{2}, which is larger than 2. Hence no (real) solutions for x.


sorry but i didnt understand this can anyone explain.
Expert Post
2 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 19093
Followers: 3404

Kudos [?]: 24872 [2] , given: 2697

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 10:11
2
This post received
KUDOS
Expert's post
hrish88 wrote:
IanStewart wrote:

I don't think we need to do any algebra here: x^2 \geq 0, so \sqrt{x^2+1} + \sqrt{x^2+2} \geq 1 + \sqrt{2}, which is larger than 2. Hence no (real) solutions for x.


sorry but i didnt understand this can anyone explain.


x^2 \geq 0, means that the lowest value of LHS is when x=0 --> \sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt2=2.4>RHS=2, hence as the lowest possible value of LHS is still greater than RHS, the equation has no real roots.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
User avatar
Joined: 09 Dec 2009
Posts: 122
Followers: 3

Kudos [?]: 21 [0], given: 19

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 10:58
I got A but solved thru a different (less clever) method. I think I may have answered correct in spite of myself though, can someone tell me if my solution is mathmatically sound or if I arrived at the correct solution by luck.

Begin by squaring both sides

(x^2+1) + (x^2 +2) = 4
drop brackets and solve
2x^2=1
x^2=1/2
x = sq root of 1/2
therefore no real roots, therefore A

feedback?
_________________

G.T.L. - GMAT, Tanning, Laundry

Round 1: 05/12/10 handling-a-grenade-thesituation-s-official-debrief-94181.html

Round 2: 07/10/10 - This time it's personal.

Expert Post
4 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 19093
Followers: 3404

Kudos [?]: 24872 [4] , given: 2697

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 11:40
4
This post received
KUDOS
Expert's post
TheSituation wrote:
I got A but solved thru a different (less clever) method. I think I may have answered correct in spite of myself though, can someone tell me if my solution is mathmatically sound or if I arrived at the correct solution by luck.

Begin by squaring both sides

(x^2+1) + (x^2 +2) = 4
drop brackets and solve
2x^2=1
x^2=1/2
x = sq root of 1/2
therefore no real roots, therefore A

feedback?


This would be the longer way, plus you'll need to square twice not once, as you made a mistake while squaring first time.

(a+b)^2=a^2+2ab+b^2, so when you square both sides you'll get: x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4 --> 2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2. At this point you should square again --> 4x^4+12x^2+8=1-4x^2+4x^4 --> 16x^2=-7, no real x satisfies this equation.

There is one more problem with your solution:

You've got (though incorrectly): x^2=\frac{1}{2} and then you concluded that this equation has no real roots, which is not right. This quadratic equation has TWO real roots: x=\frac{1}{\sqrt{2}} and x=-{\frac{1}{\sqrt{2}}}. Real roots doesn't mean that the roots must be integers, real roots means that roots must not be complex numbers, which I think we shouldn't even mention as GMAT deals ONLY with real numbers. For example x^2=-1 has no real roots and for GMAT it means that this equation has no roots, no need to consider complex roots and imaginary numbers.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 14 Aug 2009
Posts: 54
Followers: 1

Kudos [?]: 17 [0], given: 29

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 13:12
Hrish88,

Pardon my ignorance but what is RHS and LHS?

hrish88 wrote:
Bunuel wrote:
x^2 \geq 0, means that the lowest value of LHS is when x=0 --> \sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt2=2.4>RHS=2, hence as the lowest possible value of LHS is still greater than RHS, the equation has no real roots.

Hope it's clear.


Thanks Bunuel.

_________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own. I am the force; I can clear any obstacle before me or I can be lost in the maze. My choice; my responsibility; win or lose, only I hold the key to my destiny - Elaine Maxwell

1 KUDOS received
Manager
Manager
User avatar
Joined: 09 Dec 2009
Posts: 122
Followers: 3

Kudos [?]: 21 [1] , given: 19

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 13:15
1
This post received
KUDOS
TheSituation wrote:
Hope it's clear


Abudantly clear...thanks. +1

Your posts are incredibly helpful, thank you for taking the time.

Just curious, have you written the GMAT yet? If so, how did you score? If you say anything less than 790 I won't believe you..

Thanks again!!!
_________________

G.T.L. - GMAT, Tanning, Laundry

Round 1: 05/12/10 handling-a-grenade-thesituation-s-official-debrief-94181.html

Round 2: 07/10/10 - This time it's personal.

Manager
Manager
User avatar
Joined: 09 Dec 2009
Posts: 122
Followers: 3

Kudos [?]: 21 [0], given: 19

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 13:18
rbriscoe wrote:
Hrish88,

Pardon my ignorance but what is RHS and LHS?


Ha, finally a question I can help out on!!

RHS and LHS are referring to right-hand side and left-hand side of the equation.
_________________

G.T.L. - GMAT, Tanning, Laundry

Round 1: 05/12/10 handling-a-grenade-thesituation-s-official-debrief-94181.html

Round 2: 07/10/10 - This time it's personal.

Manager
Manager
avatar
Joined: 18 Jul 2009
Posts: 54
Followers: 2

Kudos [?]: 38 [0], given: 7

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 13:21
RHS means Right Hand Side and LHS means Left Hand Side of the given equation.

here RHS is 2 and LHS is \sqrt{{x^2+1}} + \sqrt{{x^2+2}}
Manager
Manager
avatar
Joined: 14 Aug 2009
Posts: 54
Followers: 1

Kudos [?]: 17 [0], given: 29

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 13:34
I am sooo lost on this question. I solved it the same as TheSiuation. I am do not understand your method below. Although I am sure it is pbvious to everyone but please clarify the rule, therum or postulate that you followed to solve the equation. Thanks in advacne.

Bunuel wrote:
TheSituation wrote:
I got A but solved thru a different (less clever) method. I think I may have answered correct in spite of myself though, can someone tell me if my solution is mathmatically sound or if I arrived at the correct solution by luck.

Begin by squaring both sides

(x^2+1) + (x^2 +2) = 4
drop brackets and solve
2x^2=1
x^2=1/2
x = sq root of 1/2
therefore no real roots, therefore A

feedback?


This would be the longer way, plus you'll need to square twice not once, as you made a mistake while squaring first time.

(a+b)^2=a^2+2ab+b^2, so when you square both sides you'll get: x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4 --> 2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2. At this point you should square again --> 4x^4+12x^2+8=1-4x^2+4x^4 --> 16x^2=-7, no real x satisfies this equation.

There is one more problem with your solution:

You've got (though incorrectly): x^2=\frac{1}{2} and then you concluded that this equation has no real roots, which is not right. This quadratic equation has TWO real roots: x=\frac{1}{\sqrt{2}} and x=-{\frac{1}{\sqrt{2}}}. Real roots doesn't mean that the roots must be integers, real roots means that roots must not be complex numbers, which I think we shouldn't even mention as GMAT deals ONLY with real numbers. For example x^2=-1 has no real roots and for GMAT it means that this equation has no roots, no need to consider complex roots and imaginary numbers.

Hope it's clear.

_________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own. I am the force; I can clear any obstacle before me or I can be lost in the maze. My choice; my responsibility; win or lose, only I hold the key to my destiny - Elaine Maxwell

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 19093
Followers: 3404

Kudos [?]: 24872 [0], given: 2697

Re: M12 Q17 [#permalink] New post 25 Feb 2010, 14:05
Expert's post
rbriscoe wrote:
I am sooo lost on this question. I solved it the same as TheSiuation. I am do not understand your method below. Although I am sure it is pbvious to everyone but please clarify the rule, therum or postulate that you followed to solve the equation. Thanks in advacne.

Bunuel wrote:
TheSituation wrote:
I got A but solved thru a different (less clever) method. I think I may have answered correct in spite of myself though, can someone tell me if my solution is mathmatically sound or if I arrived at the correct solution by luck.

Begin by squaring both sides

(x^2+1) + (x^2 +2) = 4
drop brackets and solve
2x^2=1
x^2=1/2
x = sq root of 1/2
therefore no real roots, therefore A

feedback?


This would be the longer way, plus you'll need to square twice not once, as you made a mistake while squaring first time.

(a+b)^2=a^2+2ab+b^2, so when you square both sides you'll get: x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2= 4 --> 2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2. At this point you should square again --> 4x^4+12x^2+8=1-4x^2+4x^4 --> 16x^2=-7, no real x satisfies this equation.

There is one more problem with your solution:

You've got (though incorrectly): x^2=\frac{1}{2} and then you concluded that this equation has no real roots, which is not right. This quadratic equation has TWO real roots: x=\frac{1}{\sqrt{2}} and x=-{\frac{1}{\sqrt{2}}}. Real roots doesn't mean that the roots must be integers, real roots means that roots must not be complex numbers, which I think we shouldn't even mention as GMAT deals ONLY with real numbers. For example x^2=-1 has no real roots and for GMAT it means that this equation has no roots, no need to consider complex roots and imaginary numbers.

Hope it's clear.


Frankly speaking I solved this question in different manner, as shown in my first post in this thread. The above solution is also valid, however it's quite time consuming.

We have: \sqrt{x^2+1}+\sqrt{x^2+2}=2. Now, if you choose to square both sides, then when squaring LHS \sqrt{x^2+1}+\sqrt{x^2+2}, you should apply the formula: (a+b)^2=a^2+2ab+b^2 --> (\sqrt{x^2+1})^2+2{(\sqrt{x^2+1})}{(\sqrt{x^2+2})}+(\sqrt{x^2+2})^2=x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2. When squaring RHS yuo'll get 2^2=4.

So you'll get: x^2+1+2\sqrt{(x^2+1)(x^2 +2)}+x^2+2=4, rearrange --> 2\sqrt{(x^2+1)(x^2 +2)}=1-2x^2. At this point you should square again --> (2\sqrt{(x^2+1)(x^2 +2)})^2=(1-2x^2)^2 --> 4(x^2+1)(x^2 +2)=1-4x^2+4x^4 --> 4x^4+12x^2+8=1-4x^2+4x^4, rearrange again, 4x^4 will cancel out --> 16x^2=-7 --> x^2=-\frac{7}{16}. Now, x^2 can not equal to negative number, which means that there doesn't exist an x to satisfies the given equation. So, no roots.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 26 Nov 2009
Posts: 178
Followers: 3

Kudos [?]: 53 [0], given: 5

GMAT ToolKit User GMAT Tests User
Re: M12 Q17 [#permalink] New post 26 Feb 2010, 23:30
at first glance i started substituting values and identified that x^2 cannot be greater than zero
if so then it wouldn't satisfy the equation

but i really don't what they meant as real roots and thanks bunuel for the explanation.if i had known that it is very to identify that if x^2<0 then it has no real solution
1 KUDOS received
Intern
Intern
User avatar
Joined: 16 Feb 2010
Posts: 3
Followers: 0

Kudos [?]: 8 [1] , given: 1

Re: M12 Q17 [#permalink] New post 24 Mar 2010, 03:55
1
This post received
KUDOS
don't think we need hard algebra,
just under the square should be x^2+1>0 or x^2+2>0
so,
x^2>-1 incorrect
x^2>-2 incorrect
thus no real roots... answer is A) 0
SVP
SVP
avatar
Joined: 16 Nov 2010
Posts: 1691
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 30

Kudos [?]: 286 [0], given: 36

GMAT Tests User Premium Member Reviews Badge
Re: M12 Q17 [#permalink] New post 02 Mar 2011, 18:59
I tried like this :

Let (x^2 + 1) = y

sqrt(y) + sqrt(y+1) = 2

y + y + 1 + 2(sqrt(y) * sqrt(y+1)) = 4


2y + 2(sqrt(y) * sqrt(y+1)) = 3

=> 4 * y * (y+1) = 4y^2 + 9 - 12y

=> 4y^2 + 4y = 4y^2 + 9 - 12y

=> y = 9/16

=> x^2 + 1 = 9/16 , not possible so 0 roots, answer is A.
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: M12 Q17   [#permalink] 02 Mar 2011, 18:59
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic M17 Q 17 teal 2 22 Jun 2012, 04:31
m10, q17 jainanurag78 0 06 Jul 2010, 18:51
Experts publish their posts in the topic M10 Q17 sandeepnerli 11 27 May 2010, 23:33
Verbal Test 5, Q17 11MBA 0 28 Dec 2009, 09:08
16 Experts publish their posts in the topic m01 Q17 itiskavikatha 18 18 Sep 2008, 02:14
Display posts from previous: Sort by

M12 Q17

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2    Next  [ 30 posts ] 

Moderators: Bunuel, WoundedTiger



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®.