It is currently 20 Nov 2017, 12:54

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

M12 Q17

Author Message
Director
Joined: 27 May 2008
Posts: 541

Kudos [?]: 369 [0], given: 0

Show Tags

09 Aug 2008, 07:21
4
This post was
BOOKMARKED
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

Kudos [?]: 369 [0], given: 0

GMAT Tutor
Joined: 24 Jun 2008
Posts: 1340

Kudos [?]: 2004 [10], given: 6

Show Tags

12 Aug 2008, 09:43
10
KUDOS
Expert's post
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.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Kudos [?]: 2004 [10], given: 6

Math Expert
Joined: 02 Sep 2009
Posts: 42270

Kudos [?]: 132811 [4], given: 12378

Show Tags

25 Feb 2010, 12:40
4
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.
_________________

Kudos [?]: 132811 [4], given: 12378

Director
Joined: 10 Sep 2007
Posts: 933

Kudos [?]: 355 [2], given: 0

Show Tags

09 Aug 2008, 09:09
2
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.

Kudos [?]: 355 [2], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 42270

Kudos [?]: 132811 [2], given: 12378

Show Tags

25 Feb 2010, 11:11
2
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.
_________________

Kudos [?]: 132811 [2], given: 12378

Director
Joined: 27 May 2008
Posts: 541

Kudos [?]: 369 [1], given: 0

Show Tags

09 Aug 2008, 19:15
1
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.

thanks Abhijit,
technically the language of the question is worng, should be changed to
"How many real roots does this equation have"

Kudos [?]: 369 [1], given: 0

SVP
Joined: 07 Nov 2007
Posts: 1790

Kudos [?]: 1088 [1], given: 5

Location: New York

Show Tags

14 Aug 2008, 13:09
1
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.
_________________

Smiling wins more friends than frowning

Kudos [?]: 1088 [1], given: 5

Manager
Joined: 09 Dec 2009
Posts: 120

Kudos [?]: 55 [1], given: 19

Show Tags

25 Feb 2010, 14:15
1
KUDOS
TheSituation wrote:
Hope it's clear

Abudantly clear...thanks. +1

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

Kudos [?]: 55 [1], given: 19

Math Expert
Joined: 02 Sep 2009
Posts: 42270

Kudos [?]: 132811 [1], given: 12378

Show Tags

25 Feb 2010, 15:05
1
KUDOS
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.
_________________

Kudos [?]: 132811 [1], given: 12378

Intern
Joined: 16 Feb 2010
Posts: 4

Kudos [?]: 8 [1], given: 2

Show Tags

24 Mar 2010, 04:55
1
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

Kudos [?]: 8 [1], given: 2

Senior Manager
Joined: 01 Nov 2010
Posts: 282

Kudos [?]: 87 [1], given: 44

Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)

Show Tags

03 Mar 2011, 22:52
1
KUDOS
u don't need to calculate anything.

clearly, x^2 >= 0.
sqrt 2 = 1.414
so, SQRT(x^2+1) + SQRT(x^2+2) = 2,

it doesn't have any solution beacuse left hand side is always greater than 2.
the min value of left hand side is 2.414. which is greater than 2.
_________________

kudos me if you like my post.

Attitude determine everything.
all the best and God bless you.

Kudos [?]: 87 [1], given: 44

Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 222

Kudos [?]: 64 [1], given: 16

Schools: Johnson '15

Show Tags

26 Apr 2012, 05:06
1
KUDOS
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.

this is the kind of questions i had got in real GMAT and i was stumped...took 5 mins, yet no correct answer.. .now i am slowly understanding how to answer these kinds
_________________

Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Satyameva Jayate - Truth alone triumphs

Kudos [?]: 64 [1], given: 16

Intern
Joined: 10 Aug 2012
Posts: 19

Kudos [?]: 6 [1], given: 15

Location: India
Concentration: General Management, Technology
GPA: 3.96

Show Tags

06 Mar 2013, 22:48
1
KUDOS
Given equation is

$$\sqrt{(x^2 +1)} + \sqrt{(x^2 +2)} = 2$$

As we know that $$x^2$$ is greater than zero for all values of $$x$$

So the least possible value of LHS of equation will be when$$x = 0$$

At $$x = 0$$, LHS = $$\sqrt{(0 +1)} + \sqrt{(0 +2)} >$$ RHS $$(2)$$

So equation does not have any real root.....

Kudos [?]: 6 [1], given: 15

CIO
Joined: 02 Oct 2007
Posts: 1216

Kudos [?]: 987 [0], given: 334

Show Tags

21 Aug 2008, 02:42
Thanks guys. We'll change the wording of the question.

+1 for everyone.
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
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.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 987 [0], given: 334

Intern
Joined: 16 Feb 2010
Posts: 4

Kudos [?]: 8 [0], given: 2

Show Tags

25 Feb 2010, 07:39
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.

But even answer is right, A

Kudos [?]: 8 [0], given: 2

Manager
Joined: 18 Jul 2009
Posts: 51

Kudos [?]: 143 [0], given: 7

Show Tags

25 Feb 2010, 10:42
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.

Kudos [?]: 143 [0], given: 7

Manager
Joined: 09 Dec 2009
Posts: 120

Kudos [?]: 55 [0], given: 19

Show Tags

25 Feb 2010, 11:58
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 2: 07/10/10 - This time it's personal.

Kudos [?]: 55 [0], given: 19

Manager
Joined: 14 Aug 2009
Posts: 50

Kudos [?]: 22 [0], given: 29

Show Tags

25 Feb 2010, 14:12
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

Kudos [?]: 22 [0], given: 29

Manager
Joined: 09 Dec 2009
Posts: 120

Kudos [?]: 55 [0], given: 19

Show Tags

25 Feb 2010, 14:18
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 2: 07/10/10 - This time it's personal.

Kudos [?]: 55 [0], given: 19

Manager
Joined: 18 Jul 2009
Posts: 51

Kudos [?]: 143 [0], given: 7

Show Tags

25 Feb 2010, 14:21
RHS means Right Hand Side and LHS means Left Hand Side of the given equation.

here RHS is 2 and LHS is $$\sqrt{{x^2+1}} + \sqrt{{x^2+2}}$$

Kudos [?]: 143 [0], given: 7

Re: M12 Q17   [#permalink] 25 Feb 2010, 14:21

Go to page    1   2    Next  [ 30 posts ]

Display posts from previous: Sort by

M12 Q17

Moderator: Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne 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®.