m12#17 : Retired Discussions [Locked]
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 27 Feb 2017, 20:03

### 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#17

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Manager
Joined: 26 Jul 2010
Posts: 103
Location: India
Concentration: Operations, General Management
Schools: IIMA (M)
GMAT 1: 640 Q48 V29
GMAT 2: 670 Q49 V31
WE: Supply Chain Management (Military & Defense)
Followers: 6

Kudos [?]: 26 [0], given: 6

### Show Tags

18 May 2011, 18:15
How many real roots does this equation have?

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

a. 0
b. 1
c. 2
d. 3
e. 4

you can solve the question if you wish, my question is how do you arrive to answer to this and other these type of questions.
_________________

lets start again

Director
Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
Joined: 03 Feb 2011
Posts: 920
Followers: 14

Kudos [?]: 346 [2] , given: 123

### Show Tags

18 May 2011, 18:55
2
KUDOS
This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.
Math Expert
Joined: 02 Sep 2009
Posts: 37144
Followers: 7274

Kudos [?]: 96811 [1] , given: 10786

### Show Tags

26 Jun 2012, 00:58
1
KUDOS
Expert's post
dianamao wrote:
gmat1220 wrote:
This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.

I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help

Thanks,
Diana

How did you get: 4(x^2+1)(x^2+2)+8?

How many roots does the equation $$\sqrt{x^2+1} +\sqrt{x^2+2}= 2$$ have?
A. 0
B. 1
C. 2
D. 3
E. 4

We know that $$x^2\geq 0$$ so the least value of the left hand side (LHS) of the equation is for $$x=0$$: $$\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt2\approx{2.4}$$. Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real $$x$$ can satisfy given equation.

_________________
Intern
Joined: 25 Jun 2011
Posts: 49
Location: Sydney
Followers: 0

Kudos [?]: 1 [0], given: 7

### Show Tags

26 Jun 2012, 00:46
gmat1220 wrote:
This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.

I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help

Thanks,
Diana
Intern
Joined: 25 Jun 2011
Posts: 49
Location: Sydney
Followers: 0

Kudos [?]: 1 [0], given: 7

### Show Tags

26 Jun 2012, 01:08
Bunuel wrote:
dianamao wrote:
gmat1220 wrote:
This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.

I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help

Thanks,
Diana

How did you get: 4(x^2+1)(x^2+2)+8?

How many roots does the equation $$\sqrt{x^2+1} +\sqrt{x^2+2}= 2$$ have?
A. 0
B. 1
C. 2
D. 3
E. 4

We know that $$x^2\geq 0$$ so the least value of the left hand side (LHS) of the equation is for $$x=0$$: $$\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt2\approx{2.4}$$. Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real $$x$$ can satisfy given equation.

First, I squared both sides of the equation. So I got something like (x^2+1)+2sqrt((x^2+1)(x^2+2))+(x^2+2)=4
Then I subbed relevant numbers into the discriminant formula which turns into 4(x^2+1)(x^2+2)+8. Now here, I'm thinking that 4(x^2+1)(x^2+2)+8 is definitely greater than zero, and if discriminant is greater than 0, then there are two roots, no?

P.S. I totally get your reasoning, it's just that I don't undersand why I can't reach that same answer using the discriminant rule, unless I'm not doing something right.

Thanks heaps!
Math Expert
Joined: 02 Sep 2009
Posts: 37144
Followers: 7274

Kudos [?]: 96811 [0], given: 10786

### Show Tags

26 Jun 2012, 01:31
dianamao wrote:
First, I squared both sides of the equation. So I got something like (x^2+1)+2sqrt((x^2+1)(x^2+2))+(x^2+2)=4
Then I subbed relevant numbers into the discriminant formula which turns into 4(x^2+1)(x^2+2)+8. Now here, I'm thinking that 4(x^2+1)(x^2+2)+8 is definitely greater than zero, and if discriminant is greater than 0, then there are two roots, no?

P.S. I totally get your reasoning, it's just that I don't undersand why I can't reach that same answer using the discriminant rule, unless I'm not doing something right.

Thanks heaps!

In the above reasoning there no quadratics shown. So, what is the quadratics you find the discriminant of? How did you get this quadratics?
_________________
Senior Manager
Status: Juggg..Jugggg Go!
Joined: 11 May 2012
Posts: 254
Location: India
GC Meter: A.W.E.S.O.M.E
Concentration: Entrepreneurship, General Management
GMAT 1: 620 Q46 V30
GMAT 2: 720 Q50 V38
Followers: 6

Kudos [?]: 43 [0], given: 239

### Show Tags

26 Jun 2012, 01:38
Bunuel Oh! that's some really sensible work! thanks! great job!
_________________

You haven't failed, if you haven't given up!
---
Visit my Blog www.bschooladmit.wordpress.com

Check out my other posts:
Bschool Deadlines 2013-2014 | Bschool Admission Events 2013 Start your GMAT Prep with Stacey Koprince | Get a head start in MBA finance

Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 547
Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Followers: 3

Kudos [?]: 61 [0], given: 562

### Show Tags

14 Feb 2013, 05:34
Bunuel wrote:
dianamao wrote:
First, I squared both sides of the equation. So I got something like (x^2+1)+2sqrt((x^2+1)(x^2+2))+(x^2+2)=4
Then I subbed relevant numbers into the discriminant formula which turns into 4(x^2+1)(x^2+2)+8. Now here, I'm thinking that 4(x^2+1)(x^2+2)+8 is definitely greater than zero, and if discriminant is greater than 0, then there are two roots, no?

P.S. I totally get your reasoning, it's just that I don't undersand why I can't reach that same answer using the discriminant rule, unless I'm not doing something right.

Thanks heaps!

In the above reasoning there no quadratics shown. So, what is the quadratics you find the discriminant of? How did you get this quadratics?

bro bunuel,

I did the same way .. squared.. then squared again..

in the end I got, 16 * square(x)= - 7
and now there can never be square root of a negative number , hence no roots.. and hence zero..
hope this is correct. pls let me know
_________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595

My GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992

VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1096
Location: India
GMAT 1: 410 Q35 V11
GMAT 2: 530 Q44 V20
GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)
Followers: 38

Kudos [?]: 532 [0], given: 70

### Show Tags

16 Feb 2013, 12:14
Bunuel wrote:
dianamao wrote:
gmat1220 wrote:
This equation is insane ! LHS will never be RHS. The min value of x^2 + 1 is 1 and min value of x^2 + 2 is 2

i.e 1 + 1.414 = 2.414 is never 2. So zero roots.

I understand your logic, but say I want to use the discriminant (b^2-4ac) formula to solve this question.

I get 4(x^2+1)(x^2+2)+8 which is to be obviously greater than 0, thus the equation should have two roots. But I graphed this equation and know that the equation has no roots, so I don't understand why the discriminant formula cannot be applied here. Please help

Thanks,
Diana

How did you get: 4(x^2+1)(x^2+2)+8?

How many roots does the equation $$\sqrt{x^2+1} +\sqrt{x^2+2}= 2$$ have?
A. 0
B. 1
C. 2
D. 3
E. 4

We know that $$x^2\geq 0$$ so the least value of the left hand side (LHS) of the equation is for $$x=0$$: $$\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt2\approx{2.4}$$. Now, since even the least possible value of LHS is still greater than RHS (which is 2), then no real $$x$$ can satisfy given equation.

Bunuel, Can you pls help me with the explanation:-X^2>=0, how did you get that ...Did you assume that since RHS is a positive term hence the LHS must be equal to 0...that implies x^2 must be greater than or equal to zero.

Also , can you pls post links to some the practice questions of above type...or the method used to determine the number of roots in the above cases.

Archit
Re: m12#17   [#permalink] 16 Feb 2013, 12:14
Display posts from previous: Sort by

# m12#17

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderator: Bunuel

 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®.