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# m12#17

Author Message
Manager
Joined: 26 Jul 2010
Posts: 102

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Location: India
Concentration: Operations, General Management
Schools: IIMA (M)
GMAT 1: 640 Q48 V29
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WE: Supply Chain Management (Military & Defense)

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18 May 2011, 19:15
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.
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lets start again

Kudos [?]: 32 [0], given: 6

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: 871

Kudos [?]: 396 [2], given: 123

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18 May 2011, 19:55
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.

Kudos [?]: 396 [2], given: 123

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129046 [1], given: 12187

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26 Jun 2012, 01:58
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.

_________________

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Intern
Joined: 25 Jun 2011
Posts: 47

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Location: Sydney

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26 Jun 2012, 01:46
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

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

Intern
Joined: 25 Jun 2011
Posts: 47

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

Location: Sydney

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26 Jun 2012, 02:08
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!

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

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129046 [0], given: 12187

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26 Jun 2012, 02:31
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?
_________________

Kudos [?]: 129046 [0], given: 12187

Manager
Status: Juggg..Jugggg Go!
Joined: 11 May 2012
Posts: 241

Kudos [?]: 54 [0], given: 239

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

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26 Jun 2012, 02:38
Bunuel Oh! that's some really sensible work! thanks! great job!
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Kudos [?]: 54 [0], given: 239

Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 509

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Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)

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14 Feb 2013, 06:34
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
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Kudos [?]: 72 [0], given: 562

VP
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Joined: 21 Sep 2012
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16 Feb 2013, 13:14
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

Kudos [?]: 648 [0], given: 70

Re: m12#17   [#permalink] 16 Feb 2013, 13:14
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# m12#17

Moderator: Bunuel

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