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# How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =

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Manager
Joined: 17 Aug 2018
Posts: 119
Location: India
Schools: IIMA
GMAT 1: 640 Q46 V32
How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =  [#permalink]

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Updated on: 05 Jan 2020, 04:07
9
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Difficulty:

55% (hard)

Question Stats:

59% (01:55) correct 41% (01:32) wrong based on 59 sessions

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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

M12-17

Originally posted by KaranB1 on 05 Jan 2020, 01:43.
Last edited by Bunuel on 05 Jan 2020, 04:07, edited 2 times in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 62356
Re: How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =  [#permalink]

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05 Jan 2020, 04:08
2
1
KaranB1 wrote:
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

M12-17

Official Solution:

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 $$x^2 \ge 0$$ so the least value of the left hand side (LHS) of the equation is for $$x=0$$:

$$\sqrt{x^2+1} + \sqrt{x^2+2}=1 + \sqrt{2} \approx 2.4$$. 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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Math Expert
Joined: 02 Sep 2009
Posts: 62356
Re: How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =  [#permalink]

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05 Jan 2020, 04:40
1
AnirudhaS wrote:
KaranB1 wrote:
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

M12-17

I think the question should be how many "real" roots does the equation have?

All numbers on the GMAT are real by default.
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Director
Joined: 08 Aug 2017
Posts: 708
Re: How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =  [#permalink]

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05 Jan 2020, 04:14
Very nice question.

X^2>=0

So given expression can never be equal to 2, rather it will always be greater than 2.
Ans is 0. A.
Manager
Joined: 30 Oct 2019
Posts: 198
Location: United Kingdom
Concentration: General Management, Technology
GPA: 4
Re: How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =  [#permalink]

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05 Jan 2020, 04:39
KaranB1 wrote:
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

M12-17

I think the question should be how many "real" roots does the equation have?
Intern
Joined: 22 Jun 2019
Posts: 9
Re: How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =  [#permalink]

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05 Jan 2020, 17:29
Is there a way to do this question using the discriminant?
Senior Manager
Joined: 21 Jun 2017
Posts: 416
Location: India
Concentration: Finance, Economics
Schools: IIM
GMAT 1: 620 Q47 V30
GPA: 3
WE: Corporate Finance (Commercial Banking)
Re: How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =  [#permalink]

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05 Jan 2020, 23:05
letx^2 + 1 be y
sqrt(y) + sqrt(y+1) = 2
sqrt(y) = 2 - sqrt(y+1)
squaring both sides
y= 9/16 i.e. x^2 + 1 = 9/16
x^2= -7/16
square can never be negative .
Hence, no x can satisfy the equation.
Senior Manager
Joined: 21 Jun 2017
Posts: 416
Location: India
Concentration: Finance, Economics
Schools: IIM
GMAT 1: 620 Q47 V30
GPA: 3
WE: Corporate Finance (Commercial Banking)
Re: How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =  [#permalink]

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05 Jan 2020, 23:06
letx^2 + 1 be y
sqrt(y) + sqrt(y+1) = 2
sqrt(y) = 2 - sqrt(y+1)
squaring both sides
y= 9/16 i.e. x^2 + 1 = 9/16
x^2= -7/16
square can never be negative .
Hence, no x can satisfy the equation.
Re: How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) =   [#permalink] 05 Jan 2020, 23:06
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