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655-705 (Hard)|   Algebra|      
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KaranB1
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = Updated on: 05 Jan 2020, 05:07
Originally posted by KaranB1 on 05 Jan 2020, 02:43.
Last edited by Bunuel on 05 Jan 2020, 05:07, edited 2 times in total.
Renamed the topic and edited the question.
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A
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55% (hard)

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61% (02:03) correct 39% (01:57) wrong based on 420 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

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Bunuel
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 05 Jan 2020, 05:08
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KaranB1
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
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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


Note that \(x^2 \ge 0\) for all real \(x\), so the least possible value of the left-hand side (LHS) of the equation is when \(x = 0\), which gives:

\(\sqrt{x^2+1} + \sqrt{x^2+2} = 1 + \sqrt{2} \approx{2.4}\).

Since even the smallest possible value of LHS is still greater than RHS (which is 2), there can be no real \(x\) that satisfies the given equation. Therefore, the answer is 0 roots.


Answer: A
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gvij2017
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 05 Jan 2020, 05:14
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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.
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 05 Jan 2020, 05:39
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KaranB1
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
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I think the question should be how many "real" roots does the equation have?
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 05 Jan 2020, 05:40
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AnirudhaS
KaranB1
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?
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All numbers on the GMAT are real by default.
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 05 Jan 2020, 18:29
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Is there a way to do this question using the discriminant?
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 06 Jan 2020, 00:05
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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.
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 06 Jan 2020, 00:06
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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.
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 14 Aug 2020, 11:12
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Asked: How many roots does the equation \(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\) have?

\(\sqrt{x^2 + 1} + \sqrt{x^2 + 2} = 2\)
\(\sqrt{x^2 + 1} = 2 - \sqrt{x^2 + 2}\)
\(x^2 + 1 = 4 + x^2 + 2 - 4\sqrt{x^2 + 2}\)
\(5 = 4\sqrt{x^2 + 2}\)
25 = 16(x^2 + 2)
16x^2 + 32 = 25
16x^2 = -7
No roots

IMO A
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 21 Oct 2021, 08:31
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I put forward my approach-

Let, x^2 + 1 = y

Then, sqrt(x^2 + 1) = sqrt(y) and sqrt(x^2 + 2) = sqrt(y + 1)

So, sqrt(y) + sqrt(y+1) = 2
=> sqrt(y+1) = 2- sqrt(y)
Square both sides-
=> y + 1 = 4 + y - 4sqrt(y)
=> 4sqrt(y) = 3
=> y = 9/16

Therefore, x^2 + 1 = 9/16
=> 16x^2 + 16 = 9
=> 16x^2 = -7

Since 16 is a positive integer and x^2 will also yield a positive value, there product can never be a negative number. Therefore answer is 0.
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How many roots does the equation (x^2 + 1)^(1/2) + (x^2 + 2)^(1/2) = 15 May 2025, 08:39
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