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

Question

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

Bunuel · Accepted Answer

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

gvij2017 · Answer

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.

AnirudhaS · Answer

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

Bunuel · Answer

All numbers on the GMAT are real by default.

VIVA1060 · Answer

Is there a way to do this question using the discriminant?

ShankSouljaBoi · Answer

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.

ShankSouljaBoi · Answer

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.

Kinshook · Answer

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

DeeRocket · Answer

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) =

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

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