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Tricky Square root question

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by Mo2men » Sat Dec 17, 2016 3:03 am
How many roots does the equation √(x^2+1) + √(x^2+2)=2 have?

For clarification: (x^2+1) is under one sign of square root & (x^2+2) is under one sign of square root

A. 0
B. 1
C. 2
D. 3
E. 4

OA: A

Sourc: GMATClub tests
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by GMATGuruNY » Sat Dec 17, 2016 3:46 am
Mo2men wrote:How many roots does the equation √(x^2+1) + √(x^2+2)=2 have?

For clarification: (x^2+1) is under one sign of square root & (x^2+2) is under one sign of square root

A. 0
B. 1
C. 2
D. 3
E. 4
The least possible value for x² is 0.
If we substitute x²=0 into √(x²+1) + √(x²+2)=2, we get:
√(0²+1) + √(0²+2)=2
√1 + √2 = 2
1 + (approximately 1.4) = 2
(approximately 2.4) = 2.
Since the least possible value for x² makes the left side too big, there is no viable value for x².
Thus, the equation has no roots.

The correct answer is A.
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by Jay@ManhattanReview » Tue Dec 20, 2016 9:20 pm
Mo2men wrote:How many roots does the equation √(x^2+1) + √(x^2+2)=2 have?

For clarification: (x^2+1) is under one sign of square root & (x^2+2) is under one sign of square root

A. 0
B. 1
C. 2
D. 3
E. 4

OA: A

Sourc: GMATClub tests
√(x^2+1) + √(x^2+2) = 2 can be rewritten as √(x^2+1) = 2 - √(x^2+2)

Squaring both the sides, we get, (x^2+1) = [2 - √(x^2+2)]^2

=> (x^2+1) = 4 + (x^2+2) - 4√(x^2+2)

=> 1 = 4 + 2 - 4√(x^2+2); [cancelling x^2 terms]

=> -5 = -4√(x^2+2)

=> 5/4 = √(x^2+2)

=> 25/16 = (x^2+2)

=> 25/16 - 2 = x^2

=> -7/16 = x^2

Since the square of a number cannot be negative, x would not have any root.

OA: [spoiler]A (None)[/spoiler]

-Jay

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Last edited by Jay@ManhattanReview on Wed Dec 21, 2016 3:33 am, edited 1 time in total.
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by Mo2men » Wed Dec 21, 2016 2:38 am
Jay@ManhattanReview wrote:
Mo2men wrote:How many roots does the equation √(x^2+1) + √(x^2+2)=2 have?

For clarification: (x^2+1) is under one sign of square root & (x^2+2) is under one sign of square root

A. 0
B. 1
C. 2
D. 3
E. 4

OA: A

Sourc: GMATClub tests
√(x^2+1) + √(x^2+2) = 2 can be rewritten as √(x^2+1) = 2 - √(x^2+2)

Squaring both the sides, we get, (x^2+1) = [2 - √(x^2+2)]^2

=> (x^2+1) = 4 - (x^2+2) - 4√(x^2+2)

=> 1 = 4 - 2 - 4√(x^2+2); [cancelling x^2 terms]

=> -1 = -4√(x^2+2)

=> 1/4 = √(x^2+2)

=> 1/2 = (x^2+2)

=> 1/2 - 2 = x^2

=> -3/2 = x^2

Since the square of a number cannot be negative, x would not have any root.

OA: [spoiler]A (None)[/spoiler]

-Jay

There is a mistake when when you squared [2 - √(x^2+2)].
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by Jay@ManhattanReview » Wed Dec 21, 2016 3:34 am
Mo2men wrote:
Jay@ManhattanReview wrote:
Mo2men wrote:How many roots does the equation √(x^2+1) + √(x^2+2)=2 have?

For clarification: (x^2+1) is under one sign of square root & (x^2+2) is under one sign of square root

A. 0
B. 1
C. 2
D. 3
E. 4

OA: A

Sourc: GMATClub tests
√(x^2+1) + √(x^2+2) = 2 can be rewritten as √(x^2+1) = 2 - √(x^2+2)

Squaring both the sides, we get, (x^2+1) = [2 - √(x^2+2)]^2

=> (x^2+1) = 4 - (x^2+2) - 4√(x^2+2)

=> 1 = 4 - 2 - 4√(x^2+2); [cancelling x^2 terms]

=> -1 = -4√(x^2+2)

=> 1/4 = √(x^2+2)

=> 1/2 = (x^2+2)

=> 1/2 - 2 = x^2

=> -3/2 = x^2

Since the square of a number cannot be negative, x would not have any root.

OA: [spoiler]A (None)[/spoiler]

-Jay

There is a mistake when when you squared [2 - √(x^2+2)].

You are right. I have corrected it. Thank you.
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