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by pm » Mon Jul 25, 2011 10:38 pm
Which of the following equations is true for all positive values of x and y?

1. Root x + Root y = Root x+y
2. Root x^4y^16 = x^2y^4
3.(x Root y)(y Root x)= x^2y^2
4. yRootx+ y root x = Root 4xy^2
5. (x^y)(y^y)= (xy)^2y

OA 4

Experts Can you please suggest the best way to do this?
Last edited by pm on Mon Jul 25, 2011 10:57 pm, edited 1 time in total.
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by winniethepooh » Mon Jul 25, 2011 10:52 pm
The answer is 4.

1. √x+√y = √x+√y (No change)
2. √x^4 *√y16 = x^2 * y^8
3. x√y * y√x = xy √xy
4. y√x + y√x = 2y√x = √4*√Y^2√x =√4*y^2x(answer)
5. No need to check!

Hence, 4.
Last edited by winniethepooh on Mon Jul 25, 2011 11:21 pm, edited 1 time in total.
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by pm » Mon Jul 25, 2011 11:00 pm
@Winniethepooh
How do you get xy^8 for the 2nd option ?
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by Anurag@Gurome » Mon Jul 25, 2011 11:16 pm
pm wrote:Which of the following equations is true for all positive values of x and y?

1. Root x + Root y = Root x+y
2. Root x^4y^16 = x^2y^4
3.(x Root y)(y Root x)= x^2y^2
4. yRootx+ y root x = Root 4xy^2
5. (x^y)(y^y)= (xy)^2y

OA 4

Experts Can you please suggest the best way to do this?
(1) √x + √y = √(x + y)
If x = 4, y = 9, then √x + √y = 2 + 3 = 5, and √(x + y) = √(4 + 9) = √13; not true.

(2) √(x^4)(y^16) = x² (y^4)
Now, without taking any values and taking a look on the left hand side, we get, √(x^4)(y^16) = x² * y^[16*1/2] = x² * y^8, which is not equal to the right hand side; not true.

(3) (x√y)*(y√x) = x²y²
Again, without taking any values and taking a look on the left hand side, we get, (x√y)*(y√x) = x^(3/2) * y^(3/2), which is not equal to the right hand side; not true.

(4) y√x + y√x = √(4xy²) = 2y√x
This is true as the left hand and right hand side are equal.

The correct answer is 4.
Anurag Mairal, Ph.D., MBA
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by winniethepooh » Mon Jul 25, 2011 11:19 pm
As root x^2 = x^2^1/2
and root y 16= y^16^1/2
as root indicates raised to 1/2!
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by winniethepooh » Mon Jul 25, 2011 11:20 pm
Read it wrong!
edited the post above
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