In terms of y, x =

Can you post the complete question please?
'If x >= MISSING and x = sqrt(8xy-16y^2) , then, in terms of y, x = '

x^2 - 8xy + 16y^2 = 0
(x-4y)(x-4y)=0

x-4y=0
x=4y

Is the answer not D?

alex.gellatly wrote:If x>0 and x = √(8xy-16y^2), then, in terms of y, x =

-4y
y/4
y
4y
4y^2

Thanks

Do reverse engineering. Put the options in the equation "x = √(8xy-16y^2)".
But choosing the right options to substitute is also necessary.

Since the equation has a square root, (8xy-16y^2) should turn out to be a square number. If you observe option D, suits our requirement.
Just try it.
x = √(8xy-16y^2)
substituting x = 4y.
RHS = √(8(4y)y-16y^2)
RHS = √(32y^2-16y^2)
RHS = √(16y^2)
RHS = +4y or -4y.
Since x > 0 we can say x = 4y.
[spoiler]Answer: D[/spoiler]

Brent@GMATPrepNow wrote: Set equation equal to zero: x^2 - 8xy + 16y^2 = 0
Factor left side: (x - 4y)(x - 4y) = 0
This tells us that x - 4y = 0, which means x = 4y

Thank you for your post but I'm still a little unclear with your factorization. If you add -4x and -4x you get -8x... but we need it to equal -8xy, do we not?... but here we have no y.... Am I missing something/

Thaks

Thank you for your post but I'm still a little unclear with your factorization. If you add -4x and -4x you get -8x... but we need it to equal -8xy, do we not?... but here we have no y.... Am I missing something/

Dear Alex,

The factorization goes this way -

x^2-8xy+16y^2 = 0
x^2-4xy-4xy+16y^2=0
x(x-4y)-4y(x-4y)=0
(x-4y)(x-4y)=0
i.e. (x-4y)^2=0
i.e. x-4y=0 => x=4y

OR

If you can identify that x^2-8xy+16y^2 = 0 is an expansion of (x-4y)^2 then you can direcly factorize the term.

Hope this satisfies.