In terms of y, x =

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alex.gellatly: Master | Next Rank: 500 Posts; Posts: 435; Joined: Wed Nov 16, 2011 7:27 amFollowed by: 16 members; Thanked: 48 times

by alex.gellatly » Sun Aug 26, 2012 12:25 am

If x>=0 and x = âˆš(8xy-16y^2), then, in terms of y, x =

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

Thanks
Sorry about the missing part, it has been changed!

Last edited by alex.gellatly on Sun Aug 26, 2012 7:41 pm, edited 1 time in total.

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by neelgandham » Sun Aug 26, 2012 1:56 am

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

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sun Aug 26, 2012 6:16 am

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

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

Thanks

Edited.

Note: I'm assuming that the missing portion (above) tells us that x is greater than or equal to 0.

Given: x = sqrt(8xy-16y^2)
Square both sides: x^2 = 8xy-16y^2
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

Answer = D

Cheers,
Brent

Last edited by Brent@GMATPrepNow on Sun Aug 26, 2012 5:45 pm, edited 4 times in total.

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grandh01: Master | Next Rank: 500 Posts; Posts: 265; Joined: Tue Jul 03, 2012 11:04 pmFollowed by: 6 members

by grandh01 » Sun Aug 26, 2012 4:52 pm

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

x-4y=0
x=4y

Is the answer not D?

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coolhabhi: Master | Next Rank: 500 Posts; Posts: 141; Joined: Tue Oct 04, 2011 5:17 am; Thanked: 25 times

by coolhabhi » Mon Aug 27, 2012 12:58 pm

alex.gellatly wrote:If x>0 and x = âˆš(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 = âˆš(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 = âˆš(8xy-16y^2)
substituting x = 4y.
RHS = âˆš(8(4y)y-16y^2)
RHS = âˆš(32y^2-16y^2)
RHS = âˆš(16y^2)
RHS = +4y or -4y.
Since x > 0 we can say x = 4y.
[spoiler]Answer: D[/spoiler]

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alex.gellatly: Master | Next Rank: 500 Posts; Posts: 435; Joined: Wed Nov 16, 2011 7:27 amFollowed by: 16 members; Thanked: 48 times

by alex.gellatly » Wed Aug 29, 2012 8:46 pm

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

A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231

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adthedaddy: Master | Next Rank: 500 Posts; Posts: 167; Joined: Fri Mar 09, 2012 8:35 pmFollowed by: 3 members; Thanked: 39 times

by adthedaddy » Thu Aug 30, 2012 12:04 am

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.

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