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# ­If x > y > 0, and x^2 + y^2 = 6xy, what is the value of (x-y)/(x+y)?

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Re: ­If x > y > 0, and x^2 + y^2 = 6xy, what is the value of (x-y)/(x+y)? [#permalink]
Ak2604 wrote:
Given x² + y² = 6xy

We know (x+y)² = x² + y² + 2xy

Send 2xy to the other side
(x+y)² - 2xy = x² + y²

So (x+y)² - 2xy = 6xy
(x+y)² = 8xy
x+y = √8xy

And we know
(x-y)² = x² + y² - 2xy
And
(x-y)² + 2xy = x² + y²

Then

(x-y)² + 2xy = 6xy
(x-y)² = 4xy
x-y = √4xy

So (√4xy) / (√8xy)

√xy gets cancelled

2 / √8

Multiply both num and denom by √8

2√8 / 8
√8 / 4
√2*4 / 4
2√2 / 4
√2 / 2

Option D

Posted from my mobile device

­I think you made a calculation error.
Re: ­If x > y > 0, and x^2 + y^2 = 6xy, what is the value of (x-y)/(x+y)? [#permalink]
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