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If x^2 + y^2 = xy, then (x + y)^4 =

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If x^2 + y^2 = xy, then (x + y)^4 =  [#permalink]

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New post Updated on: 09 Jul 2019, 04:06
4
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

74% (01:20) correct 26% (01:31) wrong based on 87 sessions

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If \(x^2 + y^2 = xy\), then \((x + y)^4 =\)


(A) \(xy\)

(B) \(x^2*y^2\)

(C) \(9x^2*y^2\)

(D) \((x^2 + y^2)^2\)

(E) \(x^4 + y^4\)

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Difficulty Level: 600
Spoiler: OA

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Originally posted by Bunuel on 31 May 2017, 01:45.
Last edited by SajjadAhmad on 09 Jul 2019, 04:06, edited 1 time in total.
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Re: If x^2 + y^2 = xy, then (x + y)^4 =  [#permalink]

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New post 31 May 2017, 02:04
C.
Add 2xy on both sides and rhs side becomes (a+b)^2

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Re: If x^2 + y^2 = xy, then (x + y)^4 =  [#permalink]

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New post 31 May 2017, 04:06
Given: x^2 + y^2 = xy
If we add 2xy to both sides, we get: x^2 + y^2 + 2xy = 3xy Or (x+y)^2 = 3xy

Now, (x+y)^4 = [(x+y)^2]^2 = (3xy)^2 = 9 * x^2 * y^2

Hence option C
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Re: If x^2 + y^2 = xy, then (x + y)^4 =  [#permalink]

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New post 01 Jun 2017, 17:28
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Bunuel wrote:
If \(x^2 + y^2 = xy\), then \((x + y)^4 =\)


(A) \(xy\)

(B) \(x^2*y^2\)

(C) \(9x^2*y^2\)

(D) \((x^2 + y^2)^2\)

(E) \(x^4 + y^4\)


We are given:

x^2 + y^2 = xy.

Note that (x + y)^4 = [(x + y)^2]^2 = [x^2 + 2xy + y^2]^2.

Since x^2 + y^2 = xy,

[x^2 + 2xy + y^2]^2 = [2xy + xy]^2 = (3xy)^2 = 9(x^2)(y^2).

Answer: C
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Re: If x^2 + y^2 = xy, then (x + y)^4 =  [#permalink]

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New post 28 Aug 2017, 01:13
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why is answer A.

I am getting C
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Re: If x^2 + y^2 = xy, then (x + y)^4 =  [#permalink]

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New post 28 Aug 2017, 01:56
hanyhamdani wrote:
why is answer A.

I am getting C


The correct answer is C. Edited. Thank you.
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Re: If x^2 + y^2 = xy, then (x + y)^4 =  [#permalink]

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New post 28 Aug 2017, 02:53
x^2+y^2 = xy
(x+y)^2 = x^2 + y^2 + 2xy = xy +2xy =3xy
(x+y)^4= (3xy)^2= 9x^2. y^2
option C
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Re: If x^2 + y^2 = xy, then (x + y)^4 =  [#permalink]

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New post 24 Mar 2019, 20:53
We have:

\(x^2 + y^2 = xy\)

This mean we had

\((x + y)^2 = 3xy\)

So


\(((x + y)^2)^2) = (3xy)^2\)

C
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Re: If x^2 + y^2 = xy, then (x + y)^4 =   [#permalink] 24 Mar 2019, 20:53
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