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If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =

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If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =  [#permalink]

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New post 05 Jul 2018, 04:46
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

94% (00:30) correct 6% (00:50) wrong based on 35 sessions

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Re: If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =  [#permalink]

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New post 05 Jul 2018, 05:06
x2+y2=16−2xy
Let's bring the 2xy on LHS:
X2+Y2+2xy=16

Now, as per the formula (x+y)2=X2+Y2+2xy, we have:

(X+Y)2=16; now take square both side,

(X+Y)=(16)2=256.

Answer is E.
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Re: If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =  [#permalink]

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New post 05 Jul 2018, 05:18
Bunuel wrote:
If \(x^2 + y^2 = 16 - 2xy\), then \((x + y)^4 =\)


A. 4

B. 32

C. 48

D. 64

E. 256


Given (x+y)^2=16
\((x + y)^4\)= \(((x+y)^2)^2\)=\(16^2=256\)

Ans. E
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Re: If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =  [#permalink]

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New post 05 Jul 2018, 08:15
Bunuel wrote:
If \(x^2 + y^2 = 16 - 2xy\), then \((x + y)^4 =\)


A. 4

B. 32

C. 48

D. 64

E. 256

\(x^2 + y^2 = 16 - 2xy\)

Or, \(x^2 + y^2 + 2xy = 16\)

Or, \((x + y)^2\) = \(4^2\)

So, \(( x + y)^4 = 16^2\)

Or, \(( x + y)^4 = 256\), Answer must be (E)
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Re: If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =  [#permalink]

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New post 07 Jul 2018, 18:28
Bunuel wrote:
If \(x^2 + y^2 = 16 - 2xy\), then \((x + y)^4 =\)


A. 4

B. 32

C. 48

D. 64

E. 256


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

Since x^2 + y^2 = 16 - 2xy, we can substitute 16 - 2xy for x^2 + y^2 in the expression above and obtain:

(16 - 2xy + 2xy)^2 = (16)^2 = 256

Answer: E
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If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =  [#permalink]

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New post 08 Jul 2018, 02:01
Bunuel wrote:
If \(x^2 + y^2 = 16 - 2xy\), then \((x + y)^4 =\)


A. 4

B. 32

C. 48

D. 64

E. 256



Given,

\(x^2\) + \(y^2\) = 16 - 2xy
\(x^2\) + 2xy + \(y^2\) = 16
\((x + y)^2\) = 16
\((x+y)^4\) = \(16^2\)
\((x + y)^4\) = 256

Thus the best answer is E.
If x^2 + y^2 = 16 - 2xy, then (x + y)^4 = &nbs [#permalink] 08 Jul 2018, 02:01
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