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

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Math Expert
Joined: 02 Sep 2009
Posts: 56244
If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =  [#permalink]

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05 Jul 2018, 04:46
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94% (00:49) correct 6% (00:50) wrong based on 41 sessions

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

A. 4

B. 32

C. 48

D. 64

E. 256

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Joined: 05 Jun 2018
Posts: 7
Location: India
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Re: If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =  [#permalink]

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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.

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

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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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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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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

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

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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 =   [#permalink] 08 Jul 2018, 02:01
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