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If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =

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Joined: 02 Sep 2009
Posts: 58445
If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =  [#permalink]

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07 Feb 2016, 10:07
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Question Stats:

90% (00:59) correct 10% (02:24) wrong based on 122 sessions

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If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =

A. 8
B. 11
C. 14
D. 17
E. 20

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If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =  [#permalink]

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Updated on: 10 Feb 2016, 17:15
The first thing we want to do is to write our desired solution in an alternative form. Doing so gives us $$x^2 - 2xy + y^2$$. We see that we already have a piece of information which is similar to our desired solution: $$x^2 + y^2 = 14$$. We then also see that we are provided with xy = 3. We plug in xy to get our desired form: $$x^2 - 2xy + y^2 = 14 - 2(3)$$, which becomes $$(x-y)^2 = 8$$.

Edit: Fixed my answer according to the updated question

Originally posted by Beixi88 on 07 Feb 2016, 15:19.
Last edited by Beixi88 on 10 Feb 2016, 17:15, edited 1 time in total.
Math Expert
Joined: 02 Aug 2009
Posts: 7984
Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =  [#permalink]

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07 Feb 2016, 18:43
1
Bunuel wrote:
If x^2 + y^2 = 14 and xy + 3, then (x − y)^2 =

A. 8
B. 11
C. 14
D. 17
E. 20

Hi Bunuel,
xy + 3 should be xy=3...
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Posts: 7984
Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =  [#permalink]

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07 Feb 2016, 18:45
Beixi88 wrote:
The first thing we want to do is to write our desired solution in an alternative form. Doing so gives us $$x^2 - 2xy + y^2$$. We see that we already have a piece of information which is similar to our desired solution: $$x^2 + y^2 = 14$$. We then also see that we are provided xy + 3, which becomes xy = -3. We plug in xy to get our desired form: $$x^2 - 2xy + y^2 = 14 - 2(-3)$$, which becomes $$(x-y)^2 = 20$$.

Hi
there is a typo error ..
but you cannot take xy+ 3 to mean xy=-3..
only if xy+3=0, it will mean xy=-3..
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Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =  [#permalink]

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10 Feb 2016, 11:07
chetan2u wrote:
Bunuel wrote:
If x^2 + y^2 = 14 and xy + 3, then (x − y)^2 =

A. 8
B. 11
C. 14
D. 17
E. 20

Hi Bunuel,
xy + 3 should be xy=3...

Edited. Thank you.
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Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =  [#permalink]

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28 Feb 2018, 11:10
Bunuel wrote:
If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =

A. 8
B. 11
C. 14
D. 17
E. 20

(x - y)^2 = x^2 - 2xy + y^2 = 14 - 6 = 8.

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Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =  [#permalink]

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17 Mar 2018, 00:23
Bunuel wrote:
If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =

A. 8
B. 11
C. 14
D. 17
E. 20

Expand (x − y)^2 = x^2 + y^2 - 2xy

Substitute the values, x^2 + y^2 = 14 & xy = 3

we get 14 -2(3)

8 (A)
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Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =  [#permalink]

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17 Mar 2018, 03:37
$$x^2$$ + $$y^2$$= 14

xy= 3

$$(x-y)^2$$= $$x^2$$ + $$y^2$$-2xy

= 14-2*3
= 14-6= 8

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Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 =   [#permalink] 17 Mar 2018, 03:37
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