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

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Bunuel
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If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 = [#permalink] New post   07 Feb 2016, 10:07
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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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A
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Beixi88
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If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 = [#permalink] New post   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 \((x-y)^2 = 8\).

My answer is A, 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.
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chetan2u
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Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 = [#permalink] New post   07 Feb 2016, 18:43
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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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chetan2u
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Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 = [#permalink] New post   07 Feb 2016, 18:45
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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 \((x-y)^2 = 20\).

My answer is E, 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..
rest your solution is perfect and you will get your correct answer as 14-2*3=8..
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Bunuel
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Re: If x^2 + y^2 = 14 and xy = 3, then (x − y)^2 = [#permalink] New post   10 Feb 2016, 11:07
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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] New post   28 Feb 2018, 11:10
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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.

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

Answer: A.
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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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