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# If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10

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Re: If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10  [#permalink]

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23 Jul 2014, 03:01
Kconfused wrote:
Bunuel wrote:
krit wrote:
If x^2 + y^2 = 29, what is the value of (x-y)^2

(1) xy = 10
(2) x = 5

Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2

Thanks

Given: $$x^2+y^2=29$$.
Question: $$(x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2xy=29-2xy=?$$ So, basically we should find the value of $$xy$$

(1) xy = 10. Directly gives us the value we needed. Sufficient.

(2) x = 5. Now even if we substitute the value of $$x$$ in $$x^2+y^2=29$$ we'll get two values for $$y$$: 2 and -2, hence two values for $$(x-y)^2$$: 9 and 49. Not sufficient.

Answer: A.

Hope it's clear.

And I thought we consider only the positive value of a square root!!!
Not so?

When the GMAT provides the square root sign for an even root, such as $$\sqrt{x}$$ or $$\sqrt[4]{x}$$, then the only accepted answer is the positive root. That is, $$\sqrt{4}=2$$, NOT +2 or -2.

In contrast, the equation $$x^2=4$$ has TWO solutions, +2 and -2.

Hope it's clear.
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Re: If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10  [#permalink]

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19 Sep 2016, 07:09
A is correct. Here's why:

(1) xy = 10 --> From this we know 2x5 could work or 10x1/2, but since x^2 + y^2 = 29 we know the second alternative can be eliminated. Thus we are left with +/- 2 and +/- 5.

If you plug either (both positive or both negative) into (x+y)^2, you will get the same answer.

SUFFICIENT

(2) x = 5

Plug into original equation --> 25 + y^2 = 29 --> y^2 = 4 --> y = +/- 2

NOT SUFFICIENT
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Re: If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10  [#permalink]

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11 Oct 2017, 07:28
If|x+y=15,x-y=24,solve(1)/(4)x^2-4y^2
I thank the people who are willing to teach me.

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Re: If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10  [#permalink]

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11 Oct 2017, 07:35
Christine Christina wrote:
If|x+y=15,x-y=24,solve(1)/(4)x^2-4y^2
I thank the people who are willing to teach me.

Posted from my mobile device

Welcome to the Club.

I have to say the question is not clear. Especially the highlighted part. Next, we have specific rules of posting on the forum (https://gmatclub.com/forum/rules-for-po ... 33935.html). Please read carefully and re-post accordingly. Thank you.
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Re: If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10  [#permalink]

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20 Oct 2018, 07:54
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Re: If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10   [#permalink] 20 Oct 2018, 07:54

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# If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10

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