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What is the value of (x + y)^2? [#permalink]
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04 Oct 2011, 07:37
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What is the value of (x + y)^2? (1) x^2 − xy = 28 and 3xy + y^2 = 72. (2) (x + y)^4 = 10,000
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Re: What is the value of (x + y)^2? [#permalink]
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04 Oct 2011, 09:59
How can you solve with A? I was inclining more towards E.



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Re: What is the value of (x + y)^2? [#permalink]
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04 Oct 2011, 10:12
hashjax wrote: How can you solve with A? I was inclining more towards E. Add equations in 1 x^2 − xy = 28 & 3xy + y^2 = 72. x^2+2xy+y^2=100 (x+y)^2=100 Sufficient Makes sense? kaliaabhishek wrote: D Statement 2 gives us two values. Hence insufficient
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Re: What is the value of (x + y)^2? [#permalink]
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04 Oct 2011, 14:55
Quote: Statement 2 gives us two values. Hence insufficient Actually statement 2 gives only 1 value. (x + y)^2 = 100 it cannot = 100 because the left side is a square  which means the right side must be a positive number and cannot be negative. If the square weren't there  then you can say there are 2 possible values: +100 and 100. But since the question asks for what (x+y)^2 is  we know that it must be the positive version. Therefore, either (1) or (2) would work. Answer would be (D).



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Re: What is the value of (x + y)^2? [#permalink]
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04 Oct 2011, 17:15
Let x+y = 10, then (x+y)^4 = 10000. So A it is.
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Re: What is the value of (x + y)^2? [#permalink]
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04 Oct 2011, 18:43
gmatpill wrote: Quote: Statement 2 gives us two values. Hence insufficient Actually statement 2 gives only 1 value. (x + y)^2 = 100 it cannot = 100 because the left side is a square  which means the right side must be a positive number and cannot be negative. If the square weren't there  then you can say there are 2 possible values: +100 and 100. But since the question asks for what (x+y)^2 is  we know that it must be the positive version. Therefore, either (1) or (2) would work. Answer would be (D). Yes, you're right. I missed it. Sorry all. It is indeed. Moral of the story Read the question properly
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Re: What is the value of (x + y)^2? [#permalink]
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15 Oct 2011, 10:45
"D" 1. X^2  XY = 28 Y^2 + 3XY = 72 ________________ X^2 + Y^2 + 2XY = 100 => (X + Y)^2 = 100 SUFFICIENT. 2. (X + Y)^4 = 10,000 => ((X+Y)^2)^2 = ((10)^2)^2 => (X+Y)^2 = 100 SUFFICIENT
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Re: What is the value of (x + y)^2? [#permalink]
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03 Nov 2011, 07:41
(1) Add both the eq. (2) (x + y)^4 = 10,000 means (x + y)^2 can be 100 or 100, but (x + y)^2 being a square we can have only 100. Hence Ans = D



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What is the value of (x + y)^2? [#permalink]
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What is the value of (x + y)^2? [#permalink]
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03 Sep 2016, 04:16
Bunuel , a basic question if you don't mind clarifying; For (x + y)^4 , you can simply take the square root .... but hypothetically if (x + y)^2 = 10,000 you can't simply take the square root? you'll have to FOIL it out? For example, I've seen some Qs where if (x + y)^2 = 100 .... (x + y) does not equal + 10, it's x^2 + 2xy + y^2 = 100 But (x + y)^4, you're able to simply take the square root? Just really confused about this concept. Appreciate all the help. thank you



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Re: What is the value of (x + y)^2? [#permalink]
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03 Sep 2016, 05:26
xnthic wrote: Bunuel , a basic question if you don't mind clarifying; For (x + y)^4 , you can simply take the square root .... but hypothetically if (x + y)^2 = 10,000 you can't simply take the square root? you'll have to FOIL it out? For example, I've seen some Qs where if (x + y)^2 = 100 .... (x + y) does not equal + 10, it's x^2 + 2xy + y^2 = 100 But (x + y)^4, you're able to simply take the square root? Just really confused about this concept. Appreciate all the help. thank you Let me try help you out. See, I have \((x + y)^4 = 10000\) Now, if take the fourth root of the above equation, we will get (x+y) = 10 or (x+y) =  10 So, we have two values for (x+y). But NOTICE, the question stem asks the value of \((x + y)^2\), so whichever value of (x+y) when squared will always give 100. Hence, we have a single value of (x+y) => The statement is sufficient. Even if we had \((x + y)^2 = 10000\), we would have got the value of (x+y) either equal to 100 and 100. But remember question is NOT asking you the value of (x+y), it is asking the value of \((x+y)^2\) I hope its clear now.
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Re: What is the value of (x + y)^2? [#permalink]
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04 Sep 2016, 00:26
Thanks for clarifying ! I definitely get the answer and logic.
However, related to this about a basic math concept. My question is simply: in other math situations, is it allowed to simply take square root (x + y)^2 which gives +10 OR is that not allowed and you'd have to foil out the algebraic expression?
Thanks for the help !
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Re: What is the value of (x + y)^2? [#permalink]
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04 Sep 2016, 00:35
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xnthic wrote: Thanks for clarifying ! I definitely get the answer and logic.
However, related to this about a basic math concept. My question is simply: in other math situations, is it allowed to simply take square root (x + y)^2 which gives +10 OR is that not allowed and you'd have to foil out the algebraic expression?
Thanks for the help !
Posted from my mobile device Yes, it is allowed to take the square root on both the sides of the equation. But make sure both signs(+ and ) are considered. Go through the below link to get more clarity on this concept. http://www.purplemath.com/modules/solvquad2.htm
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Re: What is the value of (x + y)^2? [#permalink]
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03 Nov 2016, 10:57
D Both options are individually sufficient 1) adding both equations, (x+y)^2=100 Sufficient 2) (x+y)^2 =100 [*can't be the negative solution as it is a square] Sufficient Sent from my HTC One E9PLUS dual sim using GMAT Club Forum mobile app



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Re: What is the value of (x + y)^2? [#permalink]
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03 Nov 2016, 12:33
What is the value of (x + y)^2?
(1) x^2 − xy = 28 and 3xy + y^2 = 72.
(2) (x + y)^4 = 10,000
(1) x^2 − xy = 28 and 3xy + y^2 = 72.
I first factored out (x+y)²=x²_2xy+y²
x²xy=28 x²=28+xy
3xy + y² = 72.
y²=723xy
if we substitute the values back into the equation we get
28+xy+2xy+723xy
From here the variables cancel out and we get =100
(2) (x + y)^4 = 10,000 Take the square root of both sides (x+y)²=100
D



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Re: What is the value of (x + y)^2? [#permalink]
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25 Dec 2017, 23:20
GMATmission wrote: What is the value of (x + y)^2?
(1) x^2 − xy = 28 and 3xy + y^2 = 72.
(2) (x + y)^4 = 10,000 We need to find (x + y)^2. And (x + y)^2 = x^2 + y^2 + 2xy (1) We are given two equations: x^2 − xy = 28 and 3xy + y^2 = 72. If we add these two equations, we get: x^2  xy + 3xy + y^2 = 28+72 OR x^2 + y^2 + 2xy = 100. This is sufficient. (2) (x + y)^4 = 10,000 Or ((x + y)^2)^2 = (100)^2 This gives us (x + y)^2 = 100 This is also sufficient. (We should note that though 10000 is also the square of 100, we cannot write 100 on RHS as LHS is (x + y)^2, and square of anything cannot be negative) Hence D answer




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