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a = x + y and b = x  y. If a^2 = b^2, what is the value of
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Updated on: 05 Jun 2013, 03:37
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a = x + y and b = x  y. If a^2 = b^2, what is the value of y? (1) √x + √y > 0 (2) √x  √y > 0 The explanation for this Q in the book says that "√x + √y > 0" indicates that x and y must be nonnegative, in order for their square roots to be real values.
I don't understand this explantion. Why can't x and y be negative?
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Originally posted by eybrj2 on 12 Jun 2012, 23:35.
Last edited by Bunuel on 05 Jun 2013, 03:37, edited 3 times in total.
EDITED THE QUESTION AND MOVED TO DS FORUM.




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a = x + y and b = x  y. If a^2 = b^2, what is the value of
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13 Jun 2012, 03:55
eybrj2 wrote: a = x + y and b = x  y. If a^2 = b^2, what is the value of y?
(1) √x + √y > 0
(2) √x  √y > 0
The explanation for this Q in the book says that "√x + √y > 0" indicates that x and y must be nonnegative, in order for their square roots to be real values.
I don't understand this explantion. Why can't x and y be negative? a = x + y and b = x  y. If a^2 = b^2, what is the value of y?Since \(a = x + y\) and \(b = x  y\) then from \(a^2 = b^2\) we have that \((x + y)^2=(xy)^2\) > \(x^2+2xy+y^2=x^22xy+y^2\) > \(4xy=0\) > \(xy=0\) > either \(x\) or \(y\) equals to zero (or both). (1) √x + √y > 0. Two cases are possible: \(x=0\) and \(y\) is ANY positive number OR \(y=0\) and \(x\) is ANY positive number. Not sufficient. (2) √x  √y > 0 > \(\sqrt{x}>\sqrt{y}\). Now, since square root function can not give negative result (\(\sqrt{some \ expression}\geq{0}\)), then \(\sqrt{x}>\sqrt{y}\geq{0}\). So, \(x>0\) and \(y=0\). Sufficient. Or another way: square \(\sqrt{x}>\sqrt{y}\) (we can safely do that since both parts of the inequality are nonnegative): \(x^2>y^2\) > \(y^2\) (square of a number) is always nonnegative, so \(x^2\) is more than some nonnegative number, which makes \(x^2\) a positive value which excludes the possibility of \(x=0\), so \(y=0\). Sufficient. Answer: B. As for your question: The GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers so even roots from negative number is undefined on the GMAT: \(\sqrt[{even}]{negative}=undefined\), for example \(\sqrt{25}=undefined\). eybrj2 wrote: if x = (2)^3
√x = 4√2
Is this correct?
If it is, is it negative number? No, that's not correct. If \(x=(2)^3=8\) then \(\sqrt{x}=\sqrt{8}=undefined\). Hope it's clear. P.S. Please read and follow: 11rulesforposting133935.html
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Re: From Advanced GMAT Quant
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12 Jun 2012, 23:39
if x = (2)^3
√x = 4√2
Is this correct?
If it is, is it negative number?



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Re: From Advanced GMAT Quant
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12 Jun 2012, 23:51
eybrj2 wrote: In chapter 9,
Q 73. a = x + y and b = x  y. If a^2 = b^2, what is the value of y?
1) √x + √y > 0
2) √x  √y > 0
The explanation for this Q in the book says that "√x + √y > 0" indicates that x and y must be nonnegative, in order for their square roots to be real values.
I don't understand this explantion. Why can't x and y be negative? Hi, If x & y are negative then √x, √y would be out of scope of GMAT. On serious note, supposedly x, y are negative. √x, √y would be imaginary quantities. √x + √y would be a point on Argand Plane (where complex numbers are represented). and you would be comparing two points (√x + √y > 0). Does this make any sense? No. Thus, it is correctly mentioned in the book that x & y are nonnegative. Let me know if you need any more assistance on this topic. Regards,



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Re: From Advanced GMAT Quant
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13 Jun 2012, 01:34
(x+y)^2 = (xy)^2 x^2 + 2xy + y^2 = x^2  2xy + y^2
2xy = 2xy 4xy = 0
either x or y must be zero, or both
1) either one of x or y could be zero. NS 2) √x > √y √x > √y >= 0
y = 0
B)



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Re: a = x + y and b = x  y. If a^2 = b^2, what is the value of
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28 Nov 2014, 14:46
Bunuel wrote: eybrj2 wrote: a = x + y and b = x  y. If a^2 = b^2, what is the value of y?
(1) √x + √y > 0
(2) √x  √y > 0
The explanation for this Q in the book says that "√x + √y > 0" indicates that x and y must be nonnegative, in order for their square roots to be real values.
I don't understand this explantion. Why can't x and y be negative? a = x + y and b = x  y. If a^2 = b^2, what is the value of y?Since \(a = x + y\) and \(b = x  y\) then from \(a^2 = b^2\) we have that \((x + y)^2=(xy)^2\) > \(x^2+2xy+y^2=x^22xy+y^2\) > \(4xy=0\) > \(xy=0\) > either \(x\) or \(y\) equals to zero (or both). (1) √x + √y > 0. Two cases are possible: \(x=0\) and \(y\) is ANY positive number OR \(y=0\) and \(x\) is ANY positive number. Not sufficient. (2) √x  √y > 0 > \(\sqrt{x}>\sqrt{y}\). Now, since square root function can not give negative result (\(\sqrt{some \ expression}\geq{0}\)), then \(\sqrt{x}>\sqrt{y}\geq{0}\). So, \(x>0\) and \(y=0\). Sufficient. Or another way: square \(\sqrt{x}>\sqrt{y}\) (we can safely do that since both parts of the inequality are nonnegative): \(x^2>y^2\) > \(y^2\) (square of a number) is always nonnegative, so \(x^2\) is more than some nonnegative number, which makes \(x^2\) a positive value which excludes the possibility of \(x=0\), so \(y=0\). Sufficient. Answer: D.? As for your question: The GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers so even roots from negative number is undefined on the GMAT: \(\sqrt[{even}]{negative}=undefined\), for example \(\sqrt{25}=undefined\). eybrj2 wrote: if x = (2)^3
√x = 4√2
Is this correct?
If it is, is it negative number? No, that's not correct. If \(x=(2)^3=8\) then \(\sqrt{x}=\sqrt{8}=undefined\). Hope it's clear. P.S. Please read and follow: 11rulesforposting133935.htmlDear Bunuel, your reasoning and answer does not match, could you please confirm that the answer is B and not D.



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Re: a = x + y and b = x  y. If a^2 = b^2, what is the value of
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29 Nov 2014, 05:16
santorasantu wrote: Bunuel wrote: eybrj2 wrote: a = x + y and b = x  y. If a^2 = b^2, what is the value of y?
(1) √x + √y > 0
(2) √x  √y > 0
The explanation for this Q in the book says that "√x + √y > 0" indicates that x and y must be nonnegative, in order for their square roots to be real values.
I don't understand this explantion. Why can't x and y be negative? a = x + y and b = x  y. If a^2 = b^2, what is the value of y?Since \(a = x + y\) and \(b = x  y\) then from \(a^2 = b^2\) we have that \((x + y)^2=(xy)^2\) > \(x^2+2xy+y^2=x^22xy+y^2\) > \(4xy=0\) > \(xy=0\) > either \(x\) or \(y\) equals to zero (or both). (1) √x + √y > 0. Two cases are possible: \(x=0\) and \(y\) is ANY positive number OR \(y=0\) and \(x\) is ANY positive number. Not sufficient. (2) √x  √y > 0 > \(\sqrt{x}>\sqrt{y}\). Now, since square root function can not give negative result (\(\sqrt{some \ expression}\geq{0}\)), then \(\sqrt{x}>\sqrt{y}\geq{0}\). So, \(x>0\) and \(y=0\). Sufficient. Or another way: square \(\sqrt{x}>\sqrt{y}\) (we can safely do that since both parts of the inequality are nonnegative): \(x^2>y^2\) > \(y^2\) (square of a number) is always nonnegative, so \(x^2\) is more than some nonnegative number, which makes \(x^2\) a positive value which excludes the possibility of \(x=0\), so \(y=0\). Sufficient. Answer: D.? As for your question: The GMAT is dealing only with Real Numbers: Integers, Fractions and Irrational Numbers so even roots from negative number is undefined on the GMAT: \(\sqrt[{even}]{negative}=undefined\), for example \(\sqrt{25}=undefined\). eybrj2 wrote: if x = (2)^3
√x = 4√2
Is this correct?
If it is, is it negative number? No, that's not correct. If \(x=(2)^3=8\) then \(\sqrt{x}=\sqrt{8}=undefined\). Hope it's clear. P.S. Please read and follow: 11rulesforposting133935.htmlDear Bunuel, your reasoning and answer does not match, could you please confirm that the answer is B and not D. Typo edited. Thank you.
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Re: a = x + y and b = x  y. If a^2 = b^2, what is the value of
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21 Mar 2015, 07:54
One general question:
If a^2 = b^2 (x+y)^2 = (xy)^2 then a = b x+y = xy
Obviously this does not work out. Where is my thinking flawed?



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a = x + y and b = x  y. If a^2 = b^2, what is the value of
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21 Mar 2015, 08:10
lucky1829 wrote: One general question:
If a^2 = b^2 (x+y)^2 = (xy)^2 then a = b x+y = xy
Obviously this does not work out. Where is my thinking flawed? Hi, you are missing on the point that if a=b, a^2 will still be equal to b^2.. so you will have a case where (x+y)=(xy)
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Re: a = x + y and b = x  y. If a^2 = b^2, what is the value of
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