Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
The explanation for this Q in the book says that "√x + √y > 0" indicates that x and y must be non-negative, 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
[#permalink]
Show Tags
13 Jun 2012, 03:55
5
6
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 non-negative, 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 \((x + y)^2=(x-y)^2\) --> \(x^2+2xy+y^2=x^2-2xy+y^2\) --> \(4xy=0\) --> \(xy=0\) --> 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 non-negative): \(x^2>y^2\) --> \(y^2\) (square of a number) is always non-negative, so \(x^2\) is more than some non-negative 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 \(\sqrt{x}=\sqrt{-8}=undefined\).
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 non-negative, 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 non-negative.
Let me know if you need any more assistance on this topic.
Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of
[#permalink]
Show Tags
28 Nov 2014, 14:46
1
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 non-negative, 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 \((x + y)^2=(x-y)^2\) --> \(x^2+2xy+y^2=x^2-2xy+y^2\) --> \(4xy=0\) --> \(xy=0\) --> 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 non-negative): \(x^2>y^2\) --> \(y^2\) (square of a number) is always non-negative, so \(x^2\) is more than some non-negative 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 \(\sqrt{x}=\sqrt{-8}=undefined\).
Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of
[#permalink]
Show Tags
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 non-negative, 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 \((x + y)^2=(x-y)^2\) --> \(x^2+2xy+y^2=x^2-2xy+y^2\) --> \(4xy=0\) --> \(xy=0\) --> 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 non-negative): \(x^2>y^2\) --> \(y^2\) (square of a number) is always non-negative, so \(x^2\) is more than some non-negative 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 \(\sqrt{x}=\sqrt{-8}=undefined\).
Re: a = x + y and b = x - y. If a^2 = b^2, what is the value of
[#permalink]
Show Tags
13 Mar 2018, 10:47
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
close
48% (02:08) correct 
