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# root(x+y)=4 and root(x)-root(y)=0

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Manager
Joined: 17 Jul 2010
Posts: 112

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02 Dec 2012, 14:29
2
6
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Difficulty:

55% (hard)

Question Stats:

59% (01:00) correct 41% (01:02) wrong based on 559 sessions

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$$\sqrt{X+Y} = 4$$ and $$\sqrt{X}-\sqrt{Y}=0$$

A. X= 2
B. X = -8
C. X= 8
D. X = -2
E. X=Y

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Kudos if you liked!

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Math Expert
Joined: 02 Sep 2009
Posts: 55271

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02 Dec 2012, 14:34
2
2
laythesmack23 wrote:
$$\sqrt{X+Y} = 4$$ and $$\sqrt{X}-\sqrt{Y}=0$$

A. X= 2
B. X = -8
C. X= 8
D. X = -2
E. X=Y

Detailed Solutions = Kudos!
Kudos if you liked!

$$\sqrt{X+Y} = 4$$ --> square both sides: $$x+y=16$$;
$$\sqrt{X}-\sqrt{Y}=0$$ --> $$\sqrt{x}=\sqrt{y}$$ --> square both sides: $$x=y$$.

So, we have that $$x+y=16$$ and $$x=y$$ --> $$x+x=16$$ --> $$x=8$$.

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Manager
Joined: 16 Feb 2012
Posts: 172
Concentration: Finance, Economics

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19 Feb 2013, 05:06
In this problem the answer could also be X=Y, isn't it?
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Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 587
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
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19 Feb 2013, 05:22
4
Stiv wrote:
In this problem the answer could also be X=Y, isn't it?

Yes, absolutely right.
One reason why only official questions should be used for practice.
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Lets Kudos!!!
Black Friday Debrief
Manager
Joined: 16 Feb 2012
Posts: 172
Concentration: Finance, Economics

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20 Feb 2013, 03:56
You have a kudos for this statement!
100% agree with you...
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Manager
Joined: 24 Jan 2013
Posts: 72

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20 Feb 2013, 17:46
1
1
Solutions provided before were partially correct:

$$\sqrt{X+Y}=4$$ --> pass square root to the other side: X+Y=16;
$$\sqrt{X}-\sqrt{Y}=0$$ --> $$\sqrt{X}=\sqrt{Y}$$ --> pass square root to the other side: X=$$(\sqrt{Y})^2$$ --> X=|Y| --> X=|+Y| or X=|-Y|

So, we have that X+Y=16 and X=|+Y| or X=|-Y| --> we don't care here if Y is positive or negative, as we take always the absolute value |Y| --> therefore this is true always 2X=16 --> X=8.

BUT X=Y is NOT true, as X=|Y|... this means X=8 and Y could be +8 or -8.

Manager
Joined: 24 Jan 2013
Posts: 72

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20 Feb 2013, 17:47
Vips0000 wrote:
Stiv wrote:
In this problem the answer could also be X=Y, isn't it?

Yes, absolutely right.
One reason why only official questions should be used for practice.

Read my post... you are not right in this case
Director
Status: Done with formalities.. and back..
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Posts: 587
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Concentration: Strategy, General Management
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20 Feb 2013, 23:29
1
johnwesley wrote:
Vips0000 wrote:
Stiv wrote:
In this problem the answer could also be X=Y, isn't it?

Yes, absolutely right.
One reason why only official questions should be used for practice.

Read my post... you are not right in this case

Really?

Check your solution once again... or rather.. steps.. its not a solution, if its not correct!

Let me point out basic mistakes!
Your solution is, x= 8, y =8 or -8
Are these value correct for $$\sqrt{x+y}=4$$
if you use x=8 and y=-8 then NO. so y=-8 is NOT a solution

Alright, lets take second equation, $$\sqrt{x}-\sqrt{y}=0$$
Do you think $$\sqrt{8}-\sqrt{-8}=0$$ ?? its NOT.

Stiv was absolutely right there and so was I
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Black Friday Debrief
Manager
Joined: 24 Jan 2013
Posts: 72

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21 Feb 2013, 01:42
Vips0000 wrote:
johnwesley wrote:
Vips0000 wrote:
Yes, absolutely right.
One reason why only official questions should be used for practice.

Read my post... you are not right in this case

Really?

Check your solution once again... or rather.. steps.. its not a solution, if its not correct!

Let me point out basic mistakes!
Your solution is, x= 8, y =8 or -8
Are these value correct for $$\sqrt{x+y}=4$$
if you use x=8 and y=-8 then NO. so y=-8 is NOT a solution

Alright, lets take second equation, $$\sqrt{x}-\sqrt{y}=0$$
Do you think $$\sqrt{8}-\sqrt{-8}=0$$ ?? its NOT.

Stiv was absolutely right there and so was I

Let me tell you (again), that you are wrong, and your response now shows I need to explain slowly:

MY solution is, x= 8, y =8 or -8 AND x=|y|
Completely CORRECT for $$\sqrt{|y|+y}=4$$
BUT you cannot say x=y, as x=|y|

In GMAT, you cannot assume a variable is either positive or negative. USE THE X OR Y AS IF THEY CAN BE POSITIVE OR NEGATIVE, NOBODY TELLS YOU THE OPPOSITE. And I have demonstrated X is always 8, and Y could be +8 or -8, as the square root of X+Y means square root of |Y|+Y. Your solution dismiss important information on the way, and mine is more complete.

Therefore, my solution (again) is CORRECT.
Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 587
Location: India
Concentration: Strategy, General Management
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21 Feb 2013, 01:56
1
johnwesley wrote:
Let me tell you (again), that you are wrong, and your response now shows I need to explain slowly:

MY solution is, x= 8, y =8 or -8 AND x=|y|
Completely CORRECT for $$\sqrt{|y|+y}=4$$
BUT you cannot say x=y, as x=|y|

In GMAT, you cannot assume a variable is either positive or negative. USE THE X OR Y AS IF THEY CAN BE POSITIVE OR NEGATIVE, NOBODY TELLS YOU THE OPPOSITE. And I have demonstrated X is always 8, and Y could be +8 or -8, as the square root of X+Y means square root of |Y|+Y. Your solution dismiss important information on the way, and mine is more complete.

Therefore, my solution (again) is CORRECT.

I'm not sure if you know the basic rules of solving an equation. The end solutions that you get should satisfy the original equation - Only then they are considered a solution.

if x=8 and y=8
both equations are satisfied.

if x=8 and y=-8
none of the equations are satisfied.

Therefore the only solution is, when x=y=8.

Regarding your solution -its absolutely wrong if you talk about Maths in general. And its a blunder, if you talk about GMAT in particular. if Y were to be -8, sqrt (Y) would be undefined on GMAT.
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Black Friday Debrief
Manager
Joined: 24 Jan 2013
Posts: 72

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21 Feb 2013, 02:19
Vips0000 wrote:
johnwesley wrote:
Let me tell you (again), that you are wrong, and your response now shows I need to explain slowly:

MY solution is, x= 8, y =8 or -8 AND x=|y|
Completely CORRECT for $$\sqrt{|y|+y}=4$$
BUT you cannot say x=y, as x=|y|

In GMAT, you cannot assume a variable is either positive or negative. USE THE X OR Y AS IF THEY CAN BE POSITIVE OR NEGATIVE, NOBODY TELLS YOU THE OPPOSITE. And I have demonstrated X is always 8, and Y could be +8 or -8, as the square root of X+Y means square root of |Y|+Y. Your solution dismiss important information on the way, and mine is more complete.

Therefore, my solution (again) is CORRECT.

I'm not sure if you know the basic rules of solving an equation. The end solutions that you get should satisfy the original equation - Only then they are considered a solution.

if x=8 and y=8
both equations are satisfied.

if x=8 and y=-8
none of the equations are satisfied.

Therefore the only solution is, when x=y=8.

Regarding your solution -its absolutely wrong if you talk about Maths in general. And its a blunder, if you talk about GMAT in particular. if Y were to be -8, sqrt (Y) would be undefined on GMAT.

I have explained twice and there is no flaw. It is simply the correct way of solving this kind of problems where nobody tells you X or Y are positive or negative.

My solution is:

Completely CORRECT for $$\sqrt{|y|+y}=4$$

Completely CORRECT for $$\sqrt{x}-\sqrt{y}=0$$, because as you say, this could be undefined... but simply do $$\sqrt{x}=\sqrt{y}$$ and then X=|Y|... magic!... your indefinition has gone while you keep all the possible outcomes of this equation. Your solution does dismiss important and basic information given by the equations.

I am still laughing you say I don't know solving equations.
Manager
Joined: 20 Feb 2009
Posts: 61
Location: chennai

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21 Feb 2013, 03:06
\sqrt{x+y}=4
squaring both sides =>
x+y=16 --- eq(1)
\sqrt{x}-\sqrt{y}=0
squaring both the sides =>
(\sqrt{x}-\sqrt{y})^2 =x-2*\sqrt{xy}+y
implies
x+y=2*\sqrt{xy} --- eq(2)
substitute eq(1) in eq(2)
2*\sqrt{xy}=16
\sqrt{xy}=8
squaring both sides the above eqn =>
xy=64
we knew that x+y=16
y=16-x
x*(16-x)=64
x^2-16x+64=0
hence
x=8

Ans (c)
Director
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21 Feb 2013, 03:27
1
johnwesley wrote:
I have explained twice and there is no flaw. It is simply the correct way of solving this kind of problems where nobody tells you X or Y are positive or negative.

My solution is:

Completely CORRECT for $$\sqrt{|y|+y}=4$$

Completely CORRECT for $$\sqrt{x}-\sqrt{y}=0$$, because as you say, this could be undefined... but simply do $$\sqrt{x}=\sqrt{y}$$ and then X=|Y|... magic!... your indefinition has gone while you keep all the possible outcomes of this equation. Your solution does dismiss important and basic information given by the equations.

I am still laughing you say I don't know solving equations.

Yup, I stand by what I said.

I would say, go back to basics; specially when I see comments like this:
"Completely CORRECT for $$\sqrt{|y|+y}=4$$"
you dont put |y| that you derived as intermediate solution in original equation and check validity of your solution. This is stupidity at best.

And since you are so adamant on proving others wrong (with comments like partially correct solution, you are wrong, again wrong etc). Here is my open challenge to you:
If one expert (Who is also a long time GMAT clubber- so we know its really an expert) says that your solution is correct- I'll get all my kudos (200) transferred to you.
Let me see whats your best bet!
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Black Friday Debrief
Manager
Joined: 24 Jan 2013
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21 Feb 2013, 05:26
Ok, I have no problem to admit my error now I see. In fact $$\sqrt{|y|+y}=4$$ is a complete non-sense for values of y<0. I really don't know how I wrote that stupidity.

All I wanted to do is point out that in GMAT, you cannot assume a variable is either positive or negative. But my calculations where wrong and I promise to be more cautious next time.

And doing numbers again:

$$x+y=16$$ and

$$\sqrt{x}=\sqrt{y}$$ meaning x=|y| (case a) or y=|x| (case b)

I could develop option a.1 (x=|y| and y>0) and option a.2 (x=|y| and y<0), b.1 (y=|x| and x>0) and option b.2 (y=|x| and x<0).

But with the equation $$x+y=16$$ we directly disregard that y<0 or that x<0.

Therefore the solutions can be x=8, and x=y

I never meant detracting your stripes or wings as GMAT master or director, only have fun answering questions. I also never meant to be adamant, but apologies if that is what it looked like, and thanks anyway for the discussion
Director
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21 Feb 2013, 20:35
johnwesley wrote:
Ok, I have no problem to admit my error now I see. In fact $$\sqrt{|y|+y}=4$$ is a complete non-sense for values of y<0. I really don't know how I wrote that stupidity.

Great that you finally saw it

Quote:
And doing numbers again:
$$x+y=16$$ and
$$\sqrt{x}=\sqrt{y}$$ meaning x=|y| (case a) or y=|x| (case b)
I could develop option a.1 (x=|y| and y>0) and option a.2 (x=|y| and y<0), b.1 (y=|x| and x>0) and option b.2 (y=|x| and x<0).
But with the equation $$x+y=16$$ we directly disregard that y<0 or that x<0.
Therefore the solutions can be x=8, and x=y

Again, it doesnt require this much work or effort. if you observe second equation once carefully, you would notice that x and y need to be greater than or equal to 0 in GMAT context (root of negative number not defined in GMAT).
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22 Feb 2013, 03:22
Finally peace, you two were quite interesting guys!!! :D Nobody will make mistake in this question again because you elaborated it to the ground! :D
Thanks to both of you, this is why I love GMATClub...
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Intern
Joined: 16 Jan 2013
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22 Feb 2013, 07:48
johnwesley wrote:
Solutions provided before were partially correct:

$$\sqrt{X+Y}=4$$ --> pass square root to the other side: X+Y=16;
$$\sqrt{X}-\sqrt{Y}=0$$ --> $$\sqrt{X}=\sqrt{Y}$$ --> pass square root to the other side: X=$$(\sqrt{Y})^2$$ --> X=|Y| --> X=|+Y| or X=|-Y|

So, we have that X+Y=16 and X=|+Y| or X=|-Y| --> we don't care here if Y is positive or negative, as we take always the absolute value |Y| --> therefore this is true always 2X=16 --> X=8.

BUT X=Y is NOT true, as X=|Y|... this means X=8 and Y could be +8 or -8.

y cannot be -8 as square_root_of(-8) is an imaginary number, if nothing is mentioned we assume the number to be as real numbers
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Joined: 06 Sep 2013
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08 Oct 2013, 13:57
feellikequitting wrote:
$$\sqrt{X+Y} = 4$$ and $$\sqrt{X}-\sqrt{Y}=0$$

A. X= 2
B. X = -8
C. X= 8
D. X = -2
E. X=Y

Detailed Solutions = Kudos!
Kudos if you liked!

Why is answer (E) given anyways? Technically it could be correct as well no? Cause x=y=8.

Nice question, thks
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08 Oct 2013, 22:00
3
1
jlgdr wrote:
feellikequitting wrote:
$$\sqrt{X+Y} = 4$$ and $$\sqrt{X}-\sqrt{Y}=0$$

A. X= 2
B. X = -8
C. X= 8
D. X = -2
E. X=Y

Detailed Solutions = Kudos!
Kudos if you liked!

Why is answer (E) given anyways? Technically it could be correct as well no? Cause x=y=8.

Nice question, thks

Actually, technically, x = y is not a solution to this set of equations. We cannot say that wherever x = y, the relations will hold. Say if x = y = 2, the first equation will not hold.

But when x = 8, you put it in the second equation, get y = 8 (so you get x = y as an intermediate result) and then you put both x = 8 and y = 8 in equation 1 which holds for these values.
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Karishma
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19 Nov 2013, 15:31
VeritasPrepKarishma wrote:
jlgdr wrote:
feellikequitting wrote:
$$\sqrt{X+Y} = 4$$ and $$\sqrt{X}-\sqrt{Y}=0$$

A. X= 2
B. X = -8
C. X= 8
D. X = -2
E. X=Y

Detailed Solutions = Kudos!
Kudos if you liked!

Why is answer (E) given anyways? Technically it could be correct as well no? Cause x=y=8.

Nice question, thks

Actually, technically, x = y is not a solution to this set of equations. We cannot say that wherever x = y, the relations will hold. Say if x = y = 2, the first equation will not hold.

But when x = 8, you put it in the second equation, get y = 8 (so you get x = y as an intermediate result) and then you put both x = 8 and y = 8 in equation 1 which holds for these values.

First I was going for E, but I remebered that X=Y "actually, technically, is not a solution". End of discussion.

Thanks VeritasPrepKarishma
Re: root(x+y)=4 and root(x)-root(y)=0   [#permalink] 19 Nov 2013, 15:31

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