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

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10 Dec 2014, 03:21
Can someone please update OA (Option E) of this question??

x = y;

Can this be changed to something x = 4?

Thanks
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28 Dec 2014, 02:17
I totally agree that X=Y is not a solution, but a equation that tells that X=Y now we need to find the value THAT satisfies the previous equation sqrt(X+Y)=4.

I an another way,After squaring both sides,it can be sen that these are two linear equations and we need to find out the point of intersection of these two lines.
one line is
X=Y and another is X+Y= 16 (after squaring) , plotting these two lines on the graph we can find the solution which will be unique.
Two straight lines with different slopes have only one point of intersection.

Solution:

As X=Y, sqrt(2X)=4
squaring both the sides
2X=16
X=8
Bingo !!!!

Paresh , I think it is nice to put such options, it tests clarity of concepts
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21 May 2017, 19:49
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

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The best solution here is as follows:

sqrt(x + y) = 4
x+y = 16

Meanwhile, we know:
sqrt(x) -sqrt(y) = 0
sqrt(x) = sqrt(y)
x = y

We can square both sides because only the positive root is valid here.

2x = 16
x = 8
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12 Jul 2017, 06:36
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.

That's not correct. If a negative number is put under a root, it is bound to give an imaginary number. Which is why, E should be correct.
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12 Jul 2017, 06:38
Bunuel wrote:
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 $$x=y$$ --> $$x+x=16$$ --> $$x=8$$.

Why isn't E correct?
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12 Jul 2017, 06:41
rekhabishop wrote:
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.

That's not correct. If a negative number is put under a root, it is bound to give an imaginary number. Which is why, E should be correct.

x and y cannot be negative because even roots from negative numbers are not defined for the GMAT. All number on the GMAT are by default real numbers.
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12 Jul 2017, 06:50
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.

I'm not getting this. How can any of the two variables or both the variables be negative? Isn't root of any negative number equal to an imaginary number?
Can you put values of x and y and show how you're claiming that x is not equal to y?

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Re: root(x+y)=4 and root(x)-root(y)=0   [#permalink] 12 Jul 2017, 06:50

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