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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
If √x -√y+√z=0, which of the following is true?  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 52% (02:38) correct 48% (02:37) wrong based on 189 sessions

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If $$√x -√y+√z=$$0, which of the following is true?

A. $$x^2+y^2+z^2=xy+xz+yz$$

B. $$x^2+y^2+z^2=2xy+2xz+2yz$$

C. $$x^2+y^2-z^2=2xy+2xz-2yz$$

D. $$x^2-y^2-z^2=xy+xz+yz$$

E. $$x^4+y^4+z^4=2xy+2xz+2yz$$

* A solution will be posted in two days.

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Re: If √x -√y+√z=0, which of the following is true?  [#permalink]

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MathRevolution wrote:
If √x -√y+√z=0, which of the following is true?

A. x^2+y^2+z^2=xy+xz+yz
B. x^2+y^2+z^2=2xy+2xz+2yz
C. x^2+y^2-z^2=2xy+2xz-2yz
D. x^2-y^2-z^2=xy+xz+yz
E. x^4+y^4+z^4=2xy+2xz+2yz

$$\sqrt{x} + \sqrt{z} = \sqrt{y}$$

$$x + z + 2\sqrt{xz} = y$$

$$x + z - y = -2\sqrt{xz}$$

$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

$$x^2 + z^2 + y^2 + 2xz - 2yz - 2xy = 4xz$$

$$x^2 + y^2 + z^2 = 2xy + 2yz + 2xz$$

Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8025
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If √x -√y+√z=0, which of the following is true?  [#permalink]

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If √x -√y+√z=0, which of the following is true?

A. x^2+y^2+z^2=xy+xz+yz
B. x^2+y^2+z^2=2xy+2xz+2yz
C. x^2+y^2-z^2=2xy+2xz-2yz
D. x^2-y^2-z^2=xy+xz+yz
E. x^4+y^4+z^4=2xy+2xz+2yz

If you square the both equations, they become x+z+2√xz=y, x+z-y=-2√xz. If you once again square the both equations, x^2+z^2+y^2+2zx-2xy-2zy=4xz is derived and eventually becomes x^2+y^2+z^2=2xy+2yz+2zx.
Thus, the answer is B.
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Re: If √x -√y+√z=0, which of the following is true?  [#permalink]

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MathRevolution wrote:
If √x -√y+√z=0, which of the following is true?

A. x^2+y^2+z^2=xy+xz+yz
B. x^2+y^2+z^2=2xy+2xz+2yz
C. x^2+y^2-z^2=2xy+2xz-2yz
D. x^2-y^2-z^2=xy+xz+yz
E. x^4+y^4+z^4=2xy+2xz+2yz

* A solution will be posted in two days.

In case one does not remembers the formula, plugging smart numbers will be highly effective.

Let $$x=z=1$$ & $$y=4$$. this satisfies our stem equation. Test the values in the options and the option where $$LHS=RHS$$ will be correct

For Option B $$LHS=1^2+4^2+1^2=18$$

$$RHS=2*1*4+2*1*1+2*4*1=18$$.

For other options, LHS will not be equal to RHS
Intern  Joined: 16 Dec 2017
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Re: If √x -√y+√z=0, which of the following is true?  [#permalink]

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MathRevolution wrote:
If √x -√y+√z=0, which of the following is true?

A. x^2+y^2+z^2=xy+xz+yz
B. x^2+y^2+z^2=2xy+2xz+2yz
C. x^2+y^2-z^2=2xy+2xz-2yz
D. x^2-y^2-z^2=xy+xz+yz
E. x^4+y^4+z^4=2xy+2xz+2yz

If you square the both equations, they become x+z+2√xz=y, x+z-y=-2√xz. If you once again square the both equations, x^2+z^2+y^2+2zx-2xy-2zy=4xz is derived and eventually becomes x^2+y^2+z^2=2xy+2yz+2zx.
Thus, the answer is B.

Thank you. Could you help me understand where the highlighted portion came from?
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Joined: 25 Feb 2013
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Location: India
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If √x -√y+√z=0, which of the following is true?  [#permalink]

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rb60006 wrote:
MathRevolution wrote:
If √x -√y+√z=0, which of the following is true?

A. x^2+y^2+z^2=xy+xz+yz
B. x^2+y^2+z^2=2xy+2xz+2yz
C. x^2+y^2-z^2=2xy+2xz-2yz
D. x^2-y^2-z^2=xy+xz+yz
E. x^4+y^4+z^4=2xy+2xz+2yz

If you square the both equations, they become x+z+2√xz=y, x+z-y=-2√xz. If you once again square the both equations, x^2+z^2+y^2+2zx-2xy-2zy=4xz is derived and eventually becomes x^2+y^2+z^2=2xy+2yz+2zx.
Thus, the answer is B.

Thank you. Could you help me understand where the highlighted portion came from?

Hi rb60006

we can rearrange the equation and write $$\sqrt{x}+\sqrt{z}=\sqrt{y}$$, so on squaring it will be of the form $$(a+b)^2=a^2+b^2+2ab$$
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Re: If √x -√y+√z=0, which of the following is true?  [#permalink]

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Is this a 700 level question?
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Joined: 25 Feb 2013
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Re: If √x -√y+√z=0, which of the following is true?  [#permalink]

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kritikalal wrote:
Is this a 700 level question?

Hi kritikalal

Check the "Tag" section just above the question. you will get all the details
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Re: If √x -√y+√z=0, which of the following is true?  [#permalink]

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_________________ Re: If √x -√y+√z=0, which of the following is true?   [#permalink] 14 May 2019, 00:20
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