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

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

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New post 28 Mar 2016, 22:43
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A
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  85% (hard)

Question Stats:

54% (02:37) correct 46% (02:23) wrong based on 222 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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New post 29 Mar 2016, 00:51
1
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


\(\sqrt{x} + \sqrt{z} = \sqrt{y}\)

\(x + z + 2\sqrt{xz} = y\)

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

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

\(x^2 + z^2 + y^2 + 2xz - 2yz - 2xy = 4xz\)

\(x^2 + y^2 + z^2 = 2xy + 2yz + 2xz\)

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

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New post 30 Mar 2016, 23:21
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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New post 25 Dec 2017, 05:49
1
1
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
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Re: If √x -√y+√z=0, which of the following is true?  [#permalink]

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New post 27 Dec 2017, 09:51
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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If √x -√y+√z=0, which of the following is true?  [#permalink]

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New post 27 Dec 2017, 09:56
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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New post 09 Jan 2018, 08:31
Is this a 700 level question?
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Re: If √x -√y+√z=0, which of the following is true?  [#permalink]

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New post 09 Jan 2018, 08:37
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? &nbs [#permalink] 09 Jan 2018, 08:37
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