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If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z?

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If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z?  [#permalink]

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New post 03 Mar 2019, 17:03
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A
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D
E

Difficulty:

  65% (hard)

Question Stats:

42% (02:20) correct 58% (02:20) wrong based on 33 sessions

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GMATH practice exercise (Quant Class 3)

If \({\left( {2x + y - z} \right)^2} + {\left( {x - y} \right)^2} + {\left( {z - 3} \right)^2} = - {\left( {3y - z} \right)^2}\) , what is the value of \(3x + 2y + z\) ?

(A) 8
(B) 9
(C) 10
(D) 11
(E) 12

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net

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Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z?  [#permalink]

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New post 03 Mar 2019, 19:08
1
(2x+y−z)^2+(x−y)^2+(z−3)^2=−(3y−z)^2
(2x+y−z)^2+(x−y)^2+(z−3)^2+(3y−z)^2=0
As sum of square is 0.so each term must be 0
2x+y-z=0
x-y=0
z-3=0
3y-z=0
solving these equation,
x=y=1
z=3
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Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z?  [#permalink]

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New post 04 Mar 2019, 05:43
fskilnik wrote:
GMATH practice exercise (Quant Class 3)

If \({\left( {2x + y - z} \right)^2} + {\left( {x - y} \right)^2} + {\left( {z - 3} \right)^2} = - {\left( {3y - z} \right)^2}\) , what is the value of \(3x + 2y + z\) ?

(A) 8
(B) 9
(C) 10
(D) 11
(E) 12

\(? = 3x + 2y + z\)

\({\left( {2x + y - z} \right)^2} + {\left( {x - y} \right)^2} + {\left( {z - 3} \right)^2} = - {\left( {3y - z} \right)^2}\,\,\,\,\,\left( * \right)\)

\(\left. \matrix{
{\left( {2x + y - z} \right)^2} + {\left( {x - y} \right)^2} + {\left( {z - 3} \right)^2}\,\,\, \ge 0\,\, \hfill \cr
- {\left( {3y - z} \right)^2}\,\, \le 0 \hfill \cr} \right\}\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,{\left( {2x + y - z} \right)^2} + {\left( {x - y} \right)^2} + {\left( {z - 3} \right)^2} = 0\,\,\,\,\,\,\left[ { = - {{\left( {3y - z} \right)}^2}} \right]\)


\({\left( {2x + y - z} \right)^2} + {\left( {x - y} \right)^2} + {\left( {z - 3} \right)^2} = 0\,\,\,\,\,\,\, \Rightarrow \,\left\{ \matrix{
\,x - y = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,x = y\,\,\, \hfill \cr
\,z - 3 = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,z = 3\,\,\,\, \hfill \cr
\,2x + y - z = 0\,\,\,\,\mathop \Rightarrow \limits^{{\rm{both}}\,{\rm{above}}} \,\,\,\,\,3x - 3 = 0\, \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left( {x,y,z} \right) = \left( {1,1,3} \right)\)


\(? = 3x + 2y + z = 3\left( 1 \right) + 2\left( 1 \right) + 3 = 8\)


The correct answer is (A).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z?  [#permalink]

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New post 04 Mar 2019, 13:48
Hello fskilnik

So its possible just to find the value of z by taking any of the roots?

Like:

\((z-3)^2 = 0\)

\(z = 3\)

And then just substitute?

Kind regards!
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If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z?  [#permalink]

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New post 04 Mar 2019, 16:19
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jfranciscocuencag wrote:
Hello fskilnik

So its possible just to find the value of z by taking any of the roots?

Like:

\((z-3)^2 = 0\)

\(z = 3\)

And then just substitute?

Kind regards!


Hi jfranciscocuencag !!

Thank you for your interest in my problem/solution.

Please be sure you understand the importance of EVEN POWERS, in the sense that (real number)^(positive even power) is always nonnegative...

That´s the reason why (for instance) \(a^2+(2b)^4+(c+3)^6 = 0\) implies \(a^2=0\) and \((2b)^4=0\) and \((c+3)^6=0\) ...

That mentioned/understood, note that there is no need to "take (square, fourth or sixth) roots" in each addend above, because a square (or a fourth or a sixth power) of a real number is zero only if the number itself is zero... (If the number is negative or positive, the square, fourth and sixth powers are positive.) In other words, we get immediately \(a=0\) and \(2b=0\) and \(c+3=0\) in the scenario created... nice, isn´t it?!

I hope everything is clear now!

Regards and success in your studies,
Fabio.

P.S.: do not forget to press "Mention this user" palette before you call someone here (in this case, me, fskilnik). Reason: I came to your question by "luck", only...
(If you press the "Mention this user" button, then the person you call receive a message... the same way I called you here... got it?!) Thank you!
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Our high-level "quant" preparation starts here: https://gmath.net

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If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z?   [#permalink] 04 Mar 2019, 16:19
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