Last visit was: 19 May 2024, 12:06 It is currently 19 May 2024, 12:06
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Tutor
Joined: 12 Oct 2010
Status:GMATH founder
Posts: 892
Own Kudos [?]: 1382 [17]
Given Kudos: 56
Send PM
Most Helpful Reply
Intern
Intern
Joined: 18 Nov 2018
Posts: 1
Own Kudos [?]: 8 [8]
Given Kudos: 73
General Discussion
Tutor
Joined: 12 Oct 2010
Status:GMATH founder
Posts: 892
Own Kudos [?]: 1382 [3]
Given Kudos: 56
Send PM
Manager
Manager
Joined: 12 Sep 2017
Posts: 239
Own Kudos [?]: 117 [1]
Given Kudos: 132
Send PM
Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z? [#permalink]
1
Kudos
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!
Tutor
Joined: 12 Oct 2010
Status:GMATH founder
Posts: 892
Own Kudos [?]: 1382 [2]
Given Kudos: 56
Send PM
Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z? [#permalink]
1
Kudos
Expert Reply
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!
User avatar
Intern
Intern
Joined: 26 Apr 2024
Posts: 5
Own Kudos [?]: 2 [0]
Given Kudos: 0
Send PM
Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z? [#permalink]
To tackle this problem, we break it down by setting each squared term in the equation to zero since the sum of squares on the left must equal the negative square on the right, which only makes sense if they're all zero. This gives us:

1. 2x + y - z = 0
2. x - y = 0
3. z - 3 = 0
4. 3y - z = 0

From these, it quickly unfolds:
x = y and z = 3
Plugging z = 3 into 3y - z = 0 simplifies to y = 1, and hence, x = 1

Finally, substituting x, y, and z into 3x + 2y + z calculates to 3(1) + 2(1) + 3 = 8

So, the answer is 8, choice (A). This method shows how breaking problems into smaller parts can clarify and simplify the solution process.
GMAT Club Bot
Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z? [#permalink]
Moderators:
Math Expert
93341 posts
Senior Moderator - Masters Forum
3137 posts