Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
May 1408:30 AM PDT
-09:30 AM PDT
Most GMAT test-takers are intimidated by the hardest GMAT Verbal questions. In this session, Target Test Prep GMAT instructor Erika Tyler-John, a 100th percentile GMAT scorer, will show you how top scorers break down challenging Verbal questions..
May 1212:00 PM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.- May 13
08:30 AM PDT
-09:30 AM PDT
In Episode 4 of our GMAT Ninja CR series, we tackle the most intimidating CR question type: Boldface & "Legalese" questions. If you've ever stared at an answer choice that reads, "The first is a consideration introduced to counter a position that... - Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
06 Jun 2017, 02:10
Kudos
Bookmarks
Knowing that x + y + z = 4
x^2 + x^2 + z^2 = 4 and
x^3 + y^3 + z^3 = 4
Solving the equation with complex numbers
I know that: 4 = x^2 + y^2 + z^2 = (x+y+z)^2 - (x*y + x*z + y*z). So (x*y + x*z + y*z) = 12
4 = x^3 + y^3 + z^3 = (x+y+z^3 - 3*(x^2*y + x^2*z + x*y^2 + x*z^2 + y^2*z + y*z^2) - 6*x*y*z.
Trying to find (x^2*y + x^2*z + x*y^2 + x*z^2 + y^2*z + y*z^2) =
x*(x*y + x*z) + y*(x*y + y*z) + z*(x*z + y*z). Knowing that x+y+z = 4, I'm using that info so:
x*(x*y + x*z + y*z) + y*(x*y + y*z + x*z) + z*(x*z + y*z + x*y) - x*y*z - x*y*z - x*y*z.
we get, (x*y + x*z + y*z)*(x+y+z) - 3*x*y*z
Once I find what x*y*z I'm getting stuck in, how do I find the x,y,z (that are allegedly complex numbers) ?
x^2 + x^2 + z^2 = 4 and
x^3 + y^3 + z^3 = 4
Solving the equation with complex numbers
I know that: 4 = x^2 + y^2 + z^2 = (x+y+z)^2 - (x*y + x*z + y*z). So (x*y + x*z + y*z) = 12
4 = x^3 + y^3 + z^3 = (x+y+z^3 - 3*(x^2*y + x^2*z + x*y^2 + x*z^2 + y^2*z + y*z^2) - 6*x*y*z.
Trying to find (x^2*y + x^2*z + x*y^2 + x*z^2 + y^2*z + y*z^2) =
x*(x*y + x*z) + y*(x*y + y*z) + z*(x*z + y*z). Knowing that x+y+z = 4, I'm using that info so:
x*(x*y + x*z + y*z) + y*(x*y + y*z + x*z) + z*(x*z + y*z + x*y) - x*y*z - x*y*z - x*y*z.
we get, (x*y + x*z + y*z)*(x+y+z) - 3*x*y*z
Once I find what x*y*z I'm getting stuck in, how do I find the x,y,z (that are allegedly complex numbers) ?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
06 Jun 2017, 10:07
Kudos
Bookmarks
mihaitudor
Dear mihaitudor,
I'm happy to respond.
First of all, I assume you know that this math is several levels beyond anything on the GMAT. There are no complex numbers on the GMAT. This is 100% off topic for the GMAT Quant. I want that to be crystal clear to everyone who reads this post.
When I first looked at this problem, I played around with algebra, and saw that this quickly led to a morass. I was intrigued by the symmetry, so I started playing around with numbers. Obviously, the numbers couldn't be three symmetrical complex numbers, each 120 degrees apart, because those would sum to zero. After a little playing around with numbers, I found the solution.
x = 2
y = 1 + i
z = 1 - i
x + y + z = 4
\(x^2\) = 4
\(y^2\) = 2i
\(z^2\) = -2i
\(x^2 + y^2 + z^2 = 4\)
\(x^3\) = 8
\(y^3\) = -2 +2i
\(z^3\) = -2 - 2i
\(x^3 + y^3 + z^3 = 4\)
It makes sense that the solution are three relatively simple complex numbers. The deviant who created this problem no doubt noticed what happened with powers of (1 + i) and (1 - i) and figured he would make this stumper of a problem.
Does all this make sense?
Mike
06 Jun 2017, 22:40
Kudos
Bookmarks
Note that this problem is too hard for GMAT. Just for anyone who enjoys solving Math.
We have
\(x+y+z=4 \quad \quad\) (1)
\(x^2+y^2+z^2=4 \quad\) (2)
\(x^3+y^3+z^3 = 4 \quad\) (3)
From (1) we have \((x+y+z)^2=16 \implies (x^2+y^2+z^2)+2(xy+yz+zx)=16 \implies xy+yz+zx=6 \quad\) (4)
Also, from (1) we have \(x+y=4-z \quad\) (5)
From (2) we have \(x^2+y^2=(x+y)^2 - 2xy=4-z^2 \)
From (5) we have \( (4-z)^2 - 2xy = 4-z^2 \implies xy=z^2-4z+6 \quad\) (6)
Now, from (3) we have
\(x^3+y^3=4-z^3 \implies (x+y)^3 - 3xy(x+y) = 4-z^3\)
from (5) and (6) we have
\((4-z)^3-3(z^2-4z+6)(4-z)=4-z^3\)
Shorten this expression, we have
\(z^3-4z^2+6z-4=0\)
We have
\(z^3-4z^2+6z-4 \\
=(z^3-4z^2+4z)+(2z-4) \\
=z(z-2)^2+2(z-2) \\
=(z-2)(z(z-2)+2) \\
=(z-2)((z-1)^2+1)\)
Hence \((z-2)((z-1)^2+1) = 0 \implies z=2\) or \((z-1)^2+1=0 \implies z=2, 1 \pm i\)
So the root of the problem is \((1+i, 1-i, 2)\)
We have
\(x+y+z=4 \quad \quad\) (1)
\(x^2+y^2+z^2=4 \quad\) (2)
\(x^3+y^3+z^3 = 4 \quad\) (3)
From (1) we have \((x+y+z)^2=16 \implies (x^2+y^2+z^2)+2(xy+yz+zx)=16 \implies xy+yz+zx=6 \quad\) (4)
Also, from (1) we have \(x+y=4-z \quad\) (5)
From (2) we have \(x^2+y^2=(x+y)^2 - 2xy=4-z^2 \)
From (5) we have \( (4-z)^2 - 2xy = 4-z^2 \implies xy=z^2-4z+6 \quad\) (6)
Now, from (3) we have
\(x^3+y^3=4-z^3 \implies (x+y)^3 - 3xy(x+y) = 4-z^3\)
from (5) and (6) we have
\((4-z)^3-3(z^2-4z+6)(4-z)=4-z^3\)
Shorten this expression, we have
\(z^3-4z^2+6z-4=0\)
We have
\(z^3-4z^2+6z-4 \\
=(z^3-4z^2+4z)+(2z-4) \\
=z(z-2)^2+2(z-2) \\
=(z-2)(z(z-2)+2) \\
=(z-2)((z-1)^2+1)\)
Hence \((z-2)((z-1)^2+1) = 0 \implies z=2\) or \((z-1)^2+1=0 \implies z=2, 1 \pm i\)
So the root of the problem is \((1+i, 1-i, 2)\)
