Hello I have a problem.

We have X+Y+Z+W=X²+Y²+Z²+W²=0

What the value of x,y,z,w?

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- May 12th 2006, 12:10 AMsopheaHello! I have a problem.
Hello I have a problem.

We have X+Y+Z+W=X²+Y²+Z²+W²=0

What the value of x,y,z,w? - May 12th 2006, 12:41 AMCaptainBlackQuote:

Originally Posted by**sophea**

Since if their sum is they must individualy all be [math0[/tex], so:

RonL

PS You should have made your post a new thread not tacked it on the the

end of an old unrelated thread - May 21st 2006, 06:26 PMsopheaQuote:

Originally Posted by**CaptainBlack**

now i have one more problem:

X².X²=2X² +[X]

Note: X-1<[X] <X - May 21st 2006, 09:55 PMCaptainBlackQuote:

Originally Posted by**sophea**

instance the inequality cannot hold if . So plug in in

turn for and then solve for . Keep any solutions for which is

equal to the assumed value used.

As an example we will work the case where we assume .

Then we want solutions of:

which is a quadratic in , so using the quadratic formula:

,

which is complex and so not an admissible solution (I will assume we want

real solutions).

Now lets try assuming , then:

and so are solutions of this equation and is consistent with

our assumption and so is a solution of the original problem.

Carrying on like this you will find all of the remaining solutions.

RonL - May 22nd 2006, 07:48 PMmpgc_acQuote:

Originally Posted by**CaptainBlack**

How smart you are!

I'm realy happy to saw your solution.

Now i have one more problem and I hope you are not feel borred because I always have more and more problem.

My problem is:

In ABC rectangle have BO bisects CBA , CO bisects ACB, MN // BC .

FInd the perimeter of AMN rectangle. Thank you advance.

my email address is "sophearot@yahoo.ca"