Thread: math problems

Hello I have a problem.
We have X+Y+Z+W=X²+Y²+Z²+W²=0
What the value of x,y,z,w?

Originally Posted by sophea

Hello I have a problem.
We have X+Y+Z+W=X²+Y²+Z²+W²=0
What the value of x,y,z,w?

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

Originally Posted by CaptainBlack

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

Ok! thank you very for you solution.
now i have one more problem:
X².X²=2X² +[X]
Note: X-1<[X] <X

Originally Posted by sophea

Ok! thank you very for you solution.
now i have one more problem:
X².X²=2X² +[X]
Note: X-1<[X] <X

There is a limited range of over which this can have a solution. For
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

Originally Posted by CaptainBlack

There is a limited range of over which this can have a solution. For
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

Dear sir!
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"