# Proof In a System of Equations

• Dec 25th 2008, 06:03 AM
Winding Function
Proof In a System of Equations
Prove that x=y=z in the equations

$x+y^2+z^4=0$
$y+z^2+x^4=0$
$z+x^2+y^4=0$

where x, y, and z are real numbers.
• Dec 26th 2008, 10:26 AM
Jester
Quote:

Originally Posted by Winding Function
Prove that x=y=z in the equations

$x+y^2+z^4=0$
$y+z^2+x^4=0$
$z+x^2+y^4=0$

where x, y, and z are real numbers.

Can you give us some indication as where this problem comes from?
• Dec 26th 2008, 07:02 PM
Winding Function
Quote:

Originally Posted by danny arrigo
Can you give us some indication as where this problem comes from?

An Alg2/Trig Text. Is it too easy/difficult?
• Dec 26th 2008, 07:30 PM
Mathstud28
Quote:

Originally Posted by Winding Function
Prove that x=y=z in the equations

$x+y^2+z^4=0$
$y+z^2+x^4=0$
$z+x^2+y^4=0$

where x, y, and z are real numbers.

I'm sorry, are you saying prove that $x=y=z$ is the only solution?
• Dec 27th 2008, 01:32 AM
susrut
Dummy bro = plz read books
x=-(y^2+z^4)

y=-(z^2+x^4)

z=-(x^2+y^4)

let us consider that x=y=z and let us assume a real value which is positive.
clearly any real number raised to even powers like 2 or 4 must give positive values
as seen from the equation a positive value cannot be equal to a negative value. so the question is wrong.

now let us assume a real negative value.
x=y=z=-a
clearly LHS of the equations are negative.
-a raised to powers of 2 or 4 will produce positive values which are much larger or smaller (in case of negative fractions) than -a; then
due to the negative sign outside bracket RHS become negative values which are much smaller or larger (in case of negative fractions) than -a

Hence we draw conclusions that the equations are wrong.
• Dec 27th 2008, 02:06 AM
flyingsquirrel
Quote:

Originally Posted by susrut
x=-(y^2+z^4)

y=-(z^2+x^4)

z=-(x^2+y^4)

[...] now let us assume a real negative value.
x=y=z=-a
clearly LHS of the equations are negative.
-a raised to powers of 2 or 4 will produce positive values which are much larger or smaller (in case of negative fractions) than -a; then
due to the negative sign outside bracket RHS become negative values which are much smaller or larger (in case of negative fractions) than -a

HENCE WE DRAW CONCLUSIONS THAT THE EQUATIONS ARE WRONG!!!!!!

The equation $-a=-(a^2+a^4)\Longleftrightarrow a(1-a-a^3)=0$ has at least one nontrivial solution because the polynomial $1-X-X^3$ has one real root... The equations are not wrong.
• Dec 27th 2008, 02:38 AM
Sean12345
susrut,

It's very rude to use a title such as "Dummy bro = plz read books".
• Dec 27th 2008, 05:52 AM
Winding Function
Quote:

Originally Posted by Mathstud28
I'm sorry, are you saying prove that $x=y=z$ is the only solution?

I'm trying to prove that for any set of solutions, (x,y,z), x=y=z.
• Dec 27th 2008, 09:24 PM
susrut
HELLO FLYINGSQUIRREL

first of all i thank you

1-X-X^3=0 has one real root which comes to be NEARLY equal to 0.68
so x=y=z~0.68 is a solution to the equations
but it is not a strong enough proof as it is a backdoor approach

SORRY TO ALL WHO FELT MY TITLE ON LAST POST ON THIS PAGE RUDE (Nerd)
• Dec 28th 2008, 12:40 AM
flyingsquirrel
Quote:

Originally Posted by susrut
1-X-X^3=0 has one real root which comes to be NEARLY equal to 0.68
so x=y=z~0.68 is a solution to the equations
but it is not a strong enough proof as it is a backdoor approach

I did not meant to prove that $x=y=z$, I just wanted to show that the three equations we are given are correct.