Prove that x=y=z in the equations

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

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

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

where x, y, and z are real numbers.

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- Dec 25th 2008, 06:03 AMWinding FunctionProof In a System of Equations
Prove that x=y=z in the equations

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

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

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

where x, y, and z are real numbers. - Dec 26th 2008, 10:26 AMJester
- Dec 26th 2008, 07:02 PMWinding Function
- Dec 26th 2008, 07:30 PMMathstud28
- Dec 27th 2008, 01:32 AMsusrutDummy 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 AMflyingsquirrel
- Dec 27th 2008, 02:38 AMSean12345
susrut,

It's very rude to use a title such as "**Dummy bro = plz read books**". - Dec 27th 2008, 05:52 AMWinding Function
- Dec 27th 2008, 09:24 PMsusrut
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

please reply

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