Prove that x=y=z in the equations

where x, y, and z are real numbers.

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- December 25th 2008, 07:03 AMWinding FunctionProof In a System of Equations
Prove that x=y=z in the equations

where x, y, and z are real numbers. - December 26th 2008, 11:26 AMJester
- December 26th 2008, 08:02 PMWinding Function
- December 26th 2008, 08:30 PMMathstud28
- December 27th 2008, 02:32 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. - December 27th 2008, 03:06 AMflyingsquirrel
- December 27th 2008, 03:38 AMSean12345
susrut,

It's very rude to use a title such as "**Dummy bro = plz read books**". - December 27th 2008, 06:52 AMWinding Function
- December 27th 2008, 10:24 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) - December 28th 2008, 01:40 AMflyingsquirrel