Proof 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.

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Originally Posted by Winding Function

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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Can you give us some indication as where this problem comes from?

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Originally Posted by danny arrigo

Can you give us some indication as where this problem comes from?

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An Alg2/Trig Text. Is it too easy/difficult?

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Originally Posted by Winding Function

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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I'm sorry, are you saying prove that $\displaystyle x=y=z$ is the only solution?

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.

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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!!!!!!

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The equation $\displaystyle -a=-(a^2+a^4)\Longleftrightarrow a(1-a-a^3)=0$ has at least one nontrivial solution because the polynomial $\displaystyle 1-X-X^3$ has one real root... The equations are not wrong.

susrut,

It's very rude to use a title such as "Dummy bro = plz read books".

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Originally Posted by Mathstud28

I'm sorry, are you saying prove that $\displaystyle x=y=z$ is the only solution?

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I'm trying to prove that for any set of solutions, (x,y,z), x=y=z.

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)

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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

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