Thread: equations help

solve for x+y+z if
19x^3+95x+57=0
19y^3+95y+57=0
19z^3+95z+57=0
x,yand z are not equal to each other

Originally Posted by kahandasa

solve for x+y+z if
19x^3+95x+57=0
19y^3+95y+57=0
19z^3+95z+57=0

All of the equations are identical, so x = y = z. Just solve the first equation for x (if possible) and then you will have y and z as well.

ya but it is very diificullt. the only way is to play with the eqation and obtain x+y+z

Hello, kahandasa!

A very strange problem . . .

. .

. .

i dont understand why x+y+z=0

Originally Posted by kahandasa

i dont understand why x+y+z=0

Viete's relations: the sum of the roots of an n-th degree polynomial equation is -1 times the coefficient of the (n-1)-th degree term divided by the coefficient of the n-th degree term. Since there is no term in the square of the variable and we have a cubic the sum of the roots is zero.

In this case we can be explicit, with soroban's notation:



etc..

Also all the roots are different since by Descartes' rule of signs there are zero positive roots and one negative root, hence exactly one real root, and as this is a polynomial with real coefficients the remaining two complex roots are a conjugate pair and hence not equal.

CB