The discussion revolves around finding the solutions to the fourth degree polynomial equation (x^4)-(3x^3)-(3x^2)+2 = 0. Participants explore methods to find the remaining roots after two complex solutions are proposed, while addressing potential errors in the initial claims about these roots.
There is disagreement regarding the validity of the proposed roots. Some participants assert that the roots are incorrect, while others initially accepted them without verification. The discussion remains unresolved regarding the correct roots of the polynomial.
Participants note that the discussion hinges on the accuracy of the initial polynomial equation and the proposed roots, which may have been miscommunicated. The mathematical steps to verify the roots are not fully explored.
Readers interested in polynomial equations, root-finding methods, and mathematical problem-solving may find this discussion relevant.
j-lee00 said:Here it is
(x^4)-(3x^3)-(3x^2)+2 = 0
Two solutions are
x = 1 - i
x = 1 + i
How can i find the other solutions, (not graphically)
Thanks
FrogPad said:General solution:
http://en.wikipedia.org/wiki/Quartic_equation
Numerical method:
http://en.wikipedia.org/wiki/Newton's_method[/QUOTE]
There's no need for that. OP already has two of the roots so all that's required is to divide the original quartic by (x^2 - 2x + 2) and factorize the resultant quadratic.
BTW. (x^2 - 2x + 2) = (x-(1+i)) (x-(1-i))
uart said:There's no need for that. OP already has two of the roots so all that's required is to divide the original quartic by (x^2 - 2x + 2) and factorize the resultant quadratic.
BTW. (x^2 - 2x + 2) = (x-(1+i)) (x-(1-i))
HallsofIvy said:One serious problem the OP has is that 1-i and 1+ i are NOT roots of
x4-3x3-3x2+2= 0!