Hi
I tried to find all values of x, real and complex in 3x^3-11x^2+25x-25=0
but I don't get the answer. I used quadratic X(3x^2-11x+25)-25=0 but that didn’t work.
Do I need to use rational roots test and long division to solve this?
Hi
I tried to find all values of x, real and complex in 3x^3-11x^2+25x-25=0
but I don't get the answer. I used quadratic X(3x^2-11x+25)-25=0 but that didn’t work.
Do I need to use rational roots test and long division to solve this?
Or you'll have to use the method by Tartaglia- del Ferro, which is a very nasty thing, but sometimes there's no other way: Cubic function - Wikipedia, the free encyclopedia
Tonio
Your question, as written, "I tried to find all values of x, real and complex in $\displaystyle 3x^3-11x^2+25x-25$" doesn't make any sense. x can be any complex number in that formula. Tonio assumed that you meant to solve $\displaystyle 3x^3- 11x^2+ 25x-25= 0$. Is that correct?
Cardanos method (below) looks like the easiest way.
I just have to divide 3x^3-11x^2+25x-25 by the leading coefficient 3 to get x^3-(11/3)x^2+(25/3)x-(25/3)=0.
That gives a=(-11/3),b=(25/3),c=(-25/3).Then solve for p,q,u.
Should this work??
For
the solutions for x are given by
where
Cardanos method gave me the following complex root solutions:
x=2.8444i+1.222
and
x=-1.7175i+1.222
These solutions look close to the answer 1+2i and 1-2i.... just need to find the real root of 5/3.
To find the real root of 5/3 should I use Cardanos method or real roots test?
Perhaps we shouldn't be so quick to jump to Tartaglia.
Actually, it would have been far simpler to do what you originally suggested: "use rational roots test and long division".
The rational roots test shows that only 5, -5, 5/3, and -5/3 could be rational roots. Trying those, we find that 5/3 is, in fact, a root. That means that (3x- 5) is a factor. Dividing $\displaystyle 3x^3- 11x^2+ 25x- 25$ by $\displaystyle 3x- 5$ gives a quotient of $\displaystyle x^2- 2x+ 5$ with no remainder.
$\displaystyle 3x^3- 11x^2+ 25x- 25= (3x- 5)(x^2- 2x+ 5)$
$\displaystyle x^2- 2x+ 5= x^2- 2x+ 1+ 4= (x- 1)^2+ 4= 0$ so
$\displaystyle (x- 1)^2= -4$ and $\displaystyle x- 1= \pm 2i$, $\displaystyle x= 1\pm 2i$.
The roots of the original equation are 5/3, 1+ 2i, and 1- 2i.
(Though, I am, frankly, impressed that you got such accurate roots using Cardano's method.)