Thursday, July 19, 2018

Higher degree polynomial with complex roots



I'm working on the following problem:



$$ r^4 - 3r^2 -4r = 0 $$



I factor out one $r$ and leaving me $ r(r^3 - 3r -4) = 0 $. One real root is $r=0$, and I'm unable to find the other ones. I tried using synthetic division but it didn't help. I tried googling synthetic division with complex root problems, but all the videos use examples that are given a complex solution in order to solve the other roots. So what could be a good approach in this problem?


Answer




Hint. Applying Cardano's formula (see the link above) to the reduced equation
$$
r^3 - 3r -4=0,
$$ one gets the real root




$$
r_1=\left(2-\sqrt{3}\right)^{1/3}+\left(2+\sqrt{3}\right)^{1/3}
$$





and the two complex roots




$$
r_{2}^{\pm}=-\frac12 \left(2- \sqrt{3}\right)^{1/3} \left(1\pm i \sqrt{3}\right)-\frac{1}{2} \left(2+\sqrt{3}\right)^{1/3}\left(1\mp i \sqrt{3}\right) .
$$



No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...