Good day all,
I have an example in my book that ask:
For which values of k, if any, does $x^3−3x^2+6x+k=0$ have 3 real roots?
I know that this is a quadratic equation and that normally I would need to take the derivative of this equation and set it equal to zero and use the quadratic formula to find the x values.
However the books states that because $6^2$ < 4 x 3 x 6, it can have no real roots. I am for some reason not able to grasps the explanation given here. If someone could point me in the right direction or explain where they are getting $6^2$ < 4 x 3 x 6, it would greatly be appreciated.
I am using the book "Mathematical Methods for Physics and Engineering" by Riley, Hobson and Bence.
Thank you in advance.
Best Answer
First off this is a cubic...
By Vieta we have $a+b+c=3$ and $ab+bc+ca=6$, where $a,b,c$ are the roots. Squaring the first gives $a^2+b^2+c^2+2(ab+bc+ca)=9$, so $a^2+b^2+c^2=-3$. This is always impossible if the roots are real.