For which values of k will $x^3−3x^2+6x+k=0$ have 3 real roots?

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.

• The text is looking at $f'(x)=3x^2-6x+6$. This quadratic has no real roots because its discriminant is $6^2-4*3*6<0$, which is what they write. As the cubic has no critical points, there can only be one real root what ever $k$ is. – lulu May 5 '16 at 18:24
• Well, almost. Any odd degree polynomial has one real root (as it has opposite signs at $\pm \infty$. So your cubic always has at least one real root. It can't have more than one because it is strictly increasing. – lulu May 5 '16 at 18:27
• As an example: $e^x$ is strictly increasing, but it has no real roots. – lulu May 5 '16 at 18:28
• Thank you for your response. I am starting to understand it more clearly now. However if the discriminant from this quadratic is negative, then how does it have only one real root? From my review shouldn't it only have one real root if the discriminant is zero? Thank you again – Nava Moore May 5 '16 at 19:48
• The quadratic has no real roots. But that's the derivative...so that's just saying that the original cubic has neither a local max nor a local min. To see the relevance for your problem: if $f(a)=0=f(b)$ for $a<b$ then the MVT tells us that there is some value $c$ such that $a<c<b$ and $f'(c)=0$. Since there are no such values $c$ we can't have two real roots. In principle you could have $0$ or $1$ real root...but as I said, every cubic has at least one real root. – lulu May 5 '16 at 19:51

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.
• Well done! A solution without the discriminant of $f$. – Peter May 5 '16 at 18:24
Rewrite the equation as $(x-1)^3+3(x-1)+k+4=0$. This function is increasing, hence it has only $1$ real root.
Let $f(x) = {x^3} - 3{x^2} + 6x + k$. Then $f'(x) = 3({x^2} - 2x + 2) = 3\left( {{{\left( {x - 1} \right)}^2} + 1} \right) > 0$. Hence $f(x)$ is a strictly increasing function. So it cannot have 3 real roots.