For which values of $k$ will $ x^3 -x^2 -8x +k = 0$have 3 real roots?

Question

I have the following equation:

$$ x^3 -x^2 -8x +k = 0$$

The question: For which values of $k$ will the cubic equation have 3 real roots?

Thank you

Answer 1

Hints:

This cubic's discriminant is

$$\Delta=64+4\cdot8^3+4k-27k^2+18\cdot8 k$$

Now just check when $\,\Delta\ge 0\;\ldots$

Answer 2

If you know the general shape of a cubic f(x) with positive leading coefficient, you will know that it has three real roots if it has a local maximum at which $f(x)\gt 0$ and a local minimum at which $f(x) \lt 0$

So with $f(x)=x^3-x^2-8x+k$ we have $f'(x)=3x^2-2x-8=(x-2)(3x+4)$

So the local maximum is at $x=-\cfrac 34$, and the minimum at $x=2$ - substitute these values in the condition for three roots to obtain constraints on $k$.

Answer 3

To expand upon @Mark Bennet's answer...

If you know the local min to be at $x=2$, then express the polynoomial as $(x-2)^2 (x-a)$ for some unknown $a$. Expand to find that the polynomial is

$$x^3-(a+4) x^2+(4 a+4) x-4 a$$

Clearly, we find $a=-3$ by comparison, so that in this configuration, $k=12$. Note that this is one extreme of the range of possible values of $k$. To find the other extreme, note that, as Mark pointed out, the maximum is at $x=-4/3$. The value of the polynomial at this maximum is in the above configuration (where the minimum lies on the $x$-axis) is

$$\left (-\frac{10}{3}\right)^2 \left ( \frac{5}{3}\right) = \frac{500}{27}$$

The other extreme for the range of possible values of $k$ occurs when this maximum lies on the $x$ axis. Thus, we subtract this maximum value from the above value of $k$ (because subtracting the constant in a polynomial just shifts the polynomial down). Therefore, the sought-after range is

$$\left\{k : k \in \left [ -\frac{176}{27},12\right]\right\}$$