The range of constant $k$ for which all the roots of equation $x^4+4x^3-8x^2\:+k\:=0$ are real.

Question

The range of constant $k$ for which all the roots of equation $x^4+4x^3-8x^2\:+k\:=0$ are real.

Options

a) $(1,3)$ b) $[1,3]$ c) $(-1,3)$ d) $(-1,3] $

Answer 1

None of them. The correct answer must be $k\in (0,3)$.

For the graph to cross the $x$-line four times, two local mins must be below it and one local max must be above it.

We will find the critical points of $f(x)=x^4+4x^3-8x^2+k$: $$\begin{align}f'(x)&=4x^3+12x^2-16x=0 \Rightarrow x_{1,2,3}=-4,0,1\\ min:&f(-4)=-128+k<0 \Rightarrow k<128\\ max:&f(0)=k>0\\ min:&f(1)=-3+k<0 \Rightarrow k<3\end{align}$$ See Desmos graph.

Answer 2

$$f(x)\equiv x^4+4x^3-8x^2\:+k\:=0$$

If this four degree polynomial has four real roots, then there is a local minimum between the first two and between the last two roots, and a local maximum between the second and third root.

The location of the extrema does not change with $k$, only their $y$-values. Thus we are looking for $k$ such that the two local minima have negative $y$-values and the local maximum has positive $y$-value.

Locate the local extrema as roots of the derivative $$ 4x^3+12x^2-16x = 4x(x^2+3x-4)=4(x+4)x(x-1),$$ i.e. the minima are at $x=-4$ and $x=1$, the maximum is at $x=0$.

The respective values of $f(x)$ are $f(-4)=-128+k$, $\quad y(0)=k \quad $, and $y(1)=-3+k$. We conclude that the desired condition is $$ 0<k<3\qquad \text{and}\qquad k<128$$