Find $a$ so that $a \in R, x^4+4x^3+ax^2+4x+1=0$ has all roots in $R$.

Question

$a \in R, x^4+4x^3+ax^2+4x+1=0$

Find $a$ so that the equation has all roots in $R$.

Firstly I thought about using Descartes' rule of signs, which says there are at most

$2$ positive root 2 negative real roots (if $a<0$)

$0$ positive roots 4 negative real roots (if $a>=0$)

Then I just went for the derivatives:

$f(x)=x^4+4x^3+ax^2+4x+1=0$ must have 4 real roots, so $f(x)'=4x^3+12x^2+2ax+4=2x^3+6x^2+ax+2=0$ must have 3 real roots, so $f(x)''=6x^2+12x+a=0$ must have 2 real roots, so $\delta=144-24a \space(>=0)$ which gives $a<=6$.

I've also tried using the AM-GM inequality, but doesn't seem to help here as $x_1+x_2+x_3+x_4=-4$ and $x_1*x_2*x_3*x_4=1$. $AM=-1$ $GM=1$ so the roots don't satisfy the conditions of $AM>=GM$.

Kind of got stuck in here, I have to get to the solution $a\in(-\infty,-10]\cup${$6$}. Could I have some hints on how to get to that, perhaps there is a factorisation I don't see? Thank you.

Answer 1

If we set $y=x+x^{-1}$ then we get the quadratic $$y^2+4y+a-2=0.\tag{*}$$ For the original to have real roots, $(*)$ must have real roots, moreover each real root of $(*)$ must satisfy $|y|\ge2$. As they add to $-4$ either both equal $-2$ or one is positive $\ge2$, etc.

Answer 2

The exceptional solution $a=6$ should perhaps jump into one's eyes (binomial coefficients). That being said, we observe that $$f(x)=(x+1)^4+(a-6) x^2$$ If $a-6$ is positive, we add two non-egative functions without a common root, thus $f(x)>0$ for all $x\in\Bbb R$.

If $a-6=0$, we happen to have a single 4fold root.

If $a-6$ is negative, it depends. First, we will have two simple real roots near $-1$, but "later" we will get two additional real roots to the right of $0$. To see when that happens, solve for the case that $f(x)$ has a local minimum at a positve $x$ and with $y=0$, i.e., $f(x)=0$ and $f'(x)=0$ have a common solution ...