Find all values of $m$ such that the equation $ mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$ has nonnegative roots.

Question

Find all values of $m$ such that the equation $$\large mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.

For an equation to have nonnegative roots, it mustn't only have negative roots.

Let $y = x^2 - x + 4$ $(y > 0)$, we have that $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = mx^2 + (2m - 1)xy + my^2$$

For the equation $$mx^2 + (2m - 1)xy + my^2 = 0$$ to have only negative roots, it must be satisfied that $\left\{ \begin{align} [(2m - 1)y]^2 - 4m^2y^2 \ge 0\\ (1 - 2m)y < 0\\ my^2 > 0 \end{align} \right.$ $\implies \left\{ \begin{align} (1 - 4m)y^2 \ge 0\\ 1 - 2m < 0\\ m > 0 \end{align} \right.$ $\implies \left\{ \begin{align} 1 - 4m \ge 0\\ m > \dfrac{1}{2}\\ m > 0 \end{align} \right.$ $\iff \left\{ \begin{align} m \le \frac{1}{4}\\ m > \dfrac{1}{2} \end{align} \right. \implies m \in \varnothing$.

Thus for $\forall m \in \mathbb R$, the equation $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.

Is the above solution correct? And should my attempt was inaccurate and yours is (hopefully) helpful, please post an answer.

Answer 1

Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$

divide both sides by $x^2$

$$0=m(x^2+16/x^2)+x+4/x+8m-1=m(x+4/x)^2+x+4/x-1$$

Now as $x\ge0,$ $$\dfrac{x+4/x}2\ge\sqrt{x\cdot4/x}=?$$

Answer 2

Express $m$ as a function of $x\ge 0$,

$$m(x) = - \frac{x(x^2-x+4)}{(x^2+4)^2}\le0\tag 1$$

It is equivalent to find the lower bound of $m(x)$ for $x>0$. Set $m'(x) = 0$ to get

$$x^4-2x^3+8x-16=(x-2)(x+2)(x^2-2x+4) = 0$$

which shows that the minimum is at $x=2$. Plug it into $(1)$ to get $m(2) = -\dfrac3{16}$. Thus, below are all values of $m$ for non-negative roots of $x$,

$$-\frac3{16} \le m \le 0$$