$ 3x^4+bx^2+3=0 $

Michal had to determine for which values of the parameter $b\in \mathbb{R}$ the equation $$ 3x^4+bx^2+3=0 $$ would have four different real roots. He proceeded as follows:

(1) He reduced the equation to a quadratic equation by means of substitution $t=x^2$: $$ 3t^2+bt+3=0. $$

(2) He expressed the roots of the quadratic equation: $$ t_{1,2}=\frac{-b\pm \sqrt{b^2-4\cdot 3\cdot 3}}{6} $$

(3) Then he returned to the substitution and put $$ x^2=t_1\mathrm{~and~} x^2=t_2. $$

(4) To have four distinct real roots of the equation, he deduced that both $t_1$ and $t_2$ must be positive and distinct, i.e., $$ t_1>0\mathrm{~and~} t_2>0 \mathrm{~and~} t_1\neq t_2 $$ (5) In order for the roots to exist and be different from each other, then the discriminant $\Delta=b^2-36$ must be positive, i.e. $$ \begin{gather} b^2>36 \cr |b|>6 \cr b\in (-\infty;-6)\cup (6;+\infty) \end{gather} $$

(6) Ensuring that $t_1>0$ and $t_2>0$ remained a challenge. Since Michal did not know how to solve this problem, he asked his fellow students for help.

Adam thinks that from $t_1>0$ and $t_2>0$ it follows $t_1t_2>0$, i.e. $$ \begin{gather} (-b+\sqrt{b^2-36})(-b-\sqrt{b^2-36})>0 \cr b^2-(b^2-36)>0 \cr 36>0. \end{gather} $$ Last inequality holds for all values $b\in \mathbb{R}$. Therefore, taking into account the result from step (5), the given equation will have four different real roots for $$ b\in (-\infty;-6) \cup (6;\infty) $$

Paula remembered that according to Vieta's formulas the roots $t_1$, $t_2$ of the quadratic equation $at^2+bt+c=0$ satisfy $$ t_1+t_2=-\frac{b}{a},~t_1t_2=\frac{c}{a}. $$ In our case $t_1+t_2=-\frac{b}{3}$, $t_1t_2=1$. As $t_1>0$ and $t_2>0$, it must hold $b<0$.

Together with the result from step (5), it follows that the given equation will have four different real roots if and only if $$ b\in (-\infty;-6) $$

Edward proceeded this way: By factoring $3t^2+bt+3$ he got: $$ \begin{gather} 3t^2+bt+3=3(t-t_1)(t-t_2) \cr t^2+\frac{b}{3}t+1=(t-t_1)(t-t_2) \cr t^2+\frac{b}{3}t+1=t^2-(t_1+t_2)t+t_1t_2 \end{gather} $$ from which it follows that $$ t_1+t_2=-\frac{b}{3}\mathrm{~and~} t_1t_2=1. $$ As $t_1>0$ and $t_2>0$, it must hold $b<0$.

If we consider the result from step (5), we conclude that the given equation will have four different real roots if and only if $$ b\in (-\infty;-6) $$ Which of the fellow students thought correctly?