Tam thức bậc hai $f\left( x \right) = {x^2} + \left( {1 - \sqrt 3 } \right)x - 8 - 5\sqrt 3$:

$\left\{ \begin{array}{l}a > 0\\\Delta  \le 0\end{array} \right..$

$\left\{ \begin{array}{l}a > 0\\\Delta  \ge 0\end{array} \right..$

$\left\{ \begin{array}{l}a > 0\\\Delta  < 0\end{array} \right..$

$\left\{ \begin{array}{l}a < 0\\\Delta  > 0\end{array} \right..$

$\left\{ \begin{array}{l}a < 0\\\Delta  \le 0\end{array} \right.$.

$\left\{ \begin{array}{l}a < 0\\\Delta  \ge 0\end{array} \right.$

$\left\{ \begin{array}{l}a > 0\\\Delta  < 0\end{array} \right.$

$\left\{ \begin{array}{l}a < 0\\\Delta  > 0\end{array} \right.$.

$f\left( x \right) > 0\,,{\rm{ }}\forall x \in \mathbb{R}$.       

$f\left( x \right) < 0\,,{\rm{ }}\forall x \in \mathbb{R}$.

$f\left( x \right)$ không đổi dấu

Tồn tại $x$ để $f\left( x \right) = 0$.

$x \in \left( {0; + \infty } \right).$        

$x \in \left( { - 2; + \infty } \right).$

$\forall x \in \mathbb{R}.$

$x \in \left( { - \infty ;2} \right).$

$0$

$1$

$2$

$3$