Problem 4. For what values of $a$ and $b$ do the equations $2 x^{3}+a x-12=0$ and $x^{2}+b x+2=0$ have two common roots?
# Solution:

Let $x_{0}$ be the common root of the two equations, that is, the following equalities hold $\left\{\begin{array}{c}2 x_{0}^{3}+a x_{0}-12=0 \\ x_{0}^{2}+b x_{0}+2=0\end{array}\right.$

Notice that $x_{0}=0$ is not a solution to the quadratic equation, and multiply both sides of this equality by $2 x_{0}$.

$$
\left\{\begin{array}{c}
2 x_{0}^{3}+a x_{0}-12=0 \\
2 x_{0}^{3}+2 b x_{0}^{2}+4 x_{0}=0
\end{array}\right.
$$

Subtract the upper equality from the lower one, we get $2 b x_{0}^{2}-a x_{0}+4 x_{0}+12=0$,

Notice that when $b=0$, the original quadratic equation has no solutions, multiply this equation by $2 b$, we get the second equality in the system:

$$
\left\{\begin{array}{c}
2 b x_{0}^{2}-a x_{0}+4 x_{0}+12=0 \\
2 b x_{0}^{2}+2 b^{2} x_{0}+4 b=0
\end{array}\right.
$$

Subtract one equality from the other.

$$
\left(2 b^{2}+a-4\right) x_{0}+4 b-12=0
$$

For the original system to have two common roots, equation (*) must have two distinct roots, which is true under the condition:

$$
\left\{\begin{array} { c } 
{ 2 b ^ { 2 } + a - 4 = 0 } \\
{ 4 b - 1 2 = 0 }
\end{array} , \text { solving the system, we get } \left\{\begin{array}{c}
a=-14 \\
b=3
\end{array}\right.\right.
$$

Substitute the obtained parameter values into the original system, perform the check:

$$
\left\{\begin{array}{c}
2 x^{3}-14 x-12=0 \\
x^{2}+3 x+2=0
\end{array}, \text { the roots of the quadratic equation }-1 \text { and }-2\right. \text { are also roots of }
$$

the first equation.

Answer: $a=-14, b=3$.