2.41 For what integer value of $p$ do the equations $3 x^{2}-4 x+p-2=0$ and $x^{2}-2 p x+5=0$ have a common root? Find this root.
### 2.41 We have

\[
\left\{\begin{array}{l}3 x^{2}-4 x+p-2=0 \\ x^{2}-2 p x+5=0\end{array}\right.
\]

Substituting the expression \( p=2-3 x^{2}+4 x \) into the second equation, we get

\[
\begin{aligned}
& 6 x^{3}-7 x^{2}-4 x+5=6 x^{3}-6 x^{2}-x^{2}-4 x+5= \\
& =6 x^{2}(x-1)-(x-1)(x+5)=(x-1)\left(6 x^{2}-x-5\right)= \\
& =(x-1)^{2}(6 x+5)
\end{aligned}
\]

i.e., \( x_{1}=1, x_{2}=-\frac{5}{6} \). From this, \( p_{1}=3, p_{2}=-\frac{41}{12} \) (contradicts the condition). Therefore, \( p=3 \). For this value of \( p \), we get the equations

\[
3 x^{2}-4 x+1=0 \text { and } x^{2}-6 x+5=0
\]

The first equation has roots \( x_{1}=1, x_{2}=\frac{1}{3} \), and the second equation has roots \( x_{1}=1, x_{2}=5 \), i.e., \( x_{1}=1 \) is a common root.

Answer: \( p=3, x=1 \).