6.127. 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.
## Solution.

Let $x_{1}$ be the common root, then

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 3 x _ { 1 } ^ { 2 } - 4 x _ { 1 } + p - 2 = 0 , } \\
{ x _ { 1 } ^ { 2 } - 2 p x _ { 1 } + 5 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3 x_{1}^{2}-4 x_{1}+p-2=0, \\
3 x_{1}^{2}-6 p x_{1}+15=0
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow-4 x_{1}+6 p x_{1}+p-2-15=0,(6 p-4) x_{1}+p-17=0,
\end{aligned}
$$

from which $x_{1}=\frac{17-p}{6 p-4} \cdot$ From the second equation of the system we have

$$
\begin{aligned}
& \left(\frac{17-p}{6 p-4}\right)^{2}-2 p\left(\frac{17-p}{6 p-4}\right)+5=0,12 p^{3}-31 p^{2}-138 p+369=0 \\
& 12 p^{3}-36 p^{2}+5 p^{2}-15 p-123 p+369=0 \\
& \left(12 p^{3}-36 p^{2}\right)+\left(5 p^{2}-15 p\right)-(123 p-369)=0 \\
& 12 p^{2}(p-3)+5 p(p-3)-123(p-3)=0 \\
& (p-3)\left(12 p^{2}+5 p-123\right)=0
\end{aligned}
$$

From this, $p-3=0, p_{1}=3$, or $12 p^{2}+5 p-123=0$, from which $p_{2}=-\frac{41}{12}$, $p_{3}=3 ; p_{2}=-\frac{41}{12}$ does not fit. Thus, $p=3$, then $x=1$.

Answer: $x=1, p=3$.