6.131. Without solving the equation $x^{2}-(2 a+1) x+a^{2}+2=0$, find the value of $a$ for which one of the roots is twice the other.
Solution.

From the condition, by Vieta's theorem, we have

$$
\left\{\begin{array} { l } 
{ x _ { 1 } + x _ { 2 } = 2 a + 1 , } \\
{ x _ { 1 } \cdot x _ { 2 } = a ^ { 2 } + 2 , } \\
{ x _ { 2 } = 2 x _ { 1 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ 3 x _ { 1 } = 2 a + 1 , } \\
{ 2 x _ { 1 } ^ { 2 } = a ^ { 2 } + 2 } \\
{ x _ { 2 } = 2 x _ { 1 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x_{1}=\frac{2 a+1}{3} \\
x_{1}^{2}=\frac{a^{2}+2}{2} \\
x_{2}=2 x_{1}
\end{array}\right.\right.\right.
$$

From here,

$$
\begin{aligned}
& \left(\frac{2 a+1}{3}\right)^{2}=\frac{a^{2}+2}{2} \Leftrightarrow \frac{4 a^{2}+4 a+1}{9}=\frac{a^{2}+2}{2} \Leftrightarrow \\
& \Leftrightarrow a^{2}-8 a+16=0 \Leftrightarrow(a-4)^{2}=0
\end{aligned}
$$

Thus, $a=4$.

Answer: $a=4$.