$2 \cdot 66$ Given that the sum of the squares of the two real roots of the equation $2 x^{2}+k x-2 k+1=0$ is $\frac{29}{4}$, then the value of $k$ is
(A) 3.
(B) -11.
(C) 3 or -11.
(D) 11.
(E) None of the above conclusions is correct.
(China Tianjin Junior High School Mathematics Competition; China Junior High School Mathematics League, 1984)
［Solution］Let the two real roots of the equation be $x_{1}, x_{2}$, then by the problem we have
$$
\left\{\begin{array}{l}
\Delta=k^{2}-4 \cdot 2(-2 k+1)>0, \\
x_{1}^{2}+x_{2}^{2}=\frac{29}{4}, \\
x_{1}+x_{2}=-\frac{k}{2}, \\
x_{1} x_{2}=\frac{1-2 k}{2} .
\end{array}\right.
$$

From (1), we get $k-8+6 \sqrt{2}$.
From (2), we have $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=\frac{29}{4}$.
From (3), (4), and the above equation, we get $k^{2}+8 k-33=0$.
Solving this, we get $k=3$ or -11. In summary, $k=3$.
Therefore, the answer is $(A)$.