G2.4 Let $f(x)=41 x^{2}-4 x+4$ and $g(x)=-2 x^{2}+x$. If $d$ is the smallest value of $k$ such that $f(x)+k g(x)=0$ has a single root, find $d$.
$$
\begin{array}{l}
41 x^{2}-4 x+4+k\left(-2 x^{2}+x\right)=0 \\
(41-2 k) x^{2}+(k-4) x+4=0
\end{array}
$$

It has a single root $\Rightarrow \Delta=0$ or $41-2 k=0$
$(k-4)^{2}-4(41-2 k)(4)=0$ or $k=\frac{41}{2}$
$k^{2}-8+16-16 \times 41+32 k=0$ or $k=\frac{41}{2}$
$k^{2}+24 k-640=0$ or $k=\frac{41}{2}$
$k=16$ or -40 or $\frac{41}{2}, d=$ the smallest value of $k=-40$