G1.4 Let $r$ and $s$ be the two distinct real roots of the equation $2\left(x^{2}+\frac{1}{x^{2}}\right)-3\left(x+\frac{1}{x}\right)=1$. If $d=r+s$, find the value of $d$.
$2\left(x^{2}+\frac{1}{x^{2}}\right)-3\left(x+\frac{1}{x}\right)=1$, real roots $r, s$. Let $t=x+\frac{1}{x}$, then $x^{2}+\frac{1}{x^{2}}=t^{2}-2$.
$$
\begin{array}{l}
2\left(t^{2}-2\right)-3 t=1 \\
2 t^{2}-3 t-5=0 \\
(2 t-5)(t+1)=0 \\
t=\frac{5}{2} \text { or }-1 \\
x+\frac{1}{x}=\frac{5}{2} \text { or } x+\frac{1}{x}=-1 \\
x=2 \text { or } \frac{1}{2} \Rightarrow r=2, s=\frac{1}{2} \Rightarrow d=r+s=\frac{5}{2}
\end{array}
$$