G3.1 If $a$ is the smallest real root of the equation $\sqrt{x(x+1)(x+2)(x+3)+1}=71$, find the value of $a$.
Let $t=x+1.5$, then the equation becomes $\sqrt{(t-1.5)(t-0.5)(t+0.5)(t+1.5)+1}=71$
$$
\begin{array}{l}
\sqrt{\left(t^{2}-\frac{9}{4}\right)\left(t^{2}-\frac{1}{4}\right)+1}=71 \\
\sqrt{t^{4}-\frac{5}{2} t^{2}+\frac{9}{16}+1}=71 \\
\Rightarrow \sqrt{t^{4}-\frac{5}{2} t^{2}+\frac{25}{16}}=71 \\
\Rightarrow \sqrt{\left(t^{2}-\frac{5}{4}\right)^{2}}=71 \\
t^{2}-\frac{5}{4}=71 \Rightarrow t^{2}=\frac{289}{4} \Rightarrow t=\frac{17}{2} \text { or } t=-\frac{17}{2} \\
x=t-1.5= \pm \frac{17}{2}-\frac{3}{2}=7 \text { or }-10 \\
a=\text { the smallest root }=-10
\end{array}
$$