What is the smallest of the roots of the equation $2(x-3 \sqrt{5})(x-5 \sqrt{3})=0$?
The equation has already been given in factored form $a(x-m)(x-n)=0$, so its roots are $m=3 \sqrt{5}$ and $n=5 \sqrt{3}$. We just need to decide which of them is the larger. This can be done in at least two ways.

We can square $m$ and $n$, obtaining $m^{2}=9 \times 5=45$ and $n^{2}=25 \times 3=75$. Since $45<75$, it follows that $3 \sqrt{5}=m=\sqrt{45}<\sqrt{75}<n=5 \sqrt{3}$.

We can also observe that $\sqrt{5}<2.3$ and $1.7<\sqrt{3}$, therefore,

$$
\begin{aligned}
& m<3 \times 2.3=6.9<8.5=5 \times 1.7<n \\
& \begin{array}{cccc}
1 & 1 & 1 & 1 \\
\hline 3 \sqrt{5} & 6.9 & 8.5 & 5 \sqrt{3}
\end{array}
\end{aligned}
$$