One, let the length of one leg and the base of an isosceles triangle be the two roots of the equation $x^{2}-6 x+a=0$. Find the range of $a$ when such a triangle exists uniquely.

When such a triangle exists uniquely. Find the range of $a$.
Let the equation $x^{2}-6 x+a=0$ have two positive roots $x_{1}, x_{2}$, and $x_{1} \leqslant x_{2}$, then
$x_{1}=3-\sqrt{9-a}, x_{2}=3+\sqrt{9-a}$. And $0<a \leqslant 9$.
(1) When $x_{1}=x_{2}$, there is only one triangle that meets the condition, which is an equilateral triangle with side $x_{1}$ $\left(=x_{2}\right)$. At this time,
$3-\sqrt{9-a}=3+\sqrt{9-a}$, i.e., $\sqrt{9-a}=0$,
$\therefore \quad a=9$.
(2) When $2 x_{1} \leqslant x_{2}$, there is also only one triangle that meets the condition, which has $x_{2}$ as the waist and $x_{1}$ as the base. At this time,
$$
6-2 \sqrt{9-a} \leqslant 3+\sqrt{9-a},
$$

Solving this gives $\quad a \leqslant 8$
Therefore, when there is only one triangle that meets the condition, the range of values for $a$ is $0<a \leqslant 8$ or $a=9$.