The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$. The following question is taken from JEE practice set.

The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$.

My Attempt:
$$D\le0\\ \implies 9a^2-4b\le0\\ \implies9a^2-36\le4b-36\\ \implies9(a-2)(a+2)\le4(b-9)$$
Now, $a+2\gt4\implies 9(a-2)(a+2)\gt36(a-2)$
Also, $b\gt b-9\implies4b\gt4(b-9)$
Thus, $4b\gt9(a-2)+a+2)\gt36(a-2)$
Therefore, $\frac{b}{a-2}\gt9$
The answer given is $18$. How to do this?
 A: We get from $9a^2\le4b$ that $a,b$ and $\frac b{a-2}$ are all positive. It is clear that raising $b$ does not violate the inequality but increases $\frac b{a-2}$, so for any $a$ we minimise $\frac b{a-2}$ by setting $b=\frac94a^2$.
Once that is fixed we then rewrite
$$\frac b{a-2}=\frac94\cdot\frac{a^2}{a-2}=\frac94\left(a+2+\frac{4}{a-2}\right)$$
and differentiate to get $$\frac94\left(1-\frac4{(a-2)^2}\right)=0\implies a=4$$
Hence the minimum of $\frac b{a-2}$ is $\frac94\cdot\frac{4^2}{4-2}=18$.
A: Another approach would be to consider that there are two cases to examine:
•  $ \ x^2 - 3ax + b \ $ has a real  "double zero" $ \ r \ \ , \ $ so $ \ 3a \ = \ 2r \ $ and $ \ b \ = \ r^2  \ \ ; \ $  or
•  the polynomial has a "complex-conjugate pair" of zeroes $ \ \rho \ \pm \ i·\sigma \ \ , \ \ \rho \ , \ \sigma \ $ real, thus, $ \ 3a \ = \ 2 \rho \ $ and $ \ b \ = \ \rho^2 + \sigma^2 \ \ . $
For the double-zero case, the imposed condition requires $ \ a \ = \ \frac23·r \ > \ 2 \ \Rightarrow \ r \ > \ 3 \ $ in the ratio $$ \ \frac{b}{a - 2} \ = \ \frac{r^2}{\frac23·r \ - \ 2} \ = \ \frac{3·r^2}{2·(r \ - \ 3)} \ \ .  $$
If we are to forgo calculus, we can look for the values of $ \ r \ $ that give "extreme" values to this ratio by solving for when it has a single value $ \ c \ \ : $
$$ \frac{3·r^2}{2·(r \ - \ 3)} \ \ = \ \ c \ \ \Rightarrow \ \ 3·r^2 \ - \ 2c·r \ + \ 6c \ \ = \ \ 0 \ \ , $$
the discriminant of which is zero for $ \ 4c^2 \ - \ 72c \ = \ 0 \ \Rightarrow \ c \ = \ 0 \ , \ 18 \ \ . \ $  Since $ \ c \ = \ 0 \ \Rightarrow \ r \ = \ 0 $ $ \Rightarrow \ a \ = \ 0 \ \ , \ $ this is excluded by the given condition, while  $ \ c \ = \ 18 \ \Rightarrow \ 3·r^2 \ - \ 36·r \ + \ 108 \ \ = \ \ 0 $ $ \Rightarrow \ r \ = \ 6 \ \Rightarrow \ a \ = \ \frac23·6 \ = \ 4  \ \ , \ $ so a minimum of $ \ \mathbf{18} \ $ is acceptable for  $ \   \frac{b}{a - 2} \ $ in the case of a double zero at $ \ 6 \ \ ( $ the polynomial becomes $ \ x^2   -   12x + 36 \ \ . \ ) $  [Alternatively, extremization by calculus has us set the derivative of the ratio equal to zero, or
$$ \frac{6·r^2 \ - \ 36·r}{4·(r \ - \ 3)^2} \ \ = \ \ 0 \ \ \Rightarrow \ \ r \ = \ 0 \ , \ 6 \ \ . \ $$
We can even "shift" the curve for the ratio by $ \ 3 \ $ units "to the right", transforming the ratio to $ \ u \ = \  r - 3 \ \rightarrow  \ \frac{3·(u + 3)^2}{2·u} \ \ , \ $ which has odd symmetry about its vertical asymptote and the extremal value $ \ 0 \ $ for $ \ u \ = \ -3 \ , \ +3 \ \rightarrow \ r \ = \ 0 \ , \ 6 \ \ . \ ] $
With  complex zeroes ,   the ratio is  $$ \frac{b}{a - 2} \ = \ \frac{\rho^2 + \sigma^2}{\frac23 \rho \ - \ 2} \ = \ \frac{3·(\rho^2 + \sigma^2)}{2·(\rho \ - \ 3)} \ \ .  $$  From this, we see that the smallest values for the ratio can be obtained for $ \ \sigma \ = \ 0 \ \ , \ $ which thereby reduces this to the real double zero case.
Hence, the minimum value 18 for the ratio is obtained for $ \ a \ = \ 4 \ \ , \ \ b \ = \ 36 \ \ . \ $  (The $ \ r \ = \ a \ = \ b \ = \ 0 \ $ result we found is a local maximum for $ \ a \ < \ 2 \ \ . \ $  )
