Question

$a^{2} + \left( 9x - 2x^{2} \right)a + \left( x^{4} - 9x^{3} + 20x^{2} \right) = 0$

$$\begin{matrix} \Rightarrow a & \ = \frac{\left( 2x^{2} - 9x \right) \pm \sqrt{4x^{4} + 81x^{2} - 36x^{3} - 4\left( x^{4} - 9x^{3} + 20x^{2} \right)}}{2} \\ a & \ = x^{2} - 4x,\ x^{2} - 5x \end{matrix}$$

for real roots of $x^{2} - 4x - a = 0,D \geq 0 \Rightarrow 16 + 4a \geq 0\ a \geq - 4$

for real roots of $x^{2} - 5x - a = 0,b \geq 0 \Rightarrow 25 + 4a \geq 0a \geq - \frac{25}{4}$

for $a > - 4$

$$\begin{matrix} x & \ = 2 - \sqrt{4 + a} \in ( - \infty,2),x = 2 + \sqrt{4 + a} \in (2,\infty) \\ x & \ = \frac{5 - \sqrt{25 + 4a}}{2} \in ( - \infty,1),x = \frac{5 + \sqrt{25 + 4a}}{2} \in (4,\infty) \\ 2 - \sqrt{4 + a} & \ = \frac{5 - \sqrt{25 + 4a}}{2} \Rightarrow a = 0 \\ 2 + \sqrt{4 + a} & \ = \frac{5 + \sqrt{25 + 4a}}{2}\text{~is impossible.~} \end{matrix}$$

For 4 distinct real roots.

$$a \in ( - 4,0) \cup (0,\infty)$$

For 3 distinct rea roots.

$$a = \{ - 4,0\}$$

For 2 distinct real roots.

$$a \in \left( - \frac{25}{4}, - 4 \right)$$

For no real roots

$$a \in \left( - \infty, - \frac{25}{4} \right)$$