Unknown parameter $a$ in inequality such that it has at least one positive or one negative solution Consider the Inequality $9^x-a\cdot 3^x-a+3\leq 0\;,$ Where $a\in \mathbb{R}$
Then The real values of $a$ so that
$(1)$ The given inequality has at least one real solution.
$(2)$ The given Inequality has at least one positive solution
$(3)$ The given Inequality has at least one negative solution.

$\bf{My\; Try::}$ For $(1)$
We can write Inequality as $$\displaystyle 9^x+3\leq a(3^x+1)\Rightarrow a\geq \frac{9^x+3}{3^x+1} = \frac{9^x-1+4}{3^x+1} = 3^x-1+\frac{4}{3^x+1}$$
So $$a\geq \underbrace{3^x+1+\frac{4}{3^x+1}}_{\bf{Using\; A.M\geq G.M}}-2\geq 4-2=2$$
So In first case we get $a\geq 2$
Now How can I solve $(2)$ and $(3)$ cases, Help required, Thanks
First denote t=3^x. Notice 9^x=t^2. Then solve inequality for t using vieta's formulas
With different requirements:
1. x is real iff t is positive.
2. x is positive iff t is greater than 1
3. x is negative iff t is less than one.
For each step add conditions of inequality for parameter a
$(3^x)^2-a(3^x)+9\le 0$. Thus 
$$(a-\sqrt{(a+6)(a+2)})/2\le 3^x\le (a+\sqrt{(a+6)(a+2)}).$$ 

Thus $(a+6)(a+2)\ge 0$.