For which values of $a$ does the function have exactly 1 root The function is : $(a-3)9^x-6.3^x+a+5=0$ and $a$ is real. I know that if its a linear or quadratic equation with discriminant = $0$ then we have 1 root but when I set $3^x = y$ and solve the discriminant of $(a-3)y^2-6y+a+5$ to be $= 0$. I don't get the answer. 
Where am I going wrong this is a high school problem and I dont't think I should be using the derivative.
 A: Hint :
Don't forget that $3^x$ i.e. $y$ is always positive.
So if you want any solution , all you need is positive root of this equation.
Note : Exactly one root must be negative else it will give two solutions or both roots must be positive and equal.
Solution :
Let $f(y)=(a-3)y^2-6y+a+5$.
Since we need exactly one positive solution, 
$f(0)<0$ and $a-3>0$ (Coefficient of $x^2$)
Or,
$f(0)>0$ and $a-3<0$ 
Or,
$f(0)<0$ and $a-3>0$ AND Discriminant =0
There is one more case , when it is no more any quadratic i.e. $a=3$.In this too you'll get one positive root.
On solving for all the conditions, you will finally get :
$a \in (-5,3] \cup$ {$4$}
A: If $a= 3$ you don't have a quadratic since the leading coefficient must be different than $0$.
In order to have one solution either the discriminant is $0$ which should be easy enough or that you have one positive and one negative solution.
If you have one negative and one positive solution then their product is negative you can apply Vieta's formulas
$$x_1x_2=\frac{a+5}{a-3}<0$$
This happens exactly when $a\in(-5,3)$,also notice when $a=-5$ you have that $y=0$ is one solution and the other is negative but $3^x\neq 0$,and when $a=3$ we have that the quadratic is actually a linear which has one solution.So summing up you get $a\in(-5,3]\cup\{4\}$
