Perfect squares always one root? I had an exam today and I was thinking about this task now, after the exam of course. 
$f(x)=a(x-b)^2 +c$  
Now, the point was to find C so that the function only has one root. Easy enough, I played with the calculator and found this. But I hate explanations like that, yes. You get a few points but far from full score. But overall I should still get an A, I hope. If $C=0$ then the expression is a perfect square and they only have one root? Is that far of?  
$a(x-b)^2= - c$  
$\frac{a(x-b)^2}{a}= - \frac{c}{a}$  
$(x-b)^2= - \frac{c}{a}$  
This also argues that c should be 0 for it to only be one root? 

We will explicitly assume that $a\ne 0$. Then, more or less as you wrote, $a(x-b)^2+c=0$ if and only if $(x-b)^2=-\frac{c}{a}$. 
Thus if $-\frac{c}{a} <0$, there is no root, since the square of a real number cannot be negative. If $-\frac{c}{a}>0$, there are two distinct roots, namely $x=b\pm\sqrt{-c/a}$. And finally, if $-\frac{c}{a}=0$, or equivalently $c=0$, there is exactly one root. 
So there is exactly one root if and only if $c=0$. 
The above is undoubtedly what you had in mind. What you actually wrote on the exam paper may not have been complete. Much of the time, a bunch of equations with little explanatory text means an incomplete solution. 
For absolute completeness let's deal with the silly case $a=0$. In that case our equation is equivalent to $c=0$. If $c\ne 0$, this has no solution. If $c=0$, the equation has infinitely many solutions, since $x$ can take on any value.
