Homework: Stuck on the last step of find the critical values Calculus section 4.4
Optimization problems
-Question: What two positive real number whose product is 19 have the smallest possible sum?
Let S be the sum of the two numbers. what is the objective functions in terms of one number, x?

-My answers so far are:
S= "$x+\frac{19}{x}$"
The interval of interest of the function is "(0, ∞)"
The numbers are __ and $\sqrt{19}$.
(type the answers, using radicals as needed)

-My work:
19=x*y
$\frac{19}{x}=y$
S=x+y
$S=x+\frac{19}{x}$
$S'=1+\frac{-19}{x^2}$
$0=1+\frac{-19}{x^2}$
$-1=\frac{-19}{x^2}$
$1=\frac{19}{x^2}$
$1=19*\frac{1}{x^2}$
$\frac{1}{19}=\frac{1}{x^2}$  This is where I'm stuck. 
Maybe I had to use $\frac{-b\pm\sqrt{b^2-4*a*c}}{a^2}$ to get a radical answer, but I can't tell where to find a, b, or c if I need to go that route.

Thank you all for your feedback, but then that still leaves the question what are the 2 numbers? I found 1 of them, $\sqrt{19}$ but other is not 0, or -$\sqrt{19}$.

That looks right to me, all you have to do is solve further
$$\frac{1}{19}=\frac{1}{x^2}$$
$$19=x^2$$
$$x=\sqrt{19}$$
$$y=19/x=\sqrt{19}$$
