Relation between quadratic inverse and it's roots Finding the inverse of a function and finding the roots appear to be a similar procedure. Note the graph below.

I took the inverse equation and simply removed the x variable and changed y= to x= and got the following:

I'm basically just wondering if this is correct, cause it seems like waste of time to find the inverse as well as the roots if they are so similar.
Is this relation true of inverse/roots of a quadratic? Why/why not?
EDIT:
It appears the reverse is true of an inverse function, by taking the non-inverse-vertex-form and removing x from y=x then y from x=y (see below graph)

 A: Consider $y=(x-1)(x-2)=x^2-3x+2$, so the roots are $x=1,2$.
We want to find $x=f(y)$, to find the inverse function, so:
\begin{align}
y&=(x-\frac32)^2-\frac14\\
y+\frac14&=(x-\frac32)^2\\
\pm\sqrt{y+\frac14}&=x-\frac32\\
x&=\frac32\pm\sqrt{y+\frac14}\tag 1
\end{align}
To see the connection between the roots of the main function and the inverse function, we can see what happens if we we start from $(1)$ and swap $x$ and $y$, and then work backwards.
\begin{align}
y&=\frac32\pm\sqrt{x+\frac14}\\
\pm\sqrt{x+\frac14}&=y-\frac32\\
x+\frac14&=(y-\frac32)^2\\
x&=(y-\frac32)^2-\frac14\\
x&=y^2-3y+\frac94-\frac14\\
x&=y^2-3y+2\\
x&=(y-1)(y-2)
\end{align}
We are right back where we started, with $x$ and $y$ swapped over. In this equation, the roots are $y=1$ and $y=2$.
This means if $y=1,2$ then $x=0$. The original equation says that if $x=1,2$ then $y=0$.
So if we insert $y=0$ into $(1)$ we get:
\begin{align}
x&=\frac32\pm\sqrt{0+\frac14}\\
x&=\frac32\pm\frac12\\
x&=1,2
\end{align}
We are trying to find $x$ when $y=0$ - the algebra doesn't change the location of the roots. So when we put $y=0$ into $(1)$, we get the roots of the original equation.
