<s>[INST] How do I get from $x - x^2 = \frac{1}{4}$ to $x =\frac{1}{2}$? I'm working on a text book problem where I need to sketch the graph of $y = 4x^2 - 4x+1$ by finding where the curve meets the $x$ axis.
To start out I set $y = 0$ then tried to isolate $x$ then,
$4x - 4x^2 = 1$
$x - x^2 =  \frac{1}{4}$
From here I want to continue algebraically to reach $x = \frac{1}{2}$ which I know is the solution from plotting the curve using an app and I can see that that makes sense.
However I am missing some concepts which allow me to turn $x - x^2 = \frac{1}{4}$ into $x = \frac{1}{2}$ and wanted some help to get unblocked.
 [/INST] Starting at $ x - x^2 = 1/4 $, subtract $ 1/4 $ to each side:
$$\mathrm{(1)} \qquad x - x^2 - 1/4 = 0 $$
Arrange the left hand side of the equation $ (1) $ in descending degree of $ x $:
$$\mathrm{(2)} \qquad -x^2 + x - 1/4 = 0 $$
I would recommend having the coefficient of the term with the highest degree of $ x $ not be negative to make the equation neater. So, after multiplying a $ -1 $ to each side:
$$\mathrm{(3)} \qquad x^2 - x + 1/4 = 0 $$
Now the left side of equation $ (3) $ is a perfect square of the form $ x^2 + 2ax + a^2 = (x + a)^2 $, where $ a = -1/2 $ since $ 2ax = -x $, or $ 2a = -1 $.
Thus, $ x^2 - x + 1/4 = (x - 1/2)^2 $, and equation $ (3) $ becomes:
$$\mathrm{(4)} \qquad (x - 1/2)^2 = 0 $$
Solving for $ x $, 
$$ x = 1/2 $$
but with a multiplicity of 2. The significance of the multiplicity arises from $ x = 1/2 $ being the $ x $-coordinate of the vertex touching (but not crossing) the $x$-axis. In the following graph below, $f(x) = -x^2 + x - 1/4$ and $g(x) = 0$.
Zooming in closely at the graph clearly shows the parabola ($f(x)$) touching the $x$-axis ($g(x)$) at $x = 1/2$.

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