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.
 A: $$x-x^2 = \frac{1}{4} \implies x^2-x+\frac{1}{4} = 0$$
Here, if you notice, you can see we have two perfect squares: $x^2$ and $\frac{1}{4}$. Also, twice the product of their square roots, which is $2\cdot x\cdot \frac{1}{2}$, gives the coefficient of the middle term. You probably know
$$a^2\pm 2ab+b^2 = (a\pm b)^2$$
Factoring the equation above in the same manner, you get
$$x^2-\color{blue}{x}+\frac{1}{4} = x^2-\color{blue}{2\cdot x\cdot\frac{1}{2}}+\left(\frac{1}{2}\right)^2 = \left(x-\frac{1}{2}\right)^2$$
$$\left(x-\frac{1}{2}\right)^2 = 0$$
$$x-\frac{1}{2} = 0$$
$$x = \frac{1}{2}$$
A: $$4x-4x^2=1\implies 4x^2-4x+1=0\implies (2x-1)^2=0\implies 2x-1=0$$
A: Hint: $$x^2-x+\frac{1}{4}=(x-\frac{1}{2})^2$$
A: $$x-x^2=\dfrac14\implies4x-4x^2=1$$
$$4x^2-4x+1=0$$
$$(2x-1)(2x-1)=0$$
$$(2x-1)^2=0$$
$$2x-1=0$$
$$x=\dfrac12$$
A: By the remarkable product formula,
$$x^2-x+\frac14=x^2-2\cdot\frac12\cdot x+\left(\frac12\right)^2=\left(x-\frac12\right)^2.$$
Hence
$$x^2-x+\frac14=0\iff x-\frac12=0.$$
A: Complete the square. We have:
$$x^2-x+\frac 14=(x-\frac12)^2-\frac14+\frac14=(x-\frac12)^2$$
A: $$x-x^2=x(1-x)\leq\left(\frac{x+1-x}{2}\right)^2=\frac{1}{4}.$$
The equality occurs for $x=1-x$ only, which gives $x=\frac{1}{2}.$
A: $x^2+1/4=x;$
Note $x>0.$
AM-GM:
$x^2+1/4 \ge 2\sqrt{x^2(1/4)}=x$;
Equality for $x^2=1/4$, hence
$x=1/2$ (discard $-1/2$ (why?)).
A: 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 $x = 1/2$.

