TITLE: Intuition behind Algebra
QUESTION [12 upvotes]: When proving, or just straight up solving equations, we often manipulate variables until we get the results we want. For example, the square root of the discriminant is simply an algebraic manipulation to get the result of the distance (or difference) between the roots. However, is there intuition behind this manipulation? I get why Algebra works in the first place but when seeing it in action, manipulations as complex as that of involved in getting the discriminant, gets me lose sense of what's actually happening within the numbers.

REPLY [4 votes]: "Every discovery contains an irrational element or a creative
  intuition."
  - Karl Popper

If we look at modern derivations of the quadratic formula, we might see algebraic manipulations that prove it, but we might not feel the intuitive essence that brings it to light. Now, the chain "motivation-intuition-discovery" is a personal experience. But just for fun, using only an algebraic approach, let us set forth to 'discover' how to solve $ax^2+bx+c = 0 \text{ , } a \ne 0$ in two different ways.
Method 1:
It makes us happy that we can solve equations like $x^2 = 2$ and when we generalize, we see two solutions for
$\tag 1 ax^2 = -c \text{ with } a \text{ and } c \text{ of opposite signs}$
namely, $x = \pm \sqrt{\frac{c}{a}}$.
In a flash of insight (change of variable), we realize that we can 'manufacture' more interesting quadratic equations with solutions. For example, 
$\tag 2 2 (x+3)^2 = 4$
has two solutions, $x = -3 \pm \sqrt2$, and (2) can be also be expressed as $2x^2 +12x +14 =0$.
With great hope, we now tell ourselves that we can somehow 'force' any quadratic equation into a '(1)/(2) pattern' and get the solution(s):
$\quad ax^2+bx+c = 0 \; \text{ iff }$
$\quad x^2+\frac{b}{a}x  = -\frac{c}{a} \; \text{ iff }$
$\quad (x+\frac{b}{2a})^2 - (\frac{b}{2a})^2   = -\frac{c}{a} \; \text{ iff }$
$\quad (x+\frac{b}{2a})^2   =  \frac{b^2}{4a^2} -\frac{c}{a} \; \text{ iff }$
$\quad x = -\frac{b}{2a}   \pm \sqrt{  \frac{b^2}{4a^2} -\frac{c}{a}} \; \text{ iff }$
$\quad x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$
This method is known as 'completing the square'.
Method 2
There is nothing in the world that you love more than factoring quadratic equations, but sometimes the solution eludes you. Then one day you say

$\quad x^2+\frac{b}{a} x+\frac{c}{a} = (x-s)(x-t) \; \text{ iff }$
$\quad -(s + t) = \frac{b}{a} \text{ AND } st = \frac{c}{a}$
Ah, two equation in two unknowns - go for it! You see a 'trick' to get things going
$\quad (s - t)^2  = (s + t)^2 - 4st =  \frac{b^2}{a^2} - \frac{4c}{a}$
So you are now one step away from finding that the distance between the two roots is equal to
$\tag 3 \frac{\sqrt{b^2 -4ac}}{|a|}$
You are now in 'mop up mode', and work out the details for your quadratic formula.
This method is known as the 'So you wanna see a factorization do you?'.