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 The quadratic formula was for me at least my first "intro" to symbols in math...instead of just a variable, there were also a's, b's and c's everywhere!  Of course when we first memorized it we were not shown a proof or given a reason why it worked, but the explanation is not that difficult (it was discovered thousands of years ago).  It is just a consequence of completing the square, and that sounds pretty geometric.  So, lets see if we can use the tools built up in the article Multiplication: a geometric interpretation to "draw out the steps" as we complete the square.The quadratic formula provides the solutions to a general equation which is quadratic in x.  Any such equation can be written in the following standard form:$\dpi{150}&space;ax^2+bx+c=0$If we were to draw this equation out pictorially, the left hand side would look something like this:Where there are a boxes of side length x.  This description makes sense so long as a is an integer, but might still be hard to deal with.  We could solve the problem by interpreting ax2 as a rectangular prism with a square base of side lengths x and height a, but this would just be an added complication since the rest of our shapes are 2 dimensional.  Instead, lets see if we cant do some algebraic simplification of our equation before attempting to solve it pictorally.First things first, instead of having an equation equal to zero, it might be more useful to have an equation where we have something=something.  That way, as we are solving it, there will be terms on each side of the equals sign to play around with.  Under our interpretation, these two terms will both be areas; so we will have two piles of shapes with the same total area, and our goal will be to shuffle them around until we can read some information off.  So, first off, we will move the constant term to the other side.Now, this is supposed to be a problem about areas, and the original trouble was caused by the coefficient a; its hard to figure out how to reason nicely about a different squares.  Before heading back to geometry then, lets divide by a, leaving us with$\dpi{150}&space;x^2+\frac{b}{a}x=-\frac{c}{a}$Here we have one possible issue: if we want the equation to represent two collections of squares and rectangles which share the same area, how are we to do this when the right hand side is negative?  It's good to keep in mind that the "area" interpretation is just a helpful visual device; the fact that its "actually" negative here is ok, we can still draw it as a positive area and then just insert a negative sign when we get back to the symbolic form.Shuffling these constants hasn't robbed the equation of any information (this new equation has the same solutions as the original), we can consider the following as our general form for the quadratic$\dpi{150}&space;x^2+\alpha&space;x=\beta$This guy has a much nicer pictorial representationOur goal here will be to take this above picture, and move/recombine/squish the shapes in question without changing their areas, until both sides of the equals sign have a square on them.  Once we have that, we will have two squares, each with equal areas, and hence with equal side lengths.  Just like in the multiplication article, we are then able to reduce our problem to a simpler one by working with the side lengths only.As a first step, lets cut the orange rectangle in half.  Leaving everything else alone, we end up with thisNow we can line the orange rectangles up alongside the blue square (this works out because the long side of each orange rectangle is side length x)The top half of the equals sign is almost a square; its just missing one corner!  To fix this, lets add a little square.  Of course, to keep the quality if we add a square to the top we need to add one to the bottom too.What size is this little square?  Well, the only reason we added it was to "fill in the hole" in the top square; and that hole has side lengths equal to the short sides of our orange rectangles.  Since the original orange rectangle had a side length $\inline&space;\alpha$, after cutting it in half these short sides must be of length $\inline&space;\alpha/2$.Now, we added our little square to the top and it'll fit perfectly, but the square dosen't line up naturally with the bottom rectangle.  That rectangle has short side of length 1, so we need to squish the red square into a rectangle with a side length of 1 (without changing its area of course).  This isn't hard to do;Now we have expressed this added area in two different ways (visualized by two different shapes), which makes it convenient to add to each side of the equation.  Pushing the pieces together;Nice!  The top half of the equals sign has been formed into a square. (In fact, if you were to step by step transcribe these pictures into symbolic expressions, all we did was "complete the square")   Let's work on the bottom half now.  To make things a little less cluttered, I'm going to fuse the purple and red parts into a single bar (whose length I will call $\inline&space;\delta$ for convenience).How do we take a given area and find the side length of a representative square?  Well, thats what the square root function is all about!  Lets "squish" this green rectangle into a square as shownPutting this back into the equation with our top square, we have reached our goalWe have been careful to perserve the equality throughout all of this, so these two squares have the same area.  This means that the side lengths on each side of the equality must be the same; and we have the equationTo get back to our solution symbolically then, we need to know our formula for $\inline&space;\dpi{100}&space;\delta$:  it is the length of our green bar, which was the sum of the lengths of the purple rectangle and the red rectangle.  Thus;Plugging this in and solving for x:This gives us the solution in terms of our form for the quadratic; but the formula is standardly presented for the "general form" with coefficients a,b,c.  We can get to this by simply substituting in the correct definitions (which can be seen towards the top of this article)$\dpi{150}&space;x=-\frac{b/a}{2}\pm\sqrt{-\frac{c}{a}+\frac{{b^2}/{a^2}}{4}}$Which can be simplified a bit to $\dpi{150}&space;x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$So there ya got it; we were able to derive the quadratic formula using only simple algebraic manipulations, the "sneaky parts" of its derivation where accomplished using pictures.