Deriving quadratic formula, Pythagorean theorem, equation of a circle

[modest kraken]

I wanna make sure I understand this before I move forward in math.

[cedar canopy]

pythagorean is easiest

lemme go get graph paper

[modest kraken]

I figured it was

[unborn cove]

triangles are fun,

so lets start

$a^2+b^2=c^2$

[regal urchin]

Styly

[cedar canopy]

so lets say u do this right?

,rotate

[cedar canopy]

so basically we rearrange the triangles to split the c^2 square into the a^2 and b^2 squares

[modest kraken]

That makes sense because it ends up being the same total volume

[cedar canopy]

yeah

you see?

alright so do you know what completing the square is?

(we're doing quadratic now)

[modest kraken]

Kinda. Visual proofs are a little bit confusing for me. I wonder how Pythagorean came up with that in the first place.

[cedar canopy]

it was actually the babylonians

so basically

we have ax^2 + bx + c = 0 right?

so first we're just gonna ignore a by redefining b as b/a and c as c/a, and since a now becomes a/a it is 1!

so we now have x^2 + bx + c = 0

notice that (x + b/2)^2 = x^2 + bx + b^2/4

so you define h = -b/2 and k as c - b^2/4

then you have (x - h)^2 + k = 0

now this can be solved for x

constructing the general formula for this method leads to the quadratic formula!

[unborn cove]

here we can see the squares and we can observe that the square of the hypotenuse can be separated into two triangles which have the same area as a and b

the sum of the two squares gives you the hypotenuse square

[modest kraken]

Oh yeah cause it's a binomial square.
Wait that is vertex form isn't it?

[cedar canopy]

yup

[modest kraken]

Oh I like that picture

That's cool

[unborn cove]

this one has different side lenghts but if you calculate the areas and sum them you can see that the small square plus the bigish one gives you the hyp one

now for the quadratic formuala

$ax^2+bx+c=0\Rightarrow x^2+\frac{b}{a}x+\frac{c}{a}=0\Rightarrow x^2+\frac{b}{a}x=-\frac{c}{a}\Rightarrow x^2+\frac{b}{a}x+(\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2$

[regal urchin]

Styly

[unborn cove]

and to finish

@modest kraken

$(x+\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2 (=) x+\frac{b}{2a}=\pm \sqrt{-\frac{c}{a}+(\frac{b}{2a})^2}(=)x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[regal urchin]

Styly

[unborn cove]

the end step is mutiplying by 2a/2a btw