Question about this guy's new quadratic solving method

Treppen

Just trying to visualise his method graphically and want to double check.

https://www.popularmechanics.com/science/math/amp30152083/solve-quadratic-equations/

So at 1:46 when he says the 2 numbers must be the same distance away from their average. Does he mean the "average" is the X value which is half way between the 2 roots?

Anyone any other solving quadratic method besides the -b , completing the square or guessing by add and multiply?

Yakuza

That. Was. Stunning(ly beautiful in its simplicity).
Mind. Blown.

Yes, as best as I can make out the average he refers to is the average of the two roots. Chosing it in the form he does makes the product (x+u)(x-u) = x²-u² which he can then solve easily. I am stunned at how simple he's made it.

timsey tiger

I really like this method. Thanks for sharing it.

Another method is completing the squares, which is typically used in deriving the -b formula.

This method is kind of the same as completing the squares, but I think that it seems much simpler to explain.

zv2

This is amazing. I'm looking at it in terms of modular arithmetic:

ax = 1 (mod m)

(a + x)/2 = u

u^2 - v^2 = 1 (mod m)

Gaoth Laidir

Where did he get the 4 to begin with? Why 4?

Michael Collins

Because if you have two numbers that sum to 8 i.e.

y + z = 8

their average must be

(y + z)/2 = 4.

The average is useful, because the average of two numbers is always exactly midway between the two.

In this case:

y............4............z
<---u---> <---u--->

So, in this example, y is 4 - u and z is 4 + u.

Proof that average of two numbers lies midway between the two numbers:

Distance of y from average:

(y + z)/2 - y = (z - y)/2

Distance of z from average:

z - (y + z)/2 = (z - y)/2

i.e. both the same.