Where does the quadratic formula come from?

NOTE:

You really don't need this: It's just for the maths purists.

Where does x=(-b+-sqrt(b^2-4ac))/(2a) come from?

We know the general formula for a quadratic equation is:

ax^2+bx+c=0

Solve using "completing the square method"

Complete the square ax^2+bx+c=0

NOTE:

In order to solve this the coefficient (or number) in front of the x^2  must be a one.

So divide all sides by a

(ax^2)/a+(bx)/a+c/a=0/a

x^2+(bx)/a+c/a=0

Remember

Is the same as

Fill in the table

NOTE:

This is the same as (x+b/(2a))^2

If you add up each area you get:

x^2+b/(2a)x+b/(2a)x+(btimesb)/(2atimes2a)

x^2+b/ax+b^2/(4a^2)

x^2+(bx)/a+c/a=0

We now have

x^2+b/ax+b^2/(4a^2)

Original number -  New number (see completing the square)

c/a-b^2/(4a^2)

So                 x^2+(bx)/a+c/a=0

Is the same as:

(x+b/(2a))^2+c/a-b^2/(4a^2)=0

(x+b/(2a))^2=-c/a+b^2/(4a^2)

x+b/(2a)=sqrt(-c/a+b^2/(4a^2))

x=-b/(2a)+-sqrt(-c/a+b^2/(4a^2))

This is actually finished but now mathematicians believe that it's better to simplify the formula even more by:

x=-b/(2a)+-sqrt(-c/a+b^2/(4a^2))

Multiply the right hand side by (2a)/(2a)    (=1)

x=(2a)/(2a)times(-b/(2a)+-sqrt(-c/a+b^2/(4a^2)))

NOTE:

On the next step (2a)/(2a)=(sqrt((2a)^2))/(2a)  and then

x=(2a)/(2a)times(-b)/(2a)+-((sqrt((2a)^2))/(2a)timessqrt((-c)/a+b^2/(4a^2)))

x=(2a)/(2a)times(-b)/(2a)+-((sqrt((2a)^2)timessqrt((-c)/a))/(2a)+(sqrt((2a)^2)times sqrt(b^2/(4a^2)))/(2a))

x=(2a)/(2a)times-b/(2a)+-sqrt((2a)^2times(-c/a)+(2a)^2times(b^2/(4a^2)))/(2a)

x=(cancel(2a))/(2a)times(-b/(cancel(2a)))+-sqrt((2a)^2times(-c/a)+(2a)^2times(b^2/(4a^2)))/(2a)

x=(-b/(2a))+-(sqrt((2a)times(2cancel(a))times(-(\ c)/cancel(\ a))+(cancel(\ 2a)timescancel(\ 2a)timesb^2)/(cancel(4)timescancel(\ a)timescancel(\ a))))/(2a)

x=(-b)/(2a)+-sqrt(-4ac+b^2)/(2a)

x=(-b+-sqrt(-4ac+b^2))/(2a)

x=(-b+-sqrt(b^2-4ac))/(2a)

This is now in the format that is recognised and used by mathematicians.