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

Complete the square and split the square into quarters using x square

Work out the equations on the square

Is the same as

This square will still be the square as the one above

Fill in the table

Now fill in the table multiplying the y by x

NOTE:

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

If you add up each area you get:

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

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

We originally had

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

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.