this is a derivation of the quadratic
formula and we are going to start with
just a quadratic equation i will start
with a very general a x squared plus b x
plus c is equal to 0 and my only
assumption here is going to be that the
value of a is not equal to 0 and I think
that's a pretty fair assumption since if
the value of a was equal to 0 this would
not be a quadratic equation it would be
a linear equation and also we are going
to complete the square here so i'm going
to make the assumption that you have a
basic knowledge of completing the square
so what i will do is begin by
rearranging some things so i am going to
move the value of C over to the other
side of the equation and i am going to
factor the a out of both sides of both
terms that are on the right side
so what that will give me is a times x
squared plus b / a x and then i'm going
to leave some space here and that is
going to be equal to negative C and you
should pause for a second here just to
make sure that what I've done is ok if i
multiply in this a times x squared and
then a times b over a x you should
recognize that i will get a x squared
plus b x plus c now I want to complete
the square on this part of the
polynomial and what we do when we're
completing the square is take that
linear term which is b over a the
coefficient of that linear term we're
going to divide it in half so i'll
multiply the a by 2 and then we
square it and that is what is going to
get added here so i'll write it as B
squared divided by four times a squared
and then I need to make sure that my
equation still holds so since i added
that to the left side i need to add it
to the right side as well but i need to
keep in mind that this has been added
inside of the parentheses and it's being
multiplied by the value of a so what I
really added to the left side of the
equation was a times b squared divided
by four times a squared
clearly i'm going to be able to cancel
the a's but recognize that what I
actually added to the left side of the
equation was this term here now if we
clean things up a little bit i now
have a times x plus b divided by two
over a squared that's the way that this
much what is inside the parentheses is
going to factor that is going to be
equal to let's see the day is going to
one of those days is going to cancel and
i'll write it in a different order so
that it's b squared divided by four
times a minus c okay then get a little
more space here i will divide both sides
by a to give me x plus b / 2 times a
squared is equal to b squared divided by
four times a squared minus c divided by
a and I'd like to take this
that's on the right side of the equation
and put it make it into one term so i
need to multiply the bottom of this by
four times a so multiply the top by four
times a and then that on the right side
will give me b squared minus 4 times ac
all / 4a squared and if you are familiar
with the quadratic formula you should
begin to see it taking shape here so x
plus b over 2a squared
now I want to get rid of that squared so
i will take the square root of both
sides of the equation which will give
the x plus b divided by 2 times a is
equal to plus or minus the square root
of b squared minus 4ac all divided by 4a
squared
I let me work on the right side here a
little bit i could make that the square
root of the top which is b squared minus
for ac divided by the square root of the
bottom the square root of 4a squared is
two times a and then I'll take the b
over 2a and move it over to the other
side so that will give me x is equal to
negative b divided by 2a plus or minus
the square root of b squared minus 4ac
all divided by 2a and conveniently those
have the same denominator and so i can
put those together and get the quadratic
formula which is x is equal to negative
b plus or minus b squared minus 4ac
with a square root all divided by 2
times a
tada
 
