Deriving the Quadratic Formula - Concept 11,248 views
The quadratic formula is one of the most important formulas that you will learn in Algebra and chances are that you have probably memorized it. But where does it come from When deriving the quadratic formula, we first start with a generic quadratic formula using coefficients a, b and c and then derive the formula by completing the square.
So hopefully by now you have seen and
are familiar with the quadratic
formula. Negative B plus or minus the square root
of B squared minus 4AC all over 2A.
You may have a song for it,
you may have some different way
of memorizing it but hopefully you've
seen it in some context.
What I want to do now is actually spend some
time and show you how this is actually derived.
So you know that it's not some crazy
formula that comes out of thin air.
Actually what ends up happening it's found
by completing the square with a general
quadratic using coefficients A, B
and C. So what we're going to
do is complete the square going to
the exact same process but instead
of doing numbers we're
going to have letters.
So the first thing we did when we were
completing the square is to move the
C over to the other side.
So we'll go ahead and do that.
X squared plus BX is
equal to negative C.
We then -- oops, why did I put a C there.
It should be X. Try that again. All right.
So we then factored out the coefficient
on our leading term.
So that means we take out the A.
Take out the A. This becomes X
squared. When we take out A from B, this is just
going to be left with B over AX and
then equals negative C.
You can always check this because when you
multiply that A back through we should
end up with B. Here A and
A cancels leaving us with B. So from
here our goal is to make this be
a perfect square.
And there's something that we have to
add in here which corresponds to this
blank spot we had here.
How we found that was by taking this middle
term and dividing it by 2. That's
what went right in this box.
So that ends up with B divided by 2A.
That term squared is what we
ended up adding over here.
So when we square B over 2A what we end
up is B squared over 4A squared.
Remember that we have to distribute
this coefficient in.
This A -- we're adding B squared over 4A
squared inside the parentheses, but
what we're really doing is adding an
A times that to that entire side.
So we added the B squared over
4A squared to this side.
Multiplying it by A. We have
to add it to the other side.
So that's plus B squared over 4A.
One of those As cancel when
we multiply it in.
So we have A times this quantity squared
is equal to negative C plus B squared
We now want to try to get our
square terms by itself.
In order to do that we need to divide
by A. Giving us X plus B over 2A
squared is equal to negative C over
A plus B squared over 4A squared.
So this A got distributed to both terms.
Leaving us right here.
So now we need to take the
square root of both sides.
The square root of a square is
just going to cancel out.
Let's go over here and write it up.
What we end up with then is
just the X plus B over 2A.
And that is going to be equal to plus or
minus the square root of this side.
Plus or minus square root of
negative CA plus B squared.
Over 4A squared.
Before I continue on, what I want to do is
actually combine things in this square
In order to combine things we
need a common denominator.
So common denominator in
this case is 4A squared.
So this term needs a 4A over 4A.
The left side stays the same.
X plus B over 2A is equal to plus or minus
square root of negative 4AC over
4A squared plus B squared
over 4A squared.
Solving for X what we have to do then
is subtract the B over 2A over.
X is equal to plus or minus this
giant ugly square root.
And for this one I'm actually
going to combine our terms.
We have our common denominator.
You can throw them together.
Negative 4AC plus B squared over 4A squared
and then we minus, make this a little
bit shorter, B over 2A.
We're almost there.
So what we have left is
this term right here.
We can distribute this radical to both
the numerator and the denominator.
I'm running out of room over here.
And lastly what we have to do is just
a little bit of rearranging.
We have plus or minus the square root,
negative 4 AC plus B squared.
We rearrange that and we end up with B
squared minus 4 AC, square root of 4
A squared is just 2A.
Where can I draw this in?
let's go over here.
What we end up with, I'm going to
take this term, put it first. Negative B
over 2A plus or minus taking the square
term writing it first B squared
minus 4AC square root of the 4A
squared is just 2A. All right.
Kind of ran out of room right here, but
hopefully you can see that we have a
So we can just throw that all together leaving
us with negative B plus or minus
the square root of B squared
minus 4AC all over 2A.
A lot of work but I wanted to take some time
and show you the quadratic formula is not some
mysterious formula. It's just found by completing a square
using the general quadratic of AX squared plus BX plus C.