Bam, Mr. Tarrou-
I just did a video showing you how to use
the “completing the square” process to
help you solve quadratic equations; those
are equations with a degree of 2, an exponent
of 2.
We’re going to now move onto the quadratic
formula, which is another method for solving
quadratic equations. And it’s sort of the
general default, the “go to” process for
solving 2nd degree equations because it always
works. Like, you might have just learned how
to “factor” to solve 2nd degree equations
(quadratics) but, you know, not all of those
are factorable, so that process, while it
can be quick, has its limitations.
Completing the square is a good process and
it pretty much always works, but the quadratic
formula is just a formula that comes back
over and over again and it’s extremely important.
But before we get to that topic, I wanted
to show you another example of why you are
learning so many ways of solving these 2nd
degree (quadratic) equations.
Well, like, why did I just cover completing
the square when I have this other process
that always works? Well, here we have a general
quadratic, standard form quadratic, ax squared
plus bx plus c equals zero. And of course,
there are no numbers in it, but I’m going
to use the “completing the square” process
to derive the quadratic formula for you, which
you are then going to use in the next video
of actually solving some equations.
So, when you complete the square, your leading
coefficient has to be one, and all the terms
with the variable x (or whatever book you
are using for the particular problem you are
working on), needs to be isolated. So we’re
going to do a couple of things, we’re going
to move c over to the other side to get the
constant away from the variable x, the independent
or domain variable.
So we have ax squared plus bx equals negative
c because we’re going to subtract c from
both sides of the equation. To get those domain
or independent variables alone—the x’s.
Now, you cannot complete the square unless
your leading coefficient is equal to one.
So now, we’re going to divide all of our
terms by a, so we have that leading coefficient
of one that we need. And we get x squared
plus b over ax. I’m going to leave some
space for completing the square, equals negative
c over a.
Now, completing the square—the process—is
you find that single degree term and what’s
in front of it is “b” but this is sort
of a derivation so it’s a little bit more
complicated than that, but we take ½ of the
coefficient of the single degree term, which
is normally just referred to as b when you
are actually solving equations using completing
the square; in this case we have b over a,
so we are going to take ½ of this coefficient
and we are going to square it, so 1 times
b is b; 2 times a is 2a. We’re going to
square that ratio and get b squared. 2 times
2 is 4. Now, it’s not 2 times 2, but it’s
2 squared, so it’s 2, times itself, which
is again 2, so it’s 4a squared.
So I’ve taken my single degree term, I’ve
multiplied by ½, which is the same as dividing
by 2 and that is going to get added to both
sides of this equation.
Okay, so here I’ve got where I stand at
the moment, and when you complete the square,
you set up a perfect square trinomial. That
is factorable by simply just doing a pattern
of square-rooting the first term, square-rooting
the last term, and keeping the sign of the
middle term.
So, this left hand side is going to factor
to be the square root of the x squared, which
is x, keep that middle sign which is positive
or plus, and the square root of b squared
is b; the square root of 4 is 2 and the square
root of a squared is equal to 2; I can do
that individually because this is just a monomial
over a monomial.
So we have, and that’s how we started off,
right, b over 2a and then, we squared it,
so we have b over 2a and that binomial is
going to be squared. I’m just doing the
memorization pattern for factoring a perfect
square trinomial. We get x plus b over 2a
squared equals negative c over a, plus b squared,
over 4a squared. Now, I’d like to write
these two terms on the right hand side of
my equation as only one term, and I need common
denominators to do that, so I’m going to
come back through here and I’m going to
multiply this by 4a. On the top and the bottom—the
numerator and denominator of this fraction,
and that is going to give me negative 4ac
and now I’m going to have a common denominator
so I’m going to write this as one fraction,
so negative 4ac because of that negative right
there, plus b squared, all over 4a squared.
So, you might want that cleaned up a little
bit, so we’re going to take x plus b over
2a squared equals negative 4ac minus b squared,
all over 4a squared. When you do the completing
the square process, after you factor that
perfect square trinomial you’ll always have
a binomial that is squared, equal to something
else so we’re going to undo that power of
2 by square-rooting both sides of the equation
and we’ll need to remember that when we
introduce our own square root, (or even root
for that matter) we need to take into account
our positive and negative answers.
So, square root the left; square root the
right, and the square root and that power
of 2 are going to cancel out, so we have x
plus b over 2a. The denominator of this fraction
is just a monomial so I can write this as
plus or minus the square root of… oops;
that’s a plus and that should be a plus
there, so that was a sign error. So we should
have, let’s see…Now I know I’m developing
the quadratic formula, so let me go ahead
and take this positive b squared that is in
the 2nd term and the negative 4ac which is
currently my first term. Now, I’m going
to just swap places. Now, when we do this,
subtraction is not commutative so I want to
make sure I move my signs with my terms. We
have a positive b squared, and minus 4ac;
you might recognize that as the discriminant
or what’s underneath the radical in the
quadratic formula. That is over the square
root of 4a squared.
I’m using the property of radicals, which
says, if I’m taking a radical, in this case
the square root of a fraction, I can square-root
the numerator and denominator separately or
separate that symbol—into—again, one for
the top and one for the bottom.
Now, you might see this b squared and think,
“Well, I can square root b squared and make
it b, I see 4, which is a perfect square,
and think, “Well, I can square-root 4 and
make it 2, but this is a binomial in the numerator
and you can’t just apply the square root
symbol to the separate terms, but you can
apply it to a monomial and apply it to each
of these two factors, but this is only one
term. The square root of 4 is equal to 2 and
the square root of a squared is equal to a,
so the square-root of 4a squared is equal
to 2a. And again, I can do that because it
was a monomial. More complicated if that were
something like 4 plus a squared, I would not
have been able to do that.
Well, let’s see, I have got a—I want to
try to solve for x. I’ve got a b over 2a
over there on the left, so I’m going to
move that b over 2a to the right hand side
of the equation. We are trying to solve for
x when we use the quadratic formula. So, we’re
going to say x is equal to…. I’m going
to subtract b over 2a from both sides of the
equation. I don’t like the way I’ve written
that. There we go. Now, I’m subtracting
b over 2a to both sides of my equation to
get x by itself. Anything subtracted by itself
is equal to zero. When you add zero, nothing
changes, so boom it’s gone. This term has
a denominator of 2a, and this term has a denominator
of 2a so they have common denominators so
we have got 2a and look at what I’ve got?
Minus b so this is negative b plus or minus
the square root of b squared, minus 4ac and
that, oops, I kind of covered up my square
root symbol there a little bit, but that is
the quadratic formula and that is how you
go from an equation in standard form, use
completing the square to help you develop
the quadratic formula, and that will solve
any quadratic equation that you like, whether
it’s factorable or not factorable. Now,
of course it won’t solve the ones that don’t
have an x intercept, but we’ll see what
happens there, as we do our examples in the
next video. I’m Mr. Tarrou…BAM
