>>This is part three of solving
quadratic equations using the
quadratic formula.
We solve this quadratic equation
using the quadratic formula,
and then we also solve it
by completing the square.
This is part three
and we're going
to continue solving
quadratic equations using the
quadratic formula.
So, here's an example;
1/6 X squared plus
X plus 1/3 quails 0.
This is in standard form and
you could say, "Well, A is 1/6,
B is 1, and C is 1/3."
But when you plug this
into the quadratic formula,
it gets pretty complicated
looking; you're going
to have some complex fractions.
So, there's an easier way
to deal with this one.
Since it's an equation, you
could multiply both sides
by the least common denominator
to eliminate the fractions.
So if you want for fun,
you could go ahead and plug
in these values of A, B and
C into the quadratic formula
and check that you're going
to get the same answer
as the way I'm going to do it
which is multiply both sides
of the equation by 6, the
least common denominator.
So if I multiply the terms on
both sides of the equation by 6,
so I have 6 times 1/6x squared
plus 6 times X plus 6 times 1/3
equals 6 times 0.
I've multiplied both sides
of the equation by 6,
the least common denominator,
to element fractions.
When I do this, I get
a much easier problem.
Now, you can't factor
this by the way.
We're practicing using
the quadratic formula,
but you wouldn't be able
to solve this by factoring.
You could solve this by
completing the square
and in fact, I think this
would be an easy one to do
because it's got a 1 in
front of the X squared
and this middle term is even
so it will be easy
to take half of it.
But we're going to go ahead
and practice using the quadratic
formula so in this case, A is 1,
B is 6, and C is 2 and hopefully
you'll agree that's a little bit
easier than plugging in 1/6, 1,
and 1/3 for the values of A, B,
and C. Alright, so
we're going to plug this
into the quadratic formula
now; X equals...alright.
So, if we kind of check
the whole thing out here,
I've got the opposite of B, so
opposite of 6 negative B plus
or minus the square root, so
plus or minus the square root
of B squared, that's 6 squared.
You don't have to put
parentheses around the 6
because it's not negative and
it won't make any difference
when it's already positive.
Minus 4AC, so that's
4 times 1 times 2.
All over 2a, 2 times 1.
So, that's plugging
in all the values
into the quadratic formula
so you have negative 6 plus
or minus and now underneath the
square root you've got 36 minus
8 all over 2.
And 36 minus 8 is 28, so you've
got the square root of 28.
Now, I'm going to
write 28 as 4 times 7
because that will make it
easier to simplify writing it
as a perfect square times 7
and you get negative
6 plus or minus.
Now you could take out...the
square root of 4 is 2
so I could take out a 2 and
also have the square root
of 7 all over 2.
Now you could split
it up at this point.
There's 2 ways to simplify this.
[ Pause ]
What you could do is factor
out the 2 out of the numerator
and that gives you
negative 3 plus
or minus the square
root of 7 all over 2.
And now the 2's a factor, you
could cancel and you're going
to get your two answers.
One is going to be negative
3 plus square root of 7
and the other is going
to be negative 3 minus
square root of 7.
You could have also taken
this step right here
[ Pause ]
and split the numerator up
so you could put negative 6
over 2 plus or minus 2
square roots of 7 over 2
and then you could just
reduce each of these,
so its negative 3 plus
or minus square root of 7
which will lead you to
the same 2 solutions;
negative 3 plus square
roots of 7
and negative 3 minus
square roots of 7.
Ok, and of course if you were
to plug in those fractions,
it would be very messy to plug
those into the quadratic formula
but if you completely simplify
it, if you just kept going
for it, you should end up with
exactly the same two solutions.
Now, I'm going to take
that problem one more time
and go ahead and do it
by completing the square
so you could see that that's
actually an easier way,
well it could be an
easier way for some people.
So, here's the problem again
and this was the solution;
we got negative 3 plus radical
7, negative 3 minus radical 7.
Let's just do this by
completing the square just
so you could see an
alternate way to do this.
So remember, you put the X
squared term and the X term
on the left side, the
constant on the other side,
and you're going to be
adding the same number
to both sides of the equations.
And we bring down...what
got to be
in parentheses, that's
got to be X...
If it's a plus 6 it's got
to be half that, so plus 3.
And then 3 squared is what
gets added to both sides.
So in other words,
X squared plus 6,
X plus 9 factors to
X plus 3 squared.
And over on the right hand
side, negative 2 plus 9 is 7.
And now we take the
square root of both sides.
Don't forget when you take
the square root of both sides,
you have to put plus
or minus in front
of the number on the right.
So, it's plus or minus
the square roots of 7.
The last thing is to
subtract 3 from both sides.
And so same answer, you would
have negative 3 plus square
roots of 7 and negative 3
minus square roots of 7.
So you might want to
compare that to what happened
when we used the quadratic
formula where we had to worry
about simplifying at the end.
Let's just look up
here a little bit.
This is what it looked
like when we were doing the
quadratic formula.
So we had a lot of
simplification here
and then we had to
factor out the 2
and do a little more simplifying
to finally get those answers
but you do get the same
solution either way.
Ok? So, you could use
the quadratic formula,
you could use completing
the square,
and if you want you could use
the quadratic formula using the
fractions of the original
quadratic equation I gave you.
So, what I wanted to
illustrate in this problem is
that if you do have a quadratic
equation with fractions,
you could multiply both
sides of the equation
by the least common denominator
in order to clear the fractions
and have easier numbers to
plug in for A, B, and C.
