>> This is Part 2 of solving
quadratic equations using the
quadratic formula, and in this
video we use the quadratic
formula to solve these
2 quadratic equations.
Alright, we're going to solve
this quadratic equation using
the quadratic formula.
The first thing we need to do
is set the equation equal to 0
so we need to subtract
4 from both sides
to have 8M squared minus
3M minus 4 equal to 0.
And now we need to identify
our variables A, B, and C,
which are the coefficients.
So we have A is equal to 8,
B is equal to negative 3,
and C is equal to
negative 4, alright?
So We want to plug that in
to the quadratic equation,
which if you know the song
you can sing it to get it,
which I'll do or you
can just write it;
it's X equals negative E plus
or minus the square route
of V squared minus
4AC all over 2A.
Okay so there's the
quadratic formula,
you don't have to sing it.
And now we're going to plug
in the values for A, B,
and C. Okay, so this big
old fraction part I'm going
to put opposite of B. Now
B is negative 3 so I plug
in negative 3 for B plus
or minus and I'm going
to do B squared, so you could
either write negative 3 times
negative 3 or put in
parenthesis negative 3 squared.
Don't forget to put
a parenthesis
around the negative 3
if you write it this way
minus 4 times A times C. Now,
A is 8 and C is negative
4, and that's all
under the square root.
And then the denominator is 2
times A so I have 2 times 8.
Okay, so that's just plugging
in the values for A, B,
and C into the quadratic
formula and now we want
to simplify the numerator
and denominator.
So I have the opposite of
negative 3 so that's 3 plus
or minus and then under the
square root I have a negative
3 squared.
Remember that means negative
3 times negative 3, 9,
and then have minus 4
times 8 times negative 4,
so that's going to be a plus
and you have 4 times 8 times 4,
32 times 4 that's 128, all
over 2 times 8 which is 16.
Almost done.
[ Silence ]
>> You need to simplify inside
that square root 9 plus 128
is 137 all over 16 and you see
if you could simplify the
square root of 137 but you can't
because there's no perfect
square that's a factor of 137,
so we've got 2 solutions.
3, -- oops, more like that --
3 plus the square root of 137
over 16 and 3 minus the
square root of 137 all over 16
and that's our answer.
Now, a lot of people get
kind of confused with all
of this mess here in the
beginning when you're filling
in all the numbers so I'm going
to show you another way
you can approach it.
I'm going to start out the
same, the same equation
after I subtract
4 from both sides
and I've identified
the coefficients A, B,
and C. What I do is
I figure out the part
that goes underneath the square
root first, so I kind of get
that all out of the way.
So B squared is going to be
negative 3 times negative 3.
Instead of writing it out as a
negative 3 parenthesis squared I
could do that in my head,
B squared is negative
3 times negative 3
or 9 minus 4 times AC, and then,
also, I just multiply A times C;
8 times negative 4,
which is negative 32.
So it's just a little bit,
you know, less of a mess
and now I need to simplify
that so that'll be 9 --
well, a negative times a
negative here that's going
to be a plus right, 128 or 137.
So by doing this when
I'm now going to plug it
into the quadratic formula I
already know that's the number
that's going to go
underneath the square root.
Alright, also, I know
in the quadratic formula
it's negative B plus
or minus the square root of B
squared minus 4 AC all over 2A,
so when I plug it in I'm going
to do negative B...I'm going
to look up here and I'm going
to say, well, B is negative 3
so negative B would be the
opposite of that positive 3 plus
or minus the square
root of, well,
that's what I already figured
out so that's what goes
underneath the square root right
here, the 137 and then
the denominator is 2A.
Okay, so I look up
here 2 times 8 is 16.
And it's just a little
bit faster.
Notice how quick it is to write
the answer so I do a little bit
of figuring out ahead of time
and I still have
the same 2 solutions
as I did it right here 3
plus radical 137 over 16
and 3 minus radical 127 over
16, so the same exact answer,
2 ways of showing your work.
Alright here's our next
example we're going to solve
by using the quadratic formula
so the first step is
set the equation to 0.
X squared minus 4X minus 8
equals 0 then we're going
to identify A, B, and C,
so remember those
are the coefficients.
So the coefficient
of X squared is 1.
The coefficient of
the X is negative 4
and the constant is negative 8,
so we're going to plug that in
to the quadratic formula the
quadratic formula is negative B
plus or minus the square root of
B squared minus 4AC all over 2A,
which, of course, you could
sing to remember that.
So what do we have X equals --
alright, so it says the opposite
of B, B is negative 4
plus or minus B squared
so that would be negative 4
squared minus 4 times A times C
which is negative 8 all over
2 times A which is 2 times 1.
Alright, so we have the opposite
of negative 4 that is 4 plus
or minus the square root of --
alright, now, inside
the square root we have
to square negative 4, negative 4
times negative 4 is 16 you have
a minus 4 times 1 times
negative 8 that's going
to be a plus 32 all over
2 times 1 which is 2.
Okay, and remember
you could have figured
out the B squared minus
4AC first and figured
out that this number underneath
the square root is going
to be 48 and that's your
next step now is to go ahead
and simplify underneath
the square root.
So I have 4 plus or minus the
square root of 48 all over 2.
Alright, now, we need to
simplify the square root of 48.
Remember that 48 is 16
times 3 so just keep
in mind the square root of
48 would be the square root
of 16 times 3 which
is the square root
of 16 times the square root of
3 so it would be 4 square root
of 3, so that's what I'm
going to do over here.
I'm going to write
this as 4 plus
or minus 4 square
roots of 3 all over 2.
Okay, our next step is to
simplify this fraction.
Now, to simplify fractions
what you do is you can factor
and cancel so that's
what we're going to do.
So we're looking at
the numerator here
and I could factor out a 4.
[ Silence ]
>> And cancel the
4 and the 2 now.
And now I've got to do the
distributive property 2 times 1
is 2 plus or minus
2 times square roots
of 3 is 2 square roots of 3
and that gives me 2 solutions;
2 plus 2 square roots of 3 and
2 minus 2 square roots of 3.
Now, you notice I started
writing this 4 plus
or minus 4 root 3 over
2 again down here.
I'm going to show you another
way since this is a fraction
with a binomial in the numerator
you could put each term
in the numerator
over the denominator,
so I have 4 over 2 plus or
minus 4 square roots of 3 over 2
and then you would cancel
each of those 4 over 2 is 2
and then this 4 over 2 is
also 2 so you have 2 plus
or minus 2 square roots
of 3 that also works.
