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>> 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.