In this video, we will use the equation 8y²
- 2y = 1. And we will want to solve it. We
will solve it, ultimately, using the quadratic
formula in part b. But before we can do that,
we need to answer part a. So in part a, we
are asked, “What are the values of a, b,
and c?” Well, before we can answer that
question, we need to make sure that our equation
is set equal to 0. So we'll take the equation
8y² - 2y = 1, and we'll set it equal to 0
by subtracting 1 from both sides, giving us
the new version of the equation 8y² - 2y
- 1 equals 0. You never want to identify your
a, b, and c values until your equation is
actually set equal to 0. You should notice
by having moved the 1 from the right side
over to the left side, it is now a -1. And
we need to have the correct sign on each of
the values. So for our particular equation,
our a value is 8, our b value is -2, and our
c value is -1. Now that the equation is set
equal to 0 and we have our a, b, and c values,
now we can solve the equation using the quadratic
formula. While we do typically state the quadratic
formula as x equals the opposite of b plus
or minus the square root of b² - 4ac all
over 2a, we do want to be particularly careful
to make sure that the variable in the equation
matches the variable that we're solving for.
What I mean by that, is that our equation
has 8y² - 2y - 1. There are no x's in the
equation. And as such, our quadratic formula
should not say “x equals.” But in this
particular case, it needs to say “y equals.”
And now that we have the quadratic formula
set up for our particular equation, we need
to substitute in 8 for a, -2 for b, and -1
for c. So our quadratic formula as we start
to solve will become y equals the opposite
of (-2) plus or minus the square root of quantity
(-2)² - 4 times 8 times -1 all over 2 times
8. I want to explicitly point out the parentheses
around the -2 in the front where we have the
opposite of -2 and the parentheses where that
-2 was squared are critical. Leaving out the
parentheses, especially where that -2 is being
squared under the radical, if you leave out
the parentheses, you fundamentally change
what it is you're trying to say. We need to
square both the negative and the 2. And so,
therefore, the parentheses are necessary.
Continuing on with our simplifying at this
point, there are three things we're going
to do as our first step. First I'm going to
simplify b in the front: the opposite of
-2. The opposite of -2 is simply a positive 2. The
next area to simplify is underneath the radical.
So we're not going to adjust or manipulate
the radical. But (-2)² is a +4. And -4 times
8 times -1 is a +32. And the final step of
simplifying that we'll do is down here with
the 2 times the 8. 2 times 8 is 16. The more
comfortable you become with the quadratic
formula, the faster you'll be able to do the
arithmetic simplifications that you need to
do to finish the work. The next step will
be to combine what is under the radical. So
we'll end up with 2 plus or minus the square
root of.... We had 4 + 32, but 4 plus 32 is
36. And so we end up with 2 plus or minus
the square root of 36 over 16. Bringing that
back up top to give us some more room here,
we have y equals-- well this radicand is a
perfect square under that square root. So
2 plus or minus the square root of 36 can
simplify down to 2 plus or minus 6 all over
16. One of the things you'll start to probably
find somewhat interesting about the quadratic
formula is the cleaner the square root, the
more work you have to do. Often the messier
the radical, the sooner you can stop your
simplifying process. But in this case, to
finish finding the solutions, where we have
y equals 2 plus or minus 6 all over 16 here's
a place where we do need to separate this
into the two possibilities and work each one
individually. So one possibility is that y
equals 2 + 6 all over 16. And the other possibility
is the y equals 2 - 6 all over 16. Just like
when we use the square root property, we ended
up with a positive or negative radical. Again,
there is a direct connection here. That's
what we have built into our quadratic formula:
a positive and a negative radical. Simplifying
the left option, y equals (2 + 6)/16. 2 +
6 is 8. And 8/16 is ½. Over on the right
side, simplifying (2 - 6)/16. 2 - 6 is -4.
And -4 over 16 is -¼. So again, we do get
two solutions, one involving a positive square
root, and then one involving a negative square
root. And we can conclude with our quick solution
set notation statement. The solution set is
the set ½, -¼. Like with any equation, we
can check our work by throwing the 1/2 back
into the y’s in the original equation, or
the negative 1/4 in for the y’s in the original
equation. Given the fact that we have fractional
answers, or we're starting to get messier
answers even then, that I would definitely
make use of a calculator to do a lot of this
checking.
