In this video, we will solve the equation
2(u + 1)(3u - 1) = 1 by using the quadratic
formula. As a starting point, we need to remember,
if we're going to use the quadratic formula,
we first have to have our equation in standard
form, meaning that it must be expanded out
and it must be equal to 0. Essentially, we
want it to look like ax² + bx + c = 0. Once
we have it in that form, we would then be
able to take that equation and solve it for
x using the quadratic formula, where x = the
opposite of b plus or minus the square root
of b² - 4ac all over 2a. So that is what
we want to do, but as a starting point, when
we look at our equation, 2(u + 1)(3u - 1)
= 1, we should notice that it's not in standard
form and it's not set equal to 0. So the first
things we'll need to do is expand out the
left side, and then subtract 1 to bring that
1 over to the left side. As a first type of
expanding, we’ll expand the two binomials
being multiplied. We’ll keep the 2 out front
but (u + 1)(3u - 1) gives us (3u² - u + 3u
- 1), if we just foil out the binomials. We
can then simplify what's inside the parenthesis
by combining like terms. So we have the 2
outside parentheses. Inside parentheses, (3u²
+ 2u - 1) close parenthesis = 1. And one more
step of simplifying the left side of the equation,
we'll distribute the 2 into the parenthesis.
2(3u² + 2u - 1) becomes 6u² + 4u - 2. And
that will equal 1. So we now have expanded
off the left side. We still need to subtract
1 from both sides to set it equal to 0. So
the equation we want to solve at this point,
which is now in standard form, is 6u² + 4u
- 3 = 0. Now that it is in standard form,
we want to make sure we recognize our a value
is 6, our b value is 4, and our c value is
-3. The other thing we need to recognize is
that we are not solving an equation for x.
So when I go to state the quadratic formula,
it will start out as “u =.” So u = the
opposite of b plus or minus the square root
of b² - 4ac all over 2a. When we substitute
in our a, b, and c from our equation, this
will become u = the opposite of 4 plus or
minus the square root of 4² - 4(6)(-3) all
over 2(6). Continuing simplifying that expression,
the -4 in the front will stay the same. In
the 2(6) in the denominator can be simplified
to 12. So we'll have -4 plus or minus a square
root all over 12. And inside that square root,
4² is 16. -4(6) is a -24. -24 times -3 is
a +72. And so our formula has now simplified
down to -4 plus or minus the square root of
16 + 72 all over 12. Cleaning up one more
step inside that radical, 16 + 72 is 88. So
we get the statement, u is equal to -4 plus
or minus the square root of 88 all over 12.
We now are in a situation where a radical
can simplify. Now 88 is not a perfect square,
but there is a perfect square that divides
into it. 4(22) is 88, so I can split apart
the square root of 88 into the square root
of 4 times the square root of 22. Anytime
we have a radical where we can simplify it,
we do need to simplify it. So continuing our
work, we have -4 plus or minus 2 root 22 all
over 12. I'm running out of room, so we'll
bring that back up over here and continue
with u =. The only way we can simplify a fraction
like this, in the particular because there
is addition and subtraction separating the
numerator, is if each of the terms up in the
numerator has a common factor with the denominator.
4 to root 22 and 12 can all be divided by
2. So we will cancel out a factor of 2 from
each of those, which will give us a statement,
u = -2 plus or minus root 22 all over 6. Since
that radical root 22 does not simplify any
further, we have done all the simplification
that we can do, and we have our solutions.
We’ll simply conclude with a solution set
statement. The solution set in this case is
-2 plus or minus root 22 all over 6.
