In this video, we will review solving quadratic 
equations by factoring, by extracting square roots, 
by completing the square, and by using the 
quadratic formula.  In this equation, we solve a 
quadratic by
factoring. We are going to solve the
equation 3x squared + 2x = 5.
Now you are tempted to just factor that right
away. Because that is what we are
supposed to do right? Factor the x out? But
no! Before you factor you have to move all terms 
to one side. We do that by subtracting 5 from 
 
both sides.
That gives us 3x squared +2x-5=0. Now we factor
this. And this factors as (3x
+5) times (x-1) = 0.
And now becuase this is equal to zero we can
apply the Zero Product Law,
letting each factor = 0. We get 3x+5=0 and
x-1=0.
Now we solve each of these equations. 3x+5=0
is solved by subtracting 5 from
both sides, using the Subtraction Property
of Equality, giving us 3x=-5. And
then dividing both sides by 3, using the Division
Property of Equality, giving
us a solution of x=-5/3. Our other equation
x-1=0 is solved simply by adding 1
to both sides, and we get x= positive 1.
We're going to do an example of solving a
quadratic by extracting square roots.
We're going to solve the quadratic equation
(2x-5) quantity squared, plus 5
equals 3. The first step in doing this type
of problem is to get your equation
in the correct form. You have to get the squared
quantity isolated on one side
of equals, and some numerical value on the
other. We do that by subtracting 5
from both sides, and we get the quantity (2x-5)
squared, equals -2.
Now we extract square roots, which essentially
is taking the square roots of
both sides of the equation. We square root
the left side, which cancels out the
square, and we square root the right side,
but we also include a +/- to account
for the negative solution. Because a negative,
squared, will square out to a
positive amount.
After we do this, we get 2x-5, the square
is canceled out, equals +/- the
square root of the right side. The square
root of -2 ends up being a complex
value of "i" times the square root of 2, where
"i" is the square root of -1.
Now we just have to solve this for x. We do
that by adding 5 to both sides,
giving us 2x = 5 +/- i square roots of 2.
Then we divide both sides by 2. And
we get x = 5 +/- i square roots of 2, divide
by 2.
In this problem, we're going to solve a quadratic
equation by completing the
square. Our quadratic is 3x squared -5x-7=0.
It's going to have a lot of
fractions in it, but this is the kind of problem
that gives students problems,
so I wanted to do one like this.
The first step is to isolate the x-squared
and x terms. We can do that by
adding 7 to both sides of the equation. This
gives us 3x squared - 5x = 7.
Our next step is to get the coefficient on
x squared equal to 1. We can do this
by dividing both sides by 3 or multiplying
both sides by 1/3. Either way, we
get x squared - 5/3 x = 7/3. This is always
how you start completing the
square. You can do these two first steps in
either order. But you always want
to get 1x squared, and you want to have the
x squared and x terms on one side
of =.
Your next step is to add this "special amount"
to both sides that is going to
make the quadratic into a perfect square.
You always take 1/2 of the
coefficient on x. In this case the coefficient
on x is 5/3, and 1/2 of 5/3 is
the same as 1/2 times 5/3, which is 1 times
5 over 2 times 3. That's 5/6.
Then we take that amount that we got and we
square it. Squaring 5/6 is the same
as 5/6 times 5/6. That makes 25 over 36. So
we add 25/36 to one side and we add
25/36 to the other side. Now in this same
step I'm going to simplify the right
side by getting a common denominator. If we
multiply 7/3 by 12/12, we get 7
times 12 is 84 over 3 times 12 is 36. We can
then add that to our 25/36 and
make it into a single fraction of 109/36.
Now, on the left side you should have a perfect
square. Because that is what
this process does. It gives you a perfect
square. We can factor our quadratic
into (x-5/6)(x-5/6). Now here's a hint: That
quantity 5/6 in the binomial is
always going to be the square root of that
"special amount" that you added. So
here we added 25/36 and the square root of
that is 5/6, it's always going to
work out that way.
Now we can rewrite this as (x - 5/6) quantity
squared, equals 109/36. We now
have an equation we can solve by extracting
square roots - it's in the form. We
take the square root of both sides. And we
get x - 5/6 = +/- the square root of
109/36. We can write the square root of 109/36
as the square root of 109 over
the square root of 36, and that square root
of 
36 turns into 6. So our
simplified form is x - 5/6 = +/- the square
root of 109 all over 36. Now, to
solve, we just add 5/6 to both sides, to get
x by itself. And we get x= 5/6 +/-
the square root of 109, over 6. And since
both terms have a denominator of 6,
we can combine them together and write as
x= (5 +/-the square root of 109),
all over 6. And if you were to solve this
with the Quadratic Formula, that is
what you would get just as well. And some
people ask, well "Why complete the
square then?" Well, completing the square
has a lot of value in Calculus and
other advanced classes. We often want to write
a quantity in a perfect square
form. And this is the way to do it. So this
procedure has a lot of applications
besides solving quadratics.
The final method we are going to look at for
solving quadratic equations is to
use the Quadratic Formula. This is probably
the easiest method to use because
it just involves plugging coefficients into
the formula x = -b +/- the square
root of (b squared - 4ac) all over 2a. One
of the things students forget,
however, is that you first have to have your
equation in the form a-x squared +
bx + c = 0.
Here in our example we have the equation (x+3)
quantity squared = x-2. We have
to do a bit of work here before we can use
the formula.
We have to multiply out all the terms on the
left. We do that with the
Distributive Property. And we get x squared
+ 6x + 9, on the left, equals x-2
on the right. Then we have to subtract x from
both sides and add 2 to both
sides. And that gives us the equation x squared
+ 5x + 11 = 0.
Now it is in the form where we can use the
quadratic formula. The coefficient
on x squared is a=1, the coefficient on x
is b=5, and the coefficient of the
constant term is just c=11. So we just plug
those values into the formula, and
we get x= -5 +/- the square root of 5 squared
- 4 times 1 times 11, all over 2.
This gives us -5 +/- the square root of -19
over 2, and the square root of -19
can be written as the square root of positive
19 times i using the definition
of i.
So this gives us an answer of x = -5 +/- the
square root of 19 times i, all
over 2.
