In this video, we're going to introduce a
new method for solving quadratic equations.
We're going to begin in part a by solving
the equation x² - 5 = 0 using the square
root property. To solve the equation x² - 5
= 0 using the square root property, we
have to begin by first isolating the x².
So we'll add 5 to both sides of the equation
to obtain x² = 5. Then this is the point
where we apply the square root property, which
says, if x² = 5, then x either equals
the square root of 5 or the negative square
root of 5. And we have our solution set. The
solution set is the set root 5, negative root
5. In part b, we are asked to solve the equation
2x² + 4x - 5 = 0 using the square root
property. The problem with this equation is
that we have this center term, the 4x term.
Because the variable x occurs twice in this
equation, once as a squared term and once
as what we would call a linear term, and x
to the first term, the square root property
cannot be applied very easily. So the best
answer we can give at this point is, we do
not yet know how to do this. It would be incorrect
to say that it is not possible to do this.
It is possible to do it, but it will be covered
in the future. So let's take a step back,
and answer part c. What is a quadratic equation?
The definition we are using for what a quadratic
equation is, is an equation that can be re-written
as ax² + bx + c equal to 0, where a is not
equal to 0. A quadratic equation may or may
not be in that form, but is one that can be
rewritten into that form. This brings us to
part d. What is the quadratic formula? The
quadratic formula is, “A formula that can
be used to solve any quadratic equation.”
More specifically, if the equation we are
trying to solve is ax² + bx + c = 0,
then the solutions are x equals the opposite
of b plus or minus the square root of b²
- 4ac, all of that over 2a where the a, the
b, and the c reference the a, and the b, and
the c from the quadratic equation that we're
solving. This formula can be used to solve
any quadratic equation that we could come
across. Now, in part e, we will return to
the equation 2x² + 4x - 5 = 0. And the
first part of part e is to address, what are
the values of a, b, and c, remembering that
a standard quadratic equation is ax² + bx
+ c = 0. Our value of a is a coefficient
on the x² term. Our value of b is a coefficient
on the x term. And our value of c is the constant.
So our values of a, b, and c are a is equal
to 2, b is equal to 4, and c is equal to -5.
It is critical that you recognize the sign
in front of a, b, and c gets carried into
the value of a, b, and c that we're going
to use. So if this has been a -4x for the
middle term, then our b value would have been
a -4. Carrying, the sign of a, b, and c will
completely impact the work we do when working
with the formula. In part ii of this, we need
to now solve the equation using the quadratic
formula. As a reminder, the quadratic formula
for the equation ax² + bx + c = 0 is
x equals the opposite of b plus or minus the
square root of b² - 4ac all over 2a. So what
we need to do is substitute into that formula
the value of a, b, and c that we have identified
in part 1. So for us, the quadratic formula
would become x equals the opposite of 4 plus
or minus the square root of 4² - 4 times
2 times -5. All of that over 2 times 2. To
simplify this, we need to follow the order
of operations. And the first work we need
to do will be underneath the radical. Under
the radical, we have 4², which is 16, and
-4 times 2 times -5 is a positive 40. In the
denominator, 2 times 2 is 4. So this will
simplify down to -4 plus or minus the square
root of 16 + 40 all over 4. Continuing our
work, we need to finish simplifying under
the radical. 16 + 40 will be 56. And so we
get -4 plus or minus the square root of 56
all over 4. Anytime we're working with radicals,
we do need to simplify them as much as possible.
56 is divisible by 4. 4 times 14 is 56. So
I can split that square root of 56 into the
square root of 4 times the square root of
14. And our expression becomes -4 plus or
minus root 4 times root 14 all over 4. Now
the purpose for the work that we just did
is a square root of 4 simplifies to 2, giving
us -4 plus or minus 2 root 14 over 4. We can
simplify this further, because both terms
up in the numerator have a factor of 2. And
there's a factor of 2 in the denominator.
So we can cancel out a 2 from each factor
in the numerator, as well as the denominator.
If we were to do this, we end up with x is
equal to -2 plus or minus root 14 all over
2. And those are our two solutions to this
particular quadratic formula: -2 plus root
14 over 2, and -2 minus root 14 over 2. Since
we did solve an equation, we're going to finish
off with the solution set notation. But we
can keep it in this compact condensed form
that the solution set is the set -2 plus or
minus root 14 over 2. It is critical that
we recognize there are two solutions in this
particular equation. If it helps you to see
this by writing this as two separate solutions,
you are free to do that, as well, in which
case the solution set would be the set -2
plus root 14 over 2, -2 minus root 14 all
over 2.
