Hello. I'm Professor Von Schmohawk
and welcome to Why U.
We have seen that single-variable
quadratic equations
can always be put into the standard form
"a x-squared + bx + c" equals zero
where a, b, and c are constants.
This equation states that the value of the
quadratic function "a x-squared + bx + c"
must be zero
so the solutions to this equation
are those values of x
that cause the quadratic function
to produce a value of zero.
These x values are called
the "zeros" or "roots" of the function.
For instance, the equation
"6 x-squared + 11x + 4" equals zero
states that the quadratic function
"6 x-squared + 11x + 4"
must have a value of zero.
So to solve this equation
we must find the x values
where the function's value is zero.
In the previous lecture, we introduced a technique
for solving quadratic equations
that involves factoring a quadratic function
into a pair of linear functions.
This technique is based on a principle called
"the zero product property"
which tells us that if we can factor a quadratic
function into a pair of linear functions
then the zeros of those linear functions
will be identical to the zeros
of the quadratic function.
Since it is a straightforward task to find
the zero of a linear function
we can always find the solutions
to the quadratic equation
if we can factor the quadratic
into a pair of linear expressions.
We showed in the previous lecture
that there are certain forms of quadratics
called "special products"
whose factors can be easily identified.
These special products include quadratics
that are the "difference of squares"
and quadratics that are "perfect squares".
We have already seen
how to identify and factor quadratics
that are the "difference of squares".
In this lecture, we will see how to identify
and factor quadratics that are "perfect squares".
Perfect squares are quadratic expressions
which can be written either as
(a + b) quantity squared
or (a - b) quantity squared.
However, there is another form that these
perfect square quadratic expressions can take.
This form can be produced
by expanding the expressions
first writing each of them
as the product of two binomials
(a + b) times (a + b)
and (a - b) times (a - b)
and then multiplying each pair of binomials.
Multiplying the first pair of binomials
"a times a" gives us a-squared
then "a times b" is ab
"b times a" is also ab
and "b times b" is b-squared.
We can then combine the ab terms
giving us 2ab.
Likewise, we can multiply
the second pair of binomials.
Multiplying the first terms gives us a-squared
then a times "negative b" is "negative ab"
"negative b" times a
once again gives us "negative ab"
and "negative b" times "negative b" is b-squared.
We can then combine
the two "negative ab" terms
giving us "negative 2ab".
So perfect square expressions of the form
(a + b) quantity squared
and (a - b) quantity squared
may also have the form
"a-squared + 2ab + b-squared"
and "a-squared - 2ab + b-squared".
When we see a quadratic expression
that fits either of these two forms
we know that the expression can be written
as a perfect square of the form
(a + b) quantity squared
or (a - b) quantity squared.
So let's look at a couple of examples
of quadratic functions
that can be written as perfect squares
and find the solutions to the corresponding
quadratic equations.
We'll start with the quadratic function
"x-squared + 6x + 9".
Since the middle term of this quadratic
is positive
this expression cannot not match
our second perfect square form
whose middle term is negative.
We can immediately see that the first and
last terms in this expression are squares
x-squared
and 3-squared.
So in this example, x corresponds
to the first squared term, a
and 3 corresponds
to the second squared term, b.
Since a and b in this example
are x and 3
the 2ab term is 2
times a, which in this example is x
times b, which is 3
or 6x
which matches the middle term
of our quadratic expression.
Therefore, this expression is in the form
"a-squared + 2ab + b-squared".
Since in this example, the a and b terms
are x and 3
this quadratic is equivalent to the quadratic
expression (x + 3) quantity squared.
So to find the zeros of the quadratic function
"x-squared + 6x + 3-squared"
or "x-squared + 6x + 9"
we can set the two linear functions that are
its factors equal to zero, to find their zeros.
Or alternatively, we can set the function
"(x + 3) quantity squared" equal to zero
and then solve for x.
When solving an equation like this
that equates an squared expression
to a constant
the "square root property" states that the
expression that is squared
is equal to the positive or negative
square root of the constant
which in this case, is zero.
So solving for x
we see that the function has the single zero,
"negative 3"
as does the original function
"x-squared + 6x + 9".
So this quadratic function has only
a single root, and thus one x-intercept.
Therefore, the solution set of the quadratic
equation "x-squared + 6x + 9" equals zero
contains only a single solution,
"negative 3".
If we wish to check our results
we can substitute this value for x
into the quadratic equation
to see if it produces a true statement.
Substituting "negative 3" for x
gives us "negative 3" squared
in the first term
which is 9
and 6 times "negative 3"
is "negative 18".
Since "9 - 18 + 9"
is zero
"negative 3" is indeed a solution
to this quadratic equation.
It is an interesting fact, that perfect square
quadratic functions always have a single root
since their factors are always two identical
linear functions with identical x-intercepts.
Now let's factor one more example
of a perfect square quadratic
"16 x-squared - 40x + 25".
To be a perfect square
this quadratic expression must fit one of
the two possible perfect square forms.
Since the middle term of this expression
is negative
this expression cannot match the
first perfect square form.
As in our first example
we see that the first and last terms
in this expression are squares
(4x) quantity squared
and 5-squared
so in this example, 4x corresponds
to the first squared term, a
and 5 corresponds
to the second squared term, b.
"Negative 2ab" is therefore "negative 2"
times a, which in this case is 4x
times b, which is 5
or "negative 40x"
which is indeed the middle term
of our quadratic expression.
So this expression has the perfect square
form "a-squared - 2ab + b-squared".
Since the a and b terms
are 4x and 5
this quadratic is equivalent to the quadratic
expression "(4x - 5) quantity squared".
Now to find the zeros of the quadratic function
"(4x) quantity squared - 40x + 5-squared"
or "16 x-squared - 40x + 25"
we can set the function
"(4x - 5) quantity squared" equal to zero
and then solve for x.
As we saw in the previous example
when solving an equation like this, that equates
an squared expression to a constant
the "square root property" states
that the expression that is squared
is equal to the positive or negative
square root of the constant
which in this case, is zero.
So solving for x
we see that the function
has the single zero, five-fourths
as does the original function
"16 x-squared - 40x + 25".
So once again, the quadratic function
has a single "root" or "zero"
and thus a single x-intercept.
Therefore, the solution set
of the quadratic equation
"16 x-squared - 40x + 25" equals zero
contains the single solution, five-fourths.
To check our results
once again we can substitute this value
into the quadratic equation
to see if it produces a true statement.
Substituting five-fourths for x
gives us five-fourths squared
in the first term
which is "25 sixteenths."
And 16 times "25 sixteenths"
is 25.
In the second term,
40 times five-fourths
is "200 fourths"
or 50.
And since "25 - 50 + 25"
is zero
we have confirmed that five-fourths
is a solution to this quadratic equation.
So when we recognize
that a quadratic expression
matches the form
"a-squared plus or minus 2ab + b-squared"
then we know that the quadratic
can be expressed as a perfect square
(a + b) quantity squared
or (a - b) quantity squared.
The advantage of course, is that the factors
of quadratics that are perfect squares
like the factors of quadratics
that are the difference of squares
can be immediately identified.
That is why quadratic expressions like this
are called "special products".
However, when a quadratic expression
is not a special product
we must use other methods
of factoring the quadratic.
In the next lecture
we will see how to use a trial and error method
of factoring quadratic expressions
called "factoring by inspection".
