Hello. I'm Professor Von Schmohawk and welcome to Why U.
We have seen that many real-world problems
can be modeled and solved using quadratic equations
and 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.
A quadratic equation like this is formed by setting a quadratic function equal to zero
in the same way that a linear equation can be formed
by setting a linear function equal to zero.
However, quadratic equations are typically not as easy to solve as linear equations.
For example, without techniques beyond what we have learned so far
the quadratic equation "6 x-squared + 11x + 4" equals zero
could be quite difficult to solve.
If we graph the quadratic function "6 x-squared + 11x + 4"
we can see that there are two values of x which cause the function's value to be zero.
But to find these x values, we must solve the quadratic equation.
In the previous lecture, we introduced an interesting trick for solving quadratic equations.
We showed that it is often possible to factor a quadratic function
into the product of two linear functions
and the zeros of those linear functions
will be identical to the zeros of the quadratic function
and therefore will be the solutions to the quadratic equation.
So given a specific quadratic function
how do we go about finding it's linear factors?
Fortunately, there are several cases of quadratic expressions called "special products"
whose factors can be easily identified.
These cases include quadratic expressions that are the "difference of squares"
and quadratic expressions that are "perfect squares".
In this lecture we will see how to factor quadratics that are the "difference of squares".
Quadratic expressions that are the "difference of squares"
can be written as two squared terms
which we will refer to here as terms a and b
where b-squared is subtracted from a-squared.
One example of this would be the quadratic expression "x-squared - 4"
since 4
is 2-squared.
First, we will show that any expression
that is the difference of squares "a-squared minus b-squared"
can be written as the product of two binomials (a + b) times (a - b).
We can prove that this is true in general
by multiplying the two binomials (a + b) and (a - b)
to see if we get the expression "a-squared minus b-squared".
Multiplying "a times a"
gives us a-squared
a times "negative b"
gives us "negative ab"
"b times a"
gives us "positive a b"
and "positive b" times "negative b"
is "negative b-squared."
Now, since "negative ab" plus "positive ab" is zero
these two terms cancel
leaving us with "a-squared minus b-squared"
which shows that quadratic expressions of the form "a-squared minus b-squared"
can be written as the product of the binomials (a + b) and (a - b).
Since in the example "x-squared minus 2-squared"
x
corresponds to the first squared term, a
and 2
corresponds to the second squared term, b
the factors (a + b) and (a - b) are (x + 2) and (x - 2).
Once we have factored a difference of squares quadratic expression
into two binomials in this way
the "zero product property" tells us that the zeros of those two factors
will also be the zeros of the quadratic function which is their product.
If we graph the quadratic function "x-squared - 4"
and graph each of its linear factors as functions of x
we see that the zeros of these two linear functions
are identical to the zeros of the quadratic function.
Finding the zeros of linear functions can be much easier
than finding the zeros of a quadratic function.
We simply set each linear function equal to zero
and solve for x.
So the "zeros" or "roots" of the quadratic function "x-squared - 4"
are the x values "negative 2" and "positive 2".
And since these two x values cause the function "x-squared - 4"
to have a value of zero
they are the solution set of the quadratic equation
"x-squared - 4" equals zero.
The zeros of quadratic functions that are the difference of squares
will always be two points located equal distances to the left and right of the origin
and so the solution set will always consist of two constants
of equal magnitude and opposite sign.
If we wish to check our results
we can now substitute these values into the quadratic equation
to see if they produces a true statement.
We start by substituting the first solution "negative 2" for x.
"Negative 2" squared
is 4
and "4 - 4"
is zero.
So this confirms that "negative 2" is a solution to this quadratic equation.
Likewise, substituting "positive 2" for x
gives us "positive 2" squared
which is 4
and once again, "4 - 4"
is zero.
So "positive 2" is also a solution
and our solution set checks out.
Another example of a difference of squares
is the quadratic expression "9 x-squared - 25".
This quadratic is the difference of the squares of the terms 3x
and 5.
So 3x
corresponds to the first squared term, a
and 5
corresponds to the second squared term, b.
Therefore, the factors (a + b) and (a - b)
are (3x + 5) and (3x - 5).
So to find the zeros of the quadratic function (3x) squared minus 5-squared"
or "9 x-squared - 25"
we can set the two linear functions that are its factors equal to zero
to find their zeros.
Setting these linear functions equal to zero and solving for x
we see that the quadratic function "9 x-squared - 25"
has zeros of "negative five-thirds" and "positive five-thirds".
Likewise, these two x values are the solution set of the quadratic equation
"9 x-squared - 25" equals zero.
To check our results, we can now substitute these values back into the quadratic equation
and see if they produces a true statement.
Substituting "negative five-thirds" for x
gives us "negative five-thirds" squared
which is "25 ninths".
9 times "25 ninths"
is 25
and "25 - 25"
is zero.
So "negative five-thirds" is a solution to this quadratic equation.
Likewise, substituting "positive five-thirds" for x
gives us "positive five-thirds" squared
which is once again, "25 ninths"
and so we get the same result as before
zero.
So positive and negative five-thirds are both solutions
and our solution set checks out.
So far we have seen two examples of quadratic expressions
that are the difference of squares.
And in both examples, all the constants in the expressions were perfect squares.
In other words, the constants were all squares of integers.
However, the constants in difference of squares expressions
don't have to be perfect squares or even integers.
For example, the quadratic expression "x-squared - 3"
is a difference of squares
even though 3 is not a perfect square.
It is the difference of x-squared
and "the square root of 3" squared.
So x
corresponds to the first squared term, a
and "the square root of 3"
corresponds to the second squared term, b.
Therefore, the factors (a + b) and (a - b)
are (x + "the square root of 3") and (x - "the square root of 3").
So these are the factors of x-squared
minus "the square root of 3" squared
or 3.
Another example would be the quadratic expression "3 x-squared - 7"
which is the difference of the square of "the square root of 3" times x
and the square of "the square root of 7".
So "the square root of 3" times x
corresponds to the first squared term, a
and "the square root of 7"
corresponds to the second squared term, b.
Therefore, the factors (a + b) and (a - b)
are "the square root of 3" times x, plus "the square root of 7"
and "the square root of 3" times x, minus "the square root of 7".
So these are the factors of the square of "the square root of 3" times x
or "3 x-squared"
minus the square of "the square root of 7"
or 7.
So the only requirement for a quadratic to be a "difference of squares"
is that it must be able to be written as x-squared multiplied by any positive constant
minus any positive constant.
When we recognize that a quadratic expression
matches the form of a difference of squares, "a-squared minus b-squared"
then we know that the expression can be factored as the product of two binomials
(a + b) and (a - b)
which then makes it a simple task to find the zeros of the quadratic function
and the solutions to the corresponding quadratic equation.
In the next lecture, we will see how to factor quadratics that are "perfect squares"