Algebra 71 – Solving Difference of Squares Quadratic Equations

Algebra 71 – Solving Difference of Squares Quadratic Equations


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”

4 thoughts on “Algebra 71 – Solving Difference of Squares Quadratic Equations

  • March 5, 2018 at 2:01 am
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    Really helpful video. Thank you.

    Reply
  • March 5, 2018 at 5:48 am
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    This man is the better explain mathematical problems. Please, continue.

    Reply
  • March 5, 2018 at 1:53 pm
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    Great animation. Great presentation. So much easier this way. Thanks!

    Reply
  • October 21, 2018 at 2:58 pm
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    Really helpful

    Reply

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