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”

Really helpful video. Thank you.

This man is the better explain mathematical problems. Please, continue.

Great animation. Great presentation. So much easier this way. Thanks!

Really helpful