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NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 1

ALGEBRA I

Lesson 1: The Distributive Property Last edited on 2/27/18

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Algebra 1

Module 4

Edited for use by the North Thurston Public Schools

ALGEBRA I

Lesson 1: Multiplying Polynomial Expressions

Example 1 The two rectangles below each represent a multiplication of two terms. Specifically, the unshaded

rectangle shows 2𝑥(𝑥) and the shaded rectangle shows 2𝑥(8).

The area of the unshaded rectangle is 2𝑥2 and the area of the shaded rectangle is 16𝑥. The total area

of the two is then 2𝑥2 + 16𝑥

If we join the two rectangles, the total area remains the same, but the product represented is a little

different.

This same idea was covered back in Module 1, Topic B. We are revisiting it here because now we want

to be able to work this process backwards --- if we know that the result is 2𝑥2 + 16𝑥, can we find what

was multiplied to create it?

But in order to be good at that, we need to be very good at both monomial and binomial multiplication.

1. Copy and complete the following monomial multiplications in your notes:

a. 4𝑥(3𝑥 + 5) b. 5𝑥(2𝑥 + 3)

2𝑥 2𝑥

By the end of this lesson, you should be able to:

multiply a monomial and a binomial

multiply two binomials

identify special products of binomials

This now represents 2𝑥(𝑥 + 8)

which means

2𝑥(𝑥 + 8) is the same as 2𝑥2 + 16𝑥

ALGEBRA I

Example 2

Using a Table as an Aid

The model used in Example 1 is known as an area model, or a generic rectangle. It is best for problems

that involve only addition, since each expression is supposed represent a measurement of length.

Without a context such as length, we can’t be sure that an algebraic expression represents a positive

quantity. But we can still mimic a generic rectangle by changing it into a table. The table serves to

remind us of the area model even though it does not represent area.

For example, the table below shows the product of (3𝑥 − 2)(𝑥 − 5). Each cell in the table represents

the multiplication of the term at the top of the column with the term at the beginning of the row:

𝑥 −5

3𝑥 3𝑥2 −15𝑥

−2 −2𝑥 +10

This table helps us to organize the multiplication of (3𝑥 − 2)(𝑥 − 5) and see the resulting terms. The

answer to the multiplication is: 3𝑥2 − 15𝑥 − 2𝑥 + 10 which simplifies to: 3𝑥2 − 17𝑥 + 10.

Without the Aid of a Table

Regardless of whether or not we use a table as an aid, multiplying of two binomials is an application of

the distributive property. Both terms of the first binomial distribute over the second. In other words:

(3𝑥 − 2)(𝑥 − 5) becomes 3𝑥(𝑥 − 5) − 2(𝑥 − 5)

Since there are 2 terms in each binomial, there are 4 multiplications that must be done. But it’s

possible to do these multiplications without separating the first binomial. The process is often

referred to by the acronym “FOIL”. This stands for First, Outside, Inside, and Last.

Here is what the process looks like. The arrows match each step of the distribution with the resulting

partial product.

(3𝑥 − 2)(𝑥 − 5)

3𝑥2

−15𝑥

−2𝑥

3𝑥2 − 15𝑥 − 2𝑥 + 10

Again, we would combine the like terms in the middle to get a final answer of:3𝑥2 − 17𝑥 + 10

inside

outside

ALGEBRA I

We leave the answer in standard order, meaning the terms are in order from highest power to lowest.

2. Copy and complete the following binomial multiplications in your notes. Leave your answer in

standard order:

a. (4𝑥 − 5)(𝑥 + 2) b. (2𝑥 + 7)(3𝑥 − 1)

Example 2 Certain types of binomial multiplication are known as “special products” because their multiplication

results always follow the same pattern. Consider these examples:

(𝑥 + 5)(𝑥 − 5) becomes

𝑥2 − 5𝑥 + 5𝑥 − 25 which becomes

𝑥2 − 25

(𝑥 + 7)(𝑥 − 7) becomes

𝑥2 − 7𝑥 + 7𝑥 − 49 which becomes

𝑥2 − 49

(3𝑥 + 5)(3𝑥 − 5) becomes

9𝑥2 − 15𝑥 + 15𝑥 − 25 which becomes

9𝑥2 − 25

(4𝑥 + 7)(4𝑥 − 7) becomes

16𝑥2 − 28𝑥 + 28𝑥 − 49 which becomes

16𝑥2 − 49

This pattern is usually referred to as “the difference of two squares” because the final answer is

always a subtraction (that’s the “difference” part of the name) of two terms that are perfect squares.

This happens whenever the two binomials are conjugate pairs, meaning that they have the same

numbers, but with an opposite sign in between them. For example:

These are conjugate pairs: These are NOT conjugate pairs:

(3𝑥 + 5)(3𝑥 − 5)

(4𝑥 + 7)(4𝑥 − 7)

(𝑥 + 3)(𝑥 − 3)

(3𝑥 + 5)(𝑥 − 5)

(4𝑥 + 7)(4𝑥 + 7)

(𝑥 + 3)(2𝑥 − 3)

When we have conjugate pairs, we can skip over the middle step of the multiplication and go directly

to a difference (subtraction) of 2 squares, like this:

(𝑥 − 9)(𝑥 + 9)

becomes

𝑥2 − 81

(𝑥 + 11)(𝑥 − 11) becomes

𝑥2 − 121

(5𝑥 + 6)(5𝑥 − 6) becomes

25𝑥2 − 36

(3𝑥 − 10)(3𝑥 + 10) becomes

9𝑥2 − 100

ALGEBRA I

3. Copy the following problems into your notes. Use the difference of two squares shortcut to

find the product of the binomials.

a. (7𝑥 + 1)(7𝑥 − 1) b. (8𝑥 − 3)(8𝑥 + 3)

Example 3 The other “special product” is created when the multiplication involves the same binomial twice,

meaning that the binomial is being squared.

Consider these examples:

(𝑥 + 5)(𝑥 + 5) becomes

𝑥2 + 5𝑥 + 5𝑥 − 25 which becomes

𝑥2 + 10𝑥 + 25

(𝑥 − 7)(𝑥 − 7) becomes

𝑥2 − 7𝑥 − 7𝑥 + 49 which becomes

𝑥2 − 14𝑥 + 49

(3𝑥 + 5)(3𝑥 + 5) becomes

9𝑥2 + 15𝑥 + 15𝑥 − 25 which becomes

9𝑥2 + 30𝑥 + 25

(4𝑥 − 7)(4𝑥 − 7) becomes

16𝑥2 − 28𝑥 − 28𝑥 − 49 which becomes

16𝑥2 − 56𝑥 + 49

You can see that the individual products are extremely similar between this and the previous example.

The only change is to the sign of one of the two middle terms. That sign change is significant because

now, instead of the two middle terms cancelling each other out, they “double up” because they are both

the same.

So the pattern here is that the binomial, squared, becomes a trinomial where the first and third terms

are perfect squares, and the middle term is twice as big as the product of the two numbers:

(3𝑥 + 5)(3𝑥 + 5) 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 9𝑥2 + 30𝑥 + 25

Often, the problem is presented as a binomial squared, rather than as two identical binomials:

(4𝑥 − 7)2 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 16𝑥2 − 56𝑥 + 49

Notice that the third term of the result is positive. That’s because (−7)2 means multiplying −7 × −7,

and two negatives multiply to create a positive result.

3𝑥, squared 5 squared 3𝑥 times 5 times 2

4𝑥, squared −7, squared 4𝑥 times −7 times 2

ALGEBRA I

CAUTION: if you use a calculator to evaluate (−7)2, be sure to put parentheses around the −7.

Otherwise, the calculator will take 7, square it, and then make it negative, which is not the same.

4. Copy the following problems into your notes. Use the binomial squared shortcut to find the

result of the multiplication.

a. (𝑥 + 6)2 b. (2𝑥 − 9)2

Problem Set

Complete the following multiplications. Use shortcuts when possible.

1. (𝑥 − 10)2 2. (5𝑥 + 8)2

3. (𝑥 − 8)(𝑥 + 8) 4. (2𝑥 − 3)(2𝑥 + 3)

5. (3𝑥 − 7)2 6. (𝑥 − 1)2

7. (2𝑥 − 9)(2𝑥 + 3) 8. (𝑥 + 5)(4𝑥 + 5)

9. (𝑥 − 10)(𝑥 + 2) 10. (𝑥 + 8)(𝑥 + 3)

11. (𝑥 + 11)(𝑥 − 5) 12. (𝑥 − 7)(𝑥 − 4)

13. (3𝑥 − 5)(2𝑥 + 7) 14. (9𝑥 − 1)(𝑥 + 3)

15. (4𝑥 − 5)(𝑥 − 2) 16. (𝑥 + 11)(2𝑥 − 7)

17. (6𝑥 + 5)(𝑥 + 2) 18. (4𝑥 + 9)(2𝑥 − 3)

NYS COMMON CORE MATHEMATICS CURRICULUM M4 Lesson 2 ALGEBRA I

Lesson 2: Multiplying and Factoring Polynomials Last edited on 6/27/17

Lesson 2: Multiplying and Factoring Polynomial Expressions

Before we begin, let’s review some vocabulary: FACTOR: a term involved in a multiplication Example: 4 is a factor of 12 𝑥𝑥 is a factor of 12𝑥𝑥 4𝑥𝑥 is a factor of 12𝑥𝑥 (𝑥𝑥 + 4) is a factor of (𝑥𝑥 + 4)(𝑥𝑥 + 3) POLYNOMIAL IN STANDARD ORDER: an algebraic expression with more than one term where the terms are in order from highest power to lowest power Example: 4𝑥𝑥2 + 12𝑥𝑥 − 8 LEADING COEFFICIENT: when a polynomial is in standard order, the leading coefficient is the coefficient of the first term (the term with the highest power) Example: the leading coefficient of 4𝑥𝑥2 + 12𝑥𝑥 − 8 is 4 TRINOMIAL: a polynomial with 3 terms, like 4𝑥𝑥2 + 12𝑥𝑥 − 8 BINOMIAL: a polynomial with 2 terms, like 4𝑥𝑥 + 12 Example 1 In the last lesson, we reviewed how to multiply binomials such as (𝑥𝑥 + 7)(𝑥𝑥 + 3), using either a table or the FOIL method. The multiplication would look like this:

or And in both methods, the final answer is: 𝑥𝑥2 + 10𝑥𝑥 + 21.

𝑥𝑥 + 7

𝑥𝑥

𝑥𝑥2 7𝑥𝑥

3𝑥𝑥 21

By the end of this lesson, you should be able to: • multiply two binomials • when given a trinomial whose leading coefficient is 1, factor into two binomials

(𝑥𝑥 + 7)(𝑥𝑥 + 3)

𝑥𝑥2⬚

+7𝑥𝑥⬚

+3𝑥𝑥⬚

The question now is: if you knew the answer was 𝑥𝑥2 + 10𝑥𝑥 + 21, could you figure out that the two binomials that created it were (𝑥𝑥 + 7)(𝑥𝑥 + 3)? Let’s see if we can find a pattern that is helpful. Consider these examples:

(𝑥𝑥 + 7)(𝑥𝑥 + 3) becomes

𝑥𝑥2 + 10𝑥𝑥 + 21

(𝑥𝑥 + 7)(𝑥𝑥 − 3) becomes

𝑥𝑥2 + 4𝑥𝑥 − 21

(𝑥𝑥 − 7)(𝑥𝑥 − 3) becomes

𝑥𝑥2 − 10𝑥𝑥 + 21

(𝑥𝑥 − 7)(𝑥𝑥 + 3) becomes

𝑥𝑥2 − 4𝑥𝑥 + 21 In each result, the first term was 𝑥𝑥2, and the size of the third term was 21. The first column, however, had the 21 as positive while the second column had it as negative. That’s because in the first column, both binomial factors had the same sign, while in the second column, one binomial had addition and the other had subtraction. The middle term showed the most variation --- it changed from 10𝑥𝑥 to 4𝑥𝑥, and sometimes it was positive while other times it was negative. This goes back to the fact that the middle term is the final result of two multiplications and one addition --- in this case, the addition of 7𝑥𝑥 and 4𝑥𝑥, but with varying signs.

1. Copy the following lists of binomial factors and trinomial results. Then match each binomial multiplication with the trinomial that would be its result.

Binomial factors Trinomial results

(𝑥𝑥 + 5)(𝑥𝑥 + 2)

(𝑥𝑥 + 5)(𝑥𝑥 − 2)

(𝑥𝑥 − 5)(𝑥𝑥 − 2)

(𝑥𝑥 − 5)(𝑥𝑥 + 2)

𝑥𝑥2 − 7𝑥𝑥 + 10

𝑥𝑥2 + 7𝑥𝑥 + 10

𝑥𝑥2 + 3𝑥𝑥 − 10

𝑥𝑥2 − 3𝑥𝑥 − 10

Example 2 Obviously, we don’t always have the option of comparing all 4 versions of how the factors might interact. Usually, we have only the trinomial and must to break it down without choices to draw from. So how do we describe this pattern in a way that we can apply it to any trinomial?

We begin by looking at the factors of the first term. In this lesson, we will work exclusively with trinomials whose leading coefficient is 1, so the first term will break apart into single letters, like this:

𝑥𝑥2 + 5𝑥𝑥 − 14

(𝑥𝑥 )(𝑥𝑥 ) We then turn our attention to the last term, the one without the 𝑥𝑥. If you recall from Example 1, the third term stayed 21 (but was sometimes positive and sometimes negative). We need to break the third term into a pair of factors:

𝑥𝑥2 + 5𝑥𝑥 − 14

(𝑥𝑥 7 )(𝑥𝑥 2) Now, keep in mind that 14 can be created by multiplying either 7 & 2 or 1 & 14. So why do we choose 7 & 2? Well, the middle term is created by adding… Can you see a way to make 5𝑥𝑥 out of 1 & 14? No. But out of 7 & 2? Sure, if the 7 is positive and the 2 is negative. So we wrap up the problem by placing those signs in the binomials:

𝑥𝑥2 + 5𝑥𝑥 − 14

(𝑥𝑥 + 7 )(𝑥𝑥 − 2) And the final answer is: (𝑥𝑥 + 5)(𝑥𝑥 − 2).

2. Copy the following problems, then use the process outlined above to finish factoring: a. 𝑥𝑥2 + 8𝑥𝑥 + 7 factors as (𝑥𝑥 + 1)(𝑥𝑥 )

b. 𝑥𝑥2 − 3𝑥𝑥 − 10 factors as (𝑥𝑥 − 5)( )

c. 𝑥𝑥2 + 2𝑥𝑥 − 15 factors as (𝑥𝑥 − 3)( )

d. 𝑥𝑥2 − 6𝑥𝑥 + 5 factors as (𝑥𝑥 − 5)( )

Example 3 Some problems are more complicated than others, depending on how many factor pairs the third term has. For example:

24 can factor using 4 & 6, 3 & 8, 2 & 12, or 1 & 24. 60 can factor using 6 & 10, 3 & 20, 2 & 30, 4 & 15, 5 & 12, or 1 & 60

How will you know which one to pick? Well, it’s a guess & check process, but there are some ways to narrow the choices. For instance, if you need to add to an odd number, you’ll need an odd and an even. So if you needed multiply to 24 but add to an odd number, you would have to be 8 & 3 or 1 & 24 because those are the

only odd/even pairs on the list of factors. In contrast, even/even pairs would add to an even number, so if you needed to multiply to 60 but add to an even number, then you would pick 6 & 10 or 2 & 30 because those are the only even/even pairs on the list of factors for 60. How will you know if you have generated all of the factor pairs possible for a certain end term? That’s a little more subtle, but involves breaking the factors down into more factors. Take, for example, 72. Most of us, when asked what multiplies to 72, would probably say 8 times 9. 72: 8 x 9 Keep in mind that 8 is the product of 2 & 4, so if we can divide something by 8, we can also divide by 2 and 4: 72: 8 x 9 2 x ______ 4 x ______ But 9 is the product of 3 & 3, so if we can divide by 9, we can also divide by 3: 72: 8 x 9 2 x ______ 4 x ______ 3 x ______ Once we have filled all the blanks (either by using a calculator or doing mental math), we check to make sure all of the factors of all of the factors are represented: 72: 8 x 9 2 x 36 4 x 18 3 x 24 Are all the factors of 18 on the list? We’ve got 2, 9, 18, and 3, but NOT 6. So we write 6 on the list: 72: 8 x 9 2 x 36 4 x 18 3 x 24 6 x ______ and the blank is 12 Now we check again: are all the factors of 12 on the list? Are all the factors of 24 on there? All the factors of 36? Well, then we have the whole list and are free to start looking for the correct pair to use. Let’s say we wanted to factor: 𝑥𝑥2 − 21𝑥𝑥 − 72. That would need: (𝑥𝑥 + 3)(𝑥𝑥 − 24). Let’s say we wanted to factor: 𝑥𝑥2 + 14𝑥𝑥 − 72. That would need: (𝑥𝑥 + 18)(𝑥𝑥 − 4) Let’s say we wanted to factor: 𝑥𝑥2 − 18𝑥𝑥 + 72. That would need: (𝑥𝑥 − 6)(𝑥𝑥 − 12).

3. Copy each trinomial into your notes, then factor into two binomials. a. 𝑥𝑥2 + 4𝑥𝑥 − 60

b. 𝑥𝑥2 − 17𝑥𝑥 + 60

c. 𝑥𝑥2 − 2𝑥𝑥 − 24

d. 𝑥𝑥2 + 11𝑥𝑥 + 24 Example 4 What if the middle term is just 𝑥𝑥? Let’s take, for example, 𝑥𝑥2 + 𝑥𝑥 − 72. What that really says is: 𝑥𝑥2 + 1𝑥𝑥 − 72. We would need a pair of factors that multiplies to 72 but adds to 1. In other words, the numbers must be back-to-back, like 4 & 5 or 6 & 7 or 8 & 9. Did you see one of those on the list from Example 3? Yep. 8 & 9. So to factor 𝑥𝑥2 + 𝑥𝑥 − 72 we would write: (𝑥𝑥 − 8)(𝑥𝑥 + 9). Does it matter what order? No. We could write: (𝑥𝑥 + 9)(𝑥𝑥 − 8) and it would still be accurate. However, if we write: (𝑥𝑥 − 9)(𝑥𝑥 + 8), that would not work because the sign on the 1𝑥𝑥 would not match the trinomial we were trying to factor. The common mistake in problems like this is to assume that because it has to add to 1, it needs to involve 1. But the reality is that the smaller the middle term becomes, the closer together the two factors need to be. So when you need to add to 1, it can sometimes be easier to list all the pairs of back-to-back numbers (2 & 3, 3 & 4, 4 & 5, etc) until you find the pair that multiplies to the 3rd term of the trinomial.

4. Copy each trinomial into your notes, then factor into two binomials: a. 𝑥𝑥2 − 𝑥𝑥 − 6

b. 𝑥𝑥2 + 𝑥𝑥 − 12

c. 𝑥𝑥2 − 𝑥𝑥 − 20

d. 𝑥𝑥2 + 𝑥𝑥 − 30 What if you have no middle term at all? Take, for example: 𝑥𝑥2 − 25. We saw in the last lesson that this is called a “difference of two squares” and it comes from multiplying conjugate pairs (binomials that use the same numbers, but opposite signs). That means 𝑥𝑥2 − 25 factors as (𝑥𝑥 + 5)(𝑥𝑥 − 5). In other words, we have to use a pair of numbers that multiplies to 25 and adds to zero.

5. Copy each trinomial, then factor it: a. 𝑥𝑥2 − 16 b. 𝑥𝑥2 − 49

Problem Set Factor these trinomials as the product of two binomials.

1. 𝑥𝑥2 − 8𝑥𝑥 − 20 2. 𝑥𝑥2 − 8𝑥𝑥 − 33

3. 𝑥𝑥2 − 8𝑥𝑥 − 9 4. 𝑥𝑥2 − 8𝑥𝑥 + 12

5. 𝑥𝑥2 + 7𝑥𝑥 + 12 6. 𝑥𝑥2 + 13𝑥𝑥 + 12

7. 𝑥𝑥2 + 4𝑥𝑥 − 12 8. 𝑥𝑥2 + 4𝑥𝑥 − 5

9. 𝑥𝑥2 + 4𝑥𝑥 − 32 10. 𝑥𝑥2 + 4𝑥𝑥 − 45

11. 𝑥𝑥2 + 4𝑥𝑥 − 60 12. 𝑥𝑥2 + 11𝑥𝑥 − 60

13. 𝑥𝑥2 − 7𝑥𝑥 − 60 14. 𝑥𝑥2 − 16𝑥𝑥 + 60

15. 𝑥𝑥2 − 23𝑥𝑥 + 60 16. 𝑥𝑥2 + 17𝑥𝑥 + 60

17. 𝑥𝑥2 + 10𝑥𝑥 + 25 18. 𝑥𝑥2 − 16𝑥𝑥 + 64

19. 𝑥𝑥2 + 4𝑥𝑥 + 4 20. 𝑥𝑥2 − 6𝑥𝑥 + 9

21. 𝑥𝑥2 + 𝑥𝑥 − 90 22. 𝑥𝑥2 + 𝑥𝑥 − 42

23. 𝑥𝑥2 − 𝑥𝑥 − 56 24. 𝑥𝑥2 − 𝑥𝑥 − 2

25. 𝑥𝑥2 − 81 26. 𝑥𝑥2 − 100

27. 𝑥𝑥2 − 36 28. 𝑥𝑥2 − 1

Lesson 3: Factoring when 𝑎 ≠ 1 Last edited on 12/30/17

Lesson 3: Factoring with a Leading Coefficient

Example 1 In Lesson 2, we saw that factoring is the reverse process of multiplication.

Consider the following example of multiplication:

(𝑥 + 3)(𝑥 + 5) 𝑥2 + 5𝑥 + 3𝑥 + 15 𝑥2 + 8𝑥 + 15.

When we compare the numbers in the factored form with the numbers in the expanded form, we see

that 15 is the product of the two numbers (3 ∙ 5), and 8 is their sum (3 + 5). This is even more obvious

when we look at the expanded form before the like terms are combined.

Now compare the expansion of this binomial product to the one above:

(2𝑥 + 3)(4𝑥 + 5) 8𝑥2 + 10𝑥 + 12𝑥 + 15 8𝑥2 + 22𝑥 + 15.

Because of the coefficients on the 𝑥’s in the binomials, the first term of the trinomial answer is no

longer simply 𝑥2. However, you can see that the last term, 15, is still the product of 3 ∙ 5.

However, we are no longer “multiplying to the end and adding to the middle”. For one thing, there is

no pair of numbers that would multiply to 15 and add to 22.

In the expression between the two arrows, before the like terms were combined, we can see the

coefficients of the linear terms (the ones with the 𝑥’s). In this case, we have 10𝑥 and 12𝑥. These

linear terms are not just the sum of the factors of 15 because each factor of 15 interacted with the

factors of the 8𝑥2.

The moral of the story is: when there are coefficients in the binomials, the middle terms are no longer

a simple sum.

1. Copy and complete the multiplications shown below in your notes.

a. (5𝑥 − 3)(2𝑥 + 1)

b. (5𝑥 − 3)(2𝑥 − 1)

c. (3𝑥 − 4)(2𝑥 − 7)

d. (3𝑥 − 4)(𝑥 − 7)

factor a trinomial whose leading coefficient is a prime number using guess & check

factor a trinomial whose leading coefficient is a prime number using a table

Example 2 Since the presence of the coefficients in the binomials makes such a difference in the middle term of

the trinomial, how do we then predict what numbers to use when we try to factor a trinomial with a

leading coefficient?

The answer depends a lot on whether you prefer guess & check, which relies fairly heavily on mental

math, or whether you prefer a process with set steps, but the steps have to be memorized.

Let’s take a look at guess & check first. If we wanted to factor this trinomial:

3𝑥2 − 13𝑥 − 10

We would start by breaking apart the first term:

3𝑥2 − 13𝑥 − 10

(3𝑥 )(𝑥 )

You can see that’s not too different than the factoring from the last lesson. We just need to include the

leading coefficient in the leading binomial. (Well, actually, it could be either binomial.)

Next, we look at the end number: 8 factors as 2 & 4 or 1 & 8. Obviously, none of those pairs adds to 10.

But even if they did, that still wouldn’t necessarily mean that we use them.

Instead, we have to consider their possible placements, and what effect that will have on the middle

term when we multiply the binomials together. Here are the options:

If it is… The outside pair is… the inside pair is… so the middle could be…

(3𝑥 2)(𝑥 5) 15𝑥 2𝑥 13𝑥 or −13𝑥

17𝑥 or −17𝑥

(3𝑥 5)(𝑥 2) 6𝑥 5𝑥 11𝑥 or −11𝑥

1𝑥 or −1𝑥

Since we want a middle term of −13𝑥 for this particular trinomial, then we choose an arrangement

that puts the 3 and 5 as the outside pair, like this:

3𝑥2 − 13𝑥 − 10

(3𝑥 2 )(𝑥 5 )

The last thing to do is decide on the signs for each binomial. We can only change signs on the 2 & 5,

not on the 3𝑥 and 𝑥. We want a middle term of −13𝑥, which will only be possible if the 3𝑥 and 5

create −15𝑥, and the 2 and 𝑥 create +2𝑥. So we end up with this:

3𝑥2 − 13𝑥 − 10

(3𝑥 + 2 )(𝑥 − 5 )

This method relies heavily on the ability to visualize what the outside/inside combination is going to

be. It’s a guess & check method, where we guess where the numbers need to go, then check the

outside/inside combination to see if it matches the middle term we need. If it does, then we’re done. If

it doesn’t, we switch the order and try again, or switch to a different pair of numbers.

There are some hints that we can take advantage of. For example, the two numbers inside the

binomials can’t share any common terms. So while we can have (3𝑥 + 2), we can’t have (3𝑥 + 6).

Also, the signs of the trinomial can help us with the signs of the binomials. If the middle term is

negative and the third term is positive, we know both binomials have subtraction, because we need to

multiply to a positive but add to a negative. If the middle term and third term are both positive, then

we know both binomials have addition.

2. Copy the trinomials into your notes and complete the factorization:

a. 2𝑥2 − 5𝑥 − 12

(2𝑥 )(𝑥 4)

b. 2𝑥2 + 9𝑥 − 18

(2𝑥 3)(𝑥 )

c. 3𝑥2 + 10𝑥 + 8

(3𝑥 4)(𝑥 )

d. 3𝑥2 − 25𝑥 + 8

(3𝑥 )(𝑥 8)

3. Take a look back at letter D. What is it about the middle term that determines the use of 8 & 1

and the placement of the 8?

Example 3 So what’s the alternative if you don’t want to guess and check?

This method goes back to the idea of the table method for multiplying binomials.

Take, for example, this problem from Lesson 1:

𝑥 −5

3𝑥 3𝑥2 −15𝑥

−2 −2𝑥 +10

If we wanted, we could work backwards from the trinomial to the table. It would start like this:

3𝑥2 − 17𝑥 + 10

3𝑥2

The next step is to figure out that the remaining two spots in the table should be.

We know the middle term is supposed to be −𝟏𝟕𝒙, but we need to split that into two separate terms.

How will we know if it’s −𝟏𝟎𝒙 and − 𝟕𝒙, or − 𝟏𝟐𝒙 and − 𝟓𝒙, or − 𝟐𝟎𝒙 and 𝟑𝒙?

Where this method is trying to go is to be able to figure out what the numbers were at the top of the

columns and the start of the rows. So there needs to be something in common between the 𝟑𝒙𝟐 and

the two empty space around it, for example. But there also needs to be something in common between

the 10 and the two empty spaces.

To figure out a pair of numbers that would have something in common with both 𝟑𝒙𝟐 and 10, what

we do is combine them --- specifically, we multiply them --- and then use that product to try and find a

pair of numbers to put into the empty spaces of the table.

So, 𝟑𝒙𝟐 times 10 is 𝟑𝟎𝒙𝟐, and we need to come up with a pair of numbers that multiplies to 𝟑𝟎 and

adds to −𝟏𝟕. The pair that makes that work is 15 & 2:

Now that we have the table filled, we look across each row to see what those terms have in common:

Next, we look down each column to see what those terms have in common:

We put the common items in the binomials: (𝟑𝒙 𝟐)(𝒙 𝟓)

(Notice that one binomial is the rows’ common terms, and the other is the columns’ common terms.)

And the last thing to do is decide on the signs: (𝟑𝒙 − 𝟐)(𝒙 − 𝟓)

We know that they both have to be subtraction because the two terms we picked to fill the table were

−𝟏𝟕𝒙 and −𝟐𝒙. That will only happen if both the 2 and the 5 have subtraction in front of them.

3𝑥2 − 17𝑥 + 10

3𝑥2 −15𝑥

−2𝑥 +10

3𝑥2 − 17𝑥 + 10

3𝑥2 −15𝑥

−2𝑥 +10

3𝑥2 − 17𝑥 + 10

3𝑥2 −15𝑥

−2𝑥 +10

𝑥 5

Problem Set

Factor the following quadratic expressions.

1. 3𝑥2 + 27𝑥 + 60 2. 2𝑥2 + 9𝑥 − 5

3. 3𝑥2 − 2𝑥 − 5 4. 2𝑥2 + 7𝑥 − 4

5. 2𝑥2 + 5𝑥 + 2 6. 5𝑥2 + 19𝑥 − 4

7. 3𝑥2 − 13𝑥 + 12 8. 5𝑥2 − 7𝑥 + 2

9. 4𝑥2 + 9𝑥 + 2 10. 4𝑥2 − 𝑥 − 3

11. 6𝑥2 − 5𝑥 + 1 12. 6𝑥2 + 𝑥 − 1

13. 6𝑥2 + 𝑥 − 2 14. 6𝑥2 + 11𝑥 − 2

15. 8𝑥2 − 2𝑥 − 3 16. 8𝑥2 − 14𝑥 + 3

Lesson 4: Special products & factors Last edited on 12/30/17

Lesson 4: Special Products and Factors

Example 1 Compare these two problems:

(𝑥 − 8)(𝑥 + 2)

𝑥2 − 8𝑥 + 2𝑥 − 16

𝑥2 − 6𝑥 − 16

(𝑥 − 4)(𝑥 + 4)

𝑥2 + 4𝑥 − 4𝑥 − 16

𝑥2 − 16

Notice that both −8 & 2 and −4 & 4 multiply to −16. But these two problems have different results

because of the way the factor pairs combine. In problem A, the −8 & 2 create like terms that are not

the same size, and combine to make −6𝑥. But in problem B, because the like terms are equal amounts

but opposite signs, they combine to 0𝑥, which drops out of the final answer.

The binomials are known as a conjugate pair and the resulting quadratic is a difference of two squares.

CONJUGATE PAIR: a pair of binomials that have the same terms, but one binomial uses addition and

the other uses subtraction.

Example: (3𝑥 − 5)(3𝑥 + 5) is a conjugate pair but (3𝑥 − 5)(𝑥 + 5) is not

(𝑥 + 2)(𝑥 − 2) is a conjugate pair but (𝑥 − 2)(𝑦 + 2) is not

DIFFERENCE OF TWO SQUARES: a quadratic binomial (two terms, and the highest power is 2) where

each term is a perfect square and the sign between them is subtraction.

Example: 9𝑥2 − 25 is a difference of two squares but 8𝑥2 − 25 is not

16𝑥2 − 1 is a difference of two squares but 16𝑥2 + 1 is not

Here are a few more examples of the same pattern:

(3𝑥 − 2)(3𝑥 + 2) becomes

9𝑥2 + 6𝑥 − 6𝑥 − 4 and then:

9𝑥2 − 4

(𝑥 − 7)(𝑥 + 7) becomes

𝑥2 + 7𝑥 − 7𝑥 − 49 and then

𝑥2 − 49

(2𝑥 + 5)(2𝑥 − 5) becomes

4𝑥2 + 10𝑥 − 10𝑥 − 25 and then

4𝑥2 − 25

In each case, the combination of like terms cancels to zero. So if we see that a problem is multiplying a

conjugate pair, we can skip the Outside and Inside pairs and go directly to the final answer.

1. Copy each multiplication in your notes. Use the shortcut to find the final answer.

a. (𝑥 + 9)(𝑥 − 9) b. (4𝑥 − 7)(4𝑥 + 7)

identify and factor a difference of two squares

identify and factor a perfect square trinomial

Example 2 The flip side of being able to go directly from a conjugate pair to a difference of two squares is that we

can also go directly from a difference of two squares back to the conjugate pair of factors.

For example:

𝑥2 − 81 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 ⇒ (𝑥 + 9)(𝑥 − 9)

Of course, with this problem, we could also ask: “what multiplies to 81 and adds to zero?”

and that would generate the exact same set of binomials.

Where this pattern is necessary is on problems where we have a leading coefficient that is a perfect

square other than 1:

25𝑥2 − 81 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 ⇒ (5𝑥 + 9)(5𝑥 − 9)

And we don’t have to go through any complicated factoring schemes. We just have to break up the 25

and the 81 into pairs of equal numbers, and then have one binomial with addition and one with

subtraction.

2. Copy the following into your notes. Then factor each into conjugate pairs.

a. 49𝑥2 − 4 b. 36𝑥2 − 1

(3𝑥 − 4)(𝑥 + 4)

becomes:

3𝑥2 + 12𝑥 − 4𝑥 − 16

and then:

3𝑥2 + 8𝑥 − 16

(3𝑥 − 4)(3𝑥 − 4)

becomes:

9𝑥2 − 12𝑥 − 12𝑥 + 16

and then:

9𝑥2 − 24𝑥 + 16

There are several significant differences between these two problems and example 1. First of all,

neither problem C nor problem D are conjugate pairs. Secondly, because of that, neither of them

resulted in a “difference of two squares” kind of answer, even though both have a subtraction and

involve at least one perfect square.

But there are also significant differences between problem C and problem D. In problem C, while the

two binomials both involve a 4, the rest of the binomials are different. On the other hand, in problem

D, it is the same binomial exactly. We’ve seen this before, in lesson 1. Having the same binomial twice

means that you are squaring that binomial: (3𝑥 − 4)(3𝑥 − 4) 𝑐𝑎𝑛 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠 (3𝑥 − 4)2.

We discussed in Lesson 1 how to use the resulting pattern to predict the results of multiplying out the

squared binomial. For example:

(4𝑥 − 7)2 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 16𝑥2 − 56𝑥 + 49

This type of result is often referred to as a perfect square trinomial.

PERFECT SQUARE TRINOMIAL: a three-term quadratic whose first and last terms are perfect squares,

and whose middle term is the product of the square roots of those terms, doubled.

Example: 16𝑥2 − 56𝑥 + 49 is a perfect square trinomial but 15𝑥2 − 56𝑥 + 49 is not

𝑥2 + 12𝑥 + 36 is a perfect square trinomial but 𝑥2 + 12𝑥 − 36 is not

4𝑥2 − 12𝑥 + 9 is a perfect square trinomial but 4𝑥2 − 6𝑥 + 9 is not

The question becomes: can we use this binomial squared pattern to predict the results of factoring a

perfect square trinomial?

Here is an example of how that would work:

Given the problem: 9𝑥2 − 24𝑥 + 16

This problem factors as:

(3𝑥 − 4)2

Let’s see another example:

Given the problem: 25𝑥2 + 20𝑥 + 4

This problem factors as:

(5𝑥 + 2)2

3. Copy each quadratic into your notes, then factor it using the perfect square trinomial pattern:

a. 49𝑥2 − 14𝑥 + 1 b. 4𝑥2 + 36𝑥 + 81

CAUTION: This works ONLY in the situations described here. If either the first or last terms are not

perfect squares, or if the middle term is not the correct amount, then this style of factoring won’t work.

4𝑥, squared −7, squared 4𝑥 times −7 times 2

3𝑥 squared −4 squared 3𝑥 times −4 times 2

5𝑥 squared 2 squared 5𝑥 times 2 times 2

Problem Set

Factor each quadratic using the patterns discussed in this lesson:

1. 25𝑥2 − 9 2. 100𝑥2 − 81

3. 36𝑥2 − 1 4. 𝑥2 − 49

5. 64𝑥2 − 25 6. 64𝑥2 − 25𝑦2

7. 100𝑥2 − 60𝑥 + 9 8. 16𝑥2 + 40𝑥 + 25

9. 4𝑥2 + 28𝑥 + 49 10. 64𝑥2 − 16𝑥 + 1

11. 𝑥2 + 10𝑥 + 25 12. 𝑥2 + 10𝑥𝑦 + 25𝑦2

Explain why each quadratic is not factorable with the patterns from this lesson:

13. 𝑥2 + 5𝑥 + 25

14. 𝑥2 + 10𝑥 − 25

15. 2𝑥2 − 20𝑥 + 25

16. 9𝑥2 − 24𝑥 + 4

17. 9𝑥2 − 24𝑥 + 15

18. 16𝑥2 + 1

19. 4𝑥2 − 3

20. 4𝑥2 − 𝑥 − 9

Lesson 5: More Factoring Last edited on 7/17/17

Lesson 5: More Factoring

𝑥𝑥2 − 9𝑥𝑥 − 10 (𝑥𝑥 − 10)(𝑥𝑥 + 1)

2𝑥𝑥2 − 9𝑥𝑥 + 10 (2𝑥𝑥 − 5)(𝑥𝑥 − 2)

These two problems have different factors because of the way the leading coefficient, 2, interacts with the rest of the factorization. One way to visualize this interaction is by using a process that is sometimes referred to as “diamond and box”. You’ve seen the box before, in the last lesson --- it’s the table that we used to break the middle term into two separate partial sums. The diamond is a way to help visualize the “outside/inside” combination of the FOIL process. If you were to use the FOIL process to distribute (2𝑥𝑥 − 5)(𝑥𝑥 − 2) you would end up with:

2𝑥𝑥2 − 4𝑥𝑥 − 5𝑥𝑥 + 10 Looking at these four terms, consider this interesting pattern:

2𝑥𝑥2 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 + 10 equals 20𝑥𝑥2 and

−4𝑥𝑥 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 5𝑥𝑥 equals 20𝑥𝑥2 The same pattern is true in example A, also: (𝑥𝑥 − 10)(𝑥𝑥 + 1) becomes 𝑥𝑥2 + 1𝑥𝑥 − 10𝑥𝑥 − 10 when distributed.

𝑥𝑥2 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 10 equals −10𝑥𝑥2 and

+1𝑥𝑥 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 10𝑥𝑥 equals −10𝑥𝑥2 Here are a few more examples of the same pattern:

(3𝑥𝑥 − 2)(2𝑥𝑥 + 7) becomes

6𝑥𝑥2 + 21𝑥𝑥 − 4𝑥𝑥 − 14 and

6𝑥𝑥2 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 14 𝑡𝑡𝑡𝑡 − 84𝑥𝑥2 +21𝑥𝑥 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 4𝑥𝑥 𝑡𝑡𝑡𝑡 − 84𝑥𝑥2

(3𝑥𝑥 − 2)(𝑥𝑥 + 7) becomes

3𝑥𝑥2 + 21𝑥𝑥 − 2𝑥𝑥 − 14 and

3𝑥𝑥2 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 14 𝑡𝑡𝑡𝑡 − 42𝑥𝑥2 +21𝑥𝑥 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 2𝑥𝑥 𝑡𝑡𝑡𝑡 − 42𝑥𝑥2

(𝑥𝑥 − 2)(𝑥𝑥 + 7) becomes

𝑥𝑥2 + 7𝑥𝑥 − 2𝑥𝑥 − 14 and 𝑥𝑥2 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 14 𝑡𝑡𝑡𝑡 − 14𝑥𝑥2

+7𝑥𝑥 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 2𝑥𝑥 𝑡𝑡𝑡𝑡 − 14𝑥𝑥2

By the end of this lesson, you should be able to: • factor a trinomial whose leading coefficient is not prime • identify and factor out a greatest common factor • factor a trinomial into a greatest common factor monomial and two binomials

As you can see, in each case, the product of the First and Last terms equals the product of the Outside and Inside terms. We can then use this pattern to predict the Outside/Inside terms, because we know they will have to be factors of the product of the First/Last terms. For example:

Given this: then the First/Last product is:

So the Outside/Inside terms need to:

and that means they must be:

6𝑥𝑥2 − 17𝑥𝑥 + 5 30𝑥𝑥2 multiply to 30𝑥𝑥2 and add to −17𝑥𝑥 −15𝑥𝑥 and −2𝑥𝑥

10𝑥𝑥2 − 𝑥𝑥 − 3 −30𝑥𝑥2 multiply to −30𝑥𝑥2 and add to −1𝑥𝑥 5𝑥𝑥 and −6𝑥𝑥

15𝑥𝑥2 + 4𝑥𝑥 − 4 −60𝑥𝑥2 multiply to −60𝑥𝑥2 and add to +4𝑥𝑥 10𝑥𝑥 and −6𝑥𝑥

So, where does a diamond play into all of this? Well, we sometimes use that structure to help organize our thoughts. Here’s how it would look for the problem: 15𝑥𝑥2 + 4𝑥𝑥 − 4 15𝑥𝑥2 + 4𝑥𝑥 − 4 Once the diamond is filled, we move on to the box: 15𝑥𝑥2 + 4𝑥𝑥 − 4 And from the box, we move on to the factors:

−60𝑥𝑥2

10𝑥𝑥 −6𝑥𝑥

+4𝑥𝑥

• Multiply first and last to fill the top corner • Write middle term in the bottom corner • Fill the two “side” corners with the #s that

multiply to top and add to bottom

−60𝑥𝑥2

10𝑥𝑥 −6𝑥𝑥

+4𝑥𝑥

15𝑥𝑥2 −6𝑥𝑥

10𝑥𝑥 −4

15𝑥𝑥2 −6𝑥𝑥

10𝑥𝑥 −4

Shared: 3𝑥𝑥

Shared: 5𝑥𝑥 −2

Final result:

15𝑥𝑥2 + 4𝑥𝑥 − 4

Factors as:

(3𝑥𝑥 + 2)(5𝑥𝑥 − 2)

1. Copy each trinomial into your notes, then use the diamond & box structure to factor each one: a. 10𝑥𝑥2 − 𝑥𝑥 − 3 b. 6𝑥𝑥2 − 17𝑥𝑥 + 3

Example 2 We started our factoring discussion by looking at problems with a leading coefficient of 1 (for example, 𝑥𝑥2 − 3𝑥𝑥 − 10). We then expanded the discussion by investing problems with a leading coefficient that is prime (for example, 2𝑥𝑥2 − 9𝑥𝑥 + 10). And this lesson covered one way to factor when the leading coefficient is not prime, or in other words, composite (for example, 6𝑥𝑥2 − 17𝑥𝑥 + 3). However, there are trinomials with a leading coefficient other than 1 where it is NOT to your advantage to use a diamond & box approach. These problems are ones where the leading coefficient is a shared factor between all of the terms of the trinomial. Compare these two problems:

2𝑥𝑥2 − 9𝑥𝑥 + 10 2𝑥𝑥2 − 8𝑥𝑥 − 10

Both of these examples have a leading coefficient that happens to be prime. But problem D’s leading coefficient is also a factor of the other two terms in the problem. This concept of a shared factor goes back to elementary school, when you first learned about the Greatest Common Factor between two numbers (for example, the GCF of 12 and 18 is 6). Here, the 2 that is the leading coefficient is also the GCF between 2, 8, and 10. Removing the GCF looks like this: 2𝑥𝑥2 − 8𝑥𝑥 − 10 𝑏𝑏𝑡𝑡𝑏𝑏𝑏𝑏𝑡𝑡𝑡𝑡𝑡𝑡 2(𝑥𝑥2 − 4𝑥𝑥 − 5) You are factoring, which means you are breaking up multiplication, and in this case, that means dividing both the 8 and the 10 by 2. Once the GCF is removed, it turns out that the remaining trinomial, 𝑥𝑥2 − 4𝑥𝑥 − 5, can factor further. There are numbers that multiply to 5 and add to 4. So the entire process looks like this:

2𝑥𝑥2 − 8𝑥𝑥 − 10 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��2(𝑥𝑥2 − 4𝑥𝑥 − 5)

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��2(𝑥𝑥 − 5)(𝑥𝑥 + 1)

2. Copy each trinomial into your notes, then factor out the GCF of each.

a. 3𝑥𝑥2 + 5𝑥𝑥 − 42 b. 4𝑥𝑥2 − 28𝑥𝑥 + 24

3. After factoring out the GCF, fully factor the remaining trinomial into two binomials. CAUTION: You could factor these problems without removing the GCF first. For example, 2𝑥𝑥2 − 8𝑥𝑥 −10 could become (2𝑥𝑥 − 10)(𝑥𝑥 + 1). However, you would then have to remove the GCF from the first parentheses in order to meet the “fully factored” requirement: 2(𝑥𝑥 − 5)(𝑥𝑥 + 1).

Example 3 Here is an example of each type of factoring discussed so far in Lessons 1 - 4:

𝑥𝑥2 + 8𝑥𝑥 − 20 𝑥𝑥2 + 𝑥𝑥 − 20 𝑥𝑥2 − 25

2𝑥𝑥2 − 3𝑥𝑥 − 5 6𝑥𝑥2 + 7𝑥𝑥 − 5 3𝑥𝑥2 − 18𝑥𝑥 + 15 Each of these problems shares two common characteristics --- they all begin with a quadratic term (power of 2) and end with a constant (a number without a variable). They all have different middle terms (consider 𝑥𝑥2 − 25 as having a middle term of 0𝑥𝑥) and different leading coefficients, and each takes a slightly different approach when it comes to breaking the trinomial back down into the binomials used to create it. However, there is one version of a quadratic that doesn’t have these two common characteristics. It doesn’t come from multiplying from two binomials, but rather from multiplying a monomial and a binomial. Compare these two problems:

(2𝑥𝑥 − 5)(𝑥𝑥 − 2) becomes

2𝑥𝑥2 − 4𝑥𝑥 − 5𝑥𝑥 + 10 which becomes 2𝑥𝑥2 − 9𝑥𝑥 + 10

2𝑥𝑥(𝑥𝑥 − 5) becomes

2𝑥𝑥2 − 10𝑥𝑥 which doesn’t have a trinomial form

Notice that problem F is not the same style as the two-term quadratic 𝑥𝑥2 − 25, which is missing a middle term. Instead, F is missing its constant --- there is no “number without an 𝑥𝑥” to end the problem. And since factoring often involves the discussion of “what multiplies to the end and adds to the middle”, not having a constant term is going to create issues. So how do we factor a quadratic like this? Well, we have to consider not “what multiplies to the end, etc”, but rather, what do the terms share --- what is their Greatest Common Factor? Take, for example, the two-term quadratic: 15𝑥𝑥2 + 10𝑥𝑥. Let’s break each term into its parts:

15𝑥𝑥2 𝑡𝑡𝑡𝑡 3 ∙ 5 ∙ 𝑥𝑥 ∙ 𝑥𝑥 10𝑥𝑥 𝑡𝑡𝑡𝑡 2 ∙ 5 ∙ 𝑥𝑥

What’s circled are the shared items, and putting those items together creates the GCF: 5𝑥𝑥 Like we did in example 2, we then factor out the GCF. What isn’t circled gets left behind as a single binomial:

15𝑥𝑥2 + 10𝑥𝑥 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��5𝑥𝑥(3𝑥𝑥 + 2)

4. Copy each quadratic into your notes, then factor out the GCF of each to leave an answer that is fully factored:

a. 8𝑥𝑥2 − 12𝑥𝑥 b. 10𝑥𝑥2 + 20𝑥𝑥 c. −6𝑥𝑥2 − 12𝑥𝑥 (hint: factor out −6𝑥𝑥) d. −𝑥𝑥2 + 3𝑥𝑥 (hint: factor out −𝑥𝑥)

Problem Set Fully factor each quadratic:

1. 5𝑥𝑥2 − 30𝑥𝑥 2. 20𝑥𝑥2 + 12𝑥𝑥

3. 25𝑥𝑥2 − 30𝑥𝑥 4. 24𝑥𝑥2 + 12𝑥𝑥

5. 𝑥𝑥2 − 10𝑥𝑥 6. 𝑥𝑥2 − 𝑥𝑥

7. 10𝑥𝑥2 − 20𝑥𝑥 − 30 8. 6𝑥𝑥2 + 18𝑥𝑥 + 12

9. 4𝑥𝑥2 + 8𝑥𝑥 − 32 10. 3𝑥𝑥2 − 3𝑥𝑥 − 18

11. 5𝑥𝑥2 − 45 12. 3𝑥𝑥2 − 60

13. 6𝑥𝑥2 + 5𝑥𝑥 − 6 14. 5𝑥𝑥2 − 6𝑥𝑥 − 8

15. 4𝑥𝑥2 + 11𝑥𝑥 + 6 16. 6𝑥𝑥2 − 11𝑥𝑥 + 5

Lesson 6: Zero Product Property Last edited on 12/30/17

Lesson 6: The Zero Product Property

Example 1 Consider the function: 𝒇(𝒙) = 𝒙𝟐 + 𝟒𝒙 − 𝟏𝟐

Let’s say you wanted to find the value of 𝒇(𝟏). That process would look like this:

𝒇(𝟏) = (𝟏)𝟐 + 𝟒(𝟏) − 𝟏𝟐

𝒇(𝟏) = 𝟏 + 𝟒 − 𝟏𝟐

𝒇(𝟏) = −𝟕

This particular trinomial factors like this: 𝒙𝟐 + 𝟒𝒙 − 𝟏𝟐 𝒃𝒆𝒄𝒐𝒎𝒆𝒔 ⇒ (𝒙 + 𝟔)(𝒙 − 𝟐)

What if we use the factored version to find the value at 1? It should come out the same as before:

𝒇(𝟏) = (𝟏 + 𝟔)(𝟏 − 𝟐)

𝒇(𝟏) = ( 𝟕 )( −𝟏 )

𝒇(𝟏) = −𝟕

And it does!

Let’s try this with a few more values:

𝒇(𝒙) = 𝒙𝟐 + 𝟒𝒙 − 𝟏𝟐 𝒇(𝒙) = (𝒙 + 𝟔)(𝒙 − 𝟐)

𝒇(𝟓) = (𝟓)𝟐 + 𝟒(𝟓) − 𝟏𝟐

𝒇(𝟓) = 𝟐𝟓 + 𝟐𝟎 − 𝟏𝟐

𝒇(𝟓) = 𝟑𝟑

𝒇(𝟓) = (𝟓 + 𝟔)(𝟓 − 𝟐)

𝒇(𝟓) = ( 𝟏𝟏 )( 𝟑 )

𝒇(𝟓) = 𝟑𝟑

𝒇(−𝟑) = (−𝟑)𝟐 + 𝟒(−𝟑) − 𝟏𝟐

𝒇(−𝟑) = 𝟗 − 𝟏𝟐 − 𝟏𝟐

𝒇(−𝟑) = −𝟏𝟓

𝒇(−𝟑) = (−𝟑 + 𝟔)(−𝟑 − 𝟐)

𝒇(−𝟑) = ( 𝟑 )( −𝟓 )

𝒇(−𝟑) = −𝟏𝟓

1. Copy and complete the following in your notes:

𝒇(𝒙) = 𝒙𝟐 + 𝟒𝒙 − 𝟏𝟐 𝒇(𝒙) = (𝒙 + 𝟔)(𝒙 − 𝟐)

𝒇(𝟎) = (𝟎)𝟐 + 𝟒(𝟎) − 𝟏𝟐

𝒇(𝟎) =

𝒇(𝟎) = (𝟎 + 𝟔)(𝟎 − 𝟐)

𝒇(𝟎) =

Of course, the value of 𝒇(𝟎) is the y-intercept, so that is a special point.

But there are two other special values that we should examine more closely: 𝒇(−𝟔) and 𝒇(𝟐).

find the 𝒙-intercepts of a quadratic function that is factorable

solve a quadratic equation involving a trinomial that is factorable

𝒇(𝒙) = 𝒙𝟐 + 𝟒𝒙 − 𝟏𝟐 𝒇(𝒙) = (𝒙 + 𝟔)(𝒙 − 𝟐)

𝒇(−𝟔) = (−𝟔)𝟐 + 𝟒(−𝟔) − 𝟏𝟐

𝒇(−𝟔) = 𝟑𝟔 − 𝟐𝟒 − 𝟏𝟐

𝒇(−𝟔) = 𝟎

𝒇(−𝟔) = (−𝟔 + 𝟔)(−𝟔 − 𝟐)

𝒇(−𝟔) = ( 𝟎 )( −𝟖 )

𝒇(−𝟔) = 𝟎

𝒇(𝟐) = (𝟐)𝟐 + 𝟒(𝟐) − 𝟏𝟐

𝒇(𝟐) = 𝟒 + 𝟖 − 𝟏𝟐

𝒇(𝟐) = 𝟎

𝒇(𝟐) = (𝟐 + 𝟔)(𝟐 − 𝟐)

𝒇(𝟐) = ( 𝟖 )( 𝟎 )

𝒇(𝟐) = 𝟎

We just found the two points, (−𝟔, 𝟎) and (𝟐, 𝟎), that are the 𝒙-intercepts of the function. On the

graph, this is where the function will pass through the 𝒙-axis. They are also the solutions to the

equation 𝒙𝟐 + 𝟒𝒙 − 𝟏𝟐 = 𝟎.

So how can we find these two points without having to guess & check? Take another look at the way

this trinomial factored: 𝒙𝟐 + 𝟒𝒙 − 𝟏𝟐 𝒃𝒆𝒄𝒂𝒎𝒆 ⇒ (𝒙 + 𝟔)(𝒙 − 𝟐)

In the factor (𝒙 + 𝟔), when we plugged in −𝟔, the factor’s value equaled zero.

In the factor (𝒙 − 𝟐), when we plugged in 𝟐, the factor’s value equaled zero.

In other words, the values of 𝒙 = −𝟔 and 𝒙 = 𝟐 were the solutions to the equations:

𝒙 + 𝟔 = 𝟎 and 𝒙 − 𝟐 = 𝟎

This means that to find the 𝒙-intercepts of a trinomial, we should factor it, and then look for the values

that would make each factor equal to zero.

Here are a few more trinomials that have been factored, and the solutions to make each factor equal

zero, and the results of substituting those values into the function:

trinomial & factors: solutions: results of substituting:

𝒙𝟐 + 𝟑𝒙 − 𝟏𝟎

(𝒙 + 𝟓)(𝒙 − 𝟐)

𝒙 = −𝟓 and

𝒙 = 𝟐

(−𝟓)𝟐 + 𝟑(−𝟓) − 𝟏𝟎⇒ 𝟐𝟓 − 𝟏𝟓 − 𝟏𝟎⇒ 𝟎

(𝟐)𝟐 + 𝟑(𝟐) − 𝟏𝟎⇒ 𝟒 + 𝟔 − 𝟏𝟎⇒ 𝟎

𝒙𝟐 − 𝟓𝒙 + 𝟔

(𝒙 − 𝟐)(𝒙 − 𝟑)

𝒙 = 𝟐 and

𝒙 = 𝟑

(𝟐)𝟐 − 𝟓(𝟐) + 𝟔⇒ 𝟒 − 𝟏𝟎 + 𝟔 ⇒ 𝟎

(𝟑)𝟐 − 𝟓(𝟑) + 𝟔⇒ 𝟗 − 𝟏𝟓 + 𝟔 ⇒ 𝟎

𝒙𝟐 − 𝟖𝒙 − 𝟗

(𝒙 − 𝟗)(𝒙 + 𝟏)

𝒙 = 𝟗 and

𝒙 = −𝟏

(𝟗)𝟐 − 𝟖(𝟗) − 𝟗⇒ 𝟖𝟏 − 𝟕𝟐 − 𝟗⇒ 𝟎

(−𝟏)𝟐 − 𝟖(−𝟏) − 𝟗 ⇒ 𝟏 + 𝟖 − 𝟗⇒ 𝟎

2. Copy each trinomial into your notes. Factor into two binomials. Then list the values that

would make each binomial equal to zero.

a. 𝒙𝟐 − 𝟓𝒙 − 𝟐𝟒 b. 𝒙𝟐 + 𝟏𝟏𝒙 + 𝟏𝟖

Example 2 When you were in elementary school, you probably learned that any number, multiplied by zero,

equals zero. Here, we are using the flip side of the same coin --- the knowledge that if a product equals

zero, then one of the two factors involved must be zero. This is called the Zero Product Property.

For a very simple example, consider the equation: 𝟓𝒏 = 𝟎

If we start with 5, multiply by an unknown number, and get an answer of zero, then the value of the

unknown variable has to be zero. In other words, the solution to the equation 𝟓𝒏 = 𝟎 is that 𝒏 = 𝟎.

If we expand the idea to cover two unknown numbers, it looks like this: 𝒎𝒏 = 𝟎

And what this would mean is that one or the other of the two variables equals zero: 𝒎 = 𝟎 𝒐𝒓 𝒏 = 𝟎

If we then express these two unknown factors as binomials, it might look like this:

(𝒙 + 𝟏)(𝒙 − 𝟑) = 𝟎

The fact that the product of the two binomials equals zero means that the value of one or the other of

the binomials equals zero. In other words,

𝒙 + 𝟏 = 𝟎 𝒐𝒓 𝒙 − 𝟑 = 𝟎

Solving these two single-step equations would then lead to the answers:

𝒙 = −𝟏 𝒐𝒓 𝒙 = 𝟑

The idea of solving for what makes each factor equal zero is the same here as it was in Example 1. It’s

just that the problem is being presented as an equation that needs to be solved, instead of a function

where we need to find the 𝒙-intercepts. However, the results of those two processes are the same.

Compare these two problems:

find the 𝒙-intercepts for:

𝒇(𝒙) = 𝒙𝟐 − 𝟒𝒙 − 𝟓

(𝒙 − 𝟓)(𝒙 + 𝟏)

the first binomial = 0 when 𝒙 = 𝟓

the second binomial = 0 when 𝒙 = −𝟏

so the 𝒙-intercepts are:

(𝟓, 𝟎) and (−𝟏, 𝟎)

solve the equation:

𝒙𝟐 − 𝟒𝒙 − 𝟓 = 𝟎

(𝒙 − 𝟓)(𝒙 + 𝟏) = 𝟎

by the zero product property, this means

𝒙 − 𝟓 = 𝟎 𝒐𝒓 𝒙 + 𝟏 = 𝟎 +𝟓 + 𝟓 −𝟏 − 𝟏

𝒙 = 𝟓 𝒐𝒓 𝒙 = −𝟏

You can see that in both cases, the values that we find for 𝒙 are the same.

Sometimes, the equation may be presented out of order, or with extra terms, like this:

𝒙𝟐 − 𝟒𝒙 = 𝟓

or 𝒙𝟐 − 𝟑𝒙 − 𝟓 = 𝒙

or 𝟐𝒙𝟐 − 𝟕𝒙 − 𝟏𝟓 = 𝒙𝟐 − 𝟑𝒙 − 𝟏𝟎

In those situations, our first job will be to put the terms on the same side, in the correct order, before

we attempt to solve it.

Remember, the Zero Product Property states that if two items multiply to a value of zero, then one of

the two items equals zero. For our purposes, this means that one side of the equation has to equal

zero, or we can’t proceed with the solution process.

So here are the steps we will take:

Put all terms on one side (keep in mind that we want the 𝒙𝟐 term to end up positive)

Combine any like terms

Factor completely

Write each factor with = 𝟎 after it

Solve for the variable

Here is an example of how this process might look:

Solve the equation:

𝟑𝒙𝟐 − 𝟒𝒙 + 𝟕 = 𝟐𝒙𝟐 + 𝟔𝒙 − 𝟐 −𝟐𝒙𝟐 − 𝟔𝒙 + 𝟐 − 𝟐𝒙𝟐 − 𝟔𝒙 + 𝟐

𝒙𝟐 − 𝟏𝟎𝒙 + 𝟗 = 𝟎

(𝒙 − 𝟗)(𝒙 − 𝟏) = 𝟎

𝒙 − 𝟗 = 𝟎 𝒐𝒓 𝒙 − 𝟏 = 𝟎

+𝟗 + 𝟗 +𝟏 + 𝟏

𝒙 = 𝟗 or 𝒙 = 𝟏

3. Copy each trinomial into your notes. Put all of the terms on the left side. Factor the trinomial.

Set each factor equal to zero, then solve.

a. 𝒙𝟐 + 𝟓𝒙 = 𝟏𝟒 b. 𝒙𝟐 + 𝟑𝒙 − 𝟒 = 𝟐𝟒

Answers to problem set on next page:

1. 𝒙 = 𝟔 𝒐𝒓 𝒙 = −𝟑 2. 𝒙 = 𝟏 𝒐𝒓 𝒙 = 𝟖 3. 𝒙 = 𝟏𝟎 𝒐𝒓 𝒙 = −𝟔 4. 𝒙 = −𝟔 𝒐𝒓 𝒙 = 𝟓

5. 𝒑 = 𝟖 𝒐𝒓 𝒑 = −𝟏 6. 𝒑 = 𝟗 𝒐𝒓 𝒑 = −𝟐 7. 𝒑 = 𝟒 𝒐𝒓 𝒑 = −𝟐 8. 𝒑 = 𝟖 𝒐𝒓 𝒑 = −𝟑

9. 𝒙 = −𝟑 𝒐𝒓 𝒙 = 𝟏 10. 𝒙 = 𝟐 𝒐𝒓 𝒙 = 𝟏 11. 𝒂 = −𝟏𝟎 𝒐𝒓 𝒂 = −𝟏 12. 𝒎 = −𝟏𝟐 𝒐𝒓 𝒎 = −𝟑

13. 𝒙 = −𝟓 𝒐𝒓 𝒙 = 𝟑 14. 𝒃 = −𝟕 𝒐𝒓 𝒃 = 𝟓 15. 𝒓 = 𝟑 𝒐𝒓 𝒓 = 𝟒 16. 𝒙 = −𝟖 𝒐𝒓 𝒙 = −𝟒

Problem Set

Factor and solve the following equations. Show all work.

1. 𝑥2 − 3𝑥 − 18 = 0 2. 𝑥2 − 9𝑥 + 8 = 0

3. 𝑥2 − 4𝑥 − 40 = 0 4. 𝑥2 + 𝑥 − 30 = 0

5. 𝑥2 + 5𝑥 − 14 = 0 6. 𝑚2 + 4𝑚 − 5 = 0

7. 𝑎2 − 10𝑎 + 16 = 0 8. 𝑝2 − 7𝑝 − 8 = 0

9. 𝑥2 − 10𝑥 + 25 = 0 10. 𝑥2 + 8𝑥 + 16 = 0

Put all the terms on the left side, then factor and solve. Show all work.

11. 𝑝2 − 7𝑝 − 8 = 10 12. 𝑝2 − 7𝑝 − 8 = −5𝑝

13. 𝑝2 − 7𝑝 − 8 = −2𝑝 + 16 14. 𝑥2 + 2𝑥 = 3

15. 𝑥2 + 2 = 3𝑥 16. 𝑎2 − 9𝑎 + 10 = 2𝑎

17. 𝑚2 + 15𝑚 + 40 = 4 18. 𝑏2 + 5𝑏 − 35 = 3𝑏

19. 6𝑟2 − 11𝑟 = 5𝑟2 − 18 20. 2𝑥2 + 11𝑥 = 𝑥2 − 𝑥 − 32

Lesson 7: Zero Product Property, day 2 Last edited on 7/17/17

Lesson 7: The Zero Product Property, Day 2

Example 1 Consider the function: 𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝟐𝟐 + 𝟒𝟒𝒙𝒙 − 𝟏𝟏𝟐𝟐 This particular trinomial factors like this: 𝒙𝒙𝟐𝟐 + 𝟒𝟒𝒙𝒙 − 𝟏𝟏𝟐𝟐

𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 �� (𝒙𝒙 + 𝟔𝟔)(𝒙𝒙 − 𝟐𝟐)

In the last lesson, we discussed how the points, (−𝟔𝟔,𝟎𝟎) and (𝟐𝟐,𝟎𝟎) are the 𝒙𝒙-intercepts of the function, and that they are also the solutions to the equation 𝒙𝒙𝟐𝟐 + 𝟒𝟒𝒙𝒙 − 𝟏𝟏𝟐𝟐 = 𝟎𝟎. So 𝒙𝒙-intercepts are the same as the 𝒙𝒙 values at which the function equals zero. We solve that equation using this process:

• Put all terms on one side (keep in mind that we want the 𝒙𝒙𝟐𝟐 term to end up positive) • Combine any like terms • Factor completely • Write each factor with = 𝟎𝟎 after it • Solve for the variable

This particular trinomial has a leading coefficient of 1, and factors into two binomials. But we’ve had several lessons with trinomials that required different types of factoring. So the question becomes: How does this solution process transfer to these other types of trinomials? Compare these two problems:

𝒙𝒙𝟐𝟐 − 𝟔𝟔𝒙𝒙 − 𝟏𝟏𝟔𝟔 = 𝟎𝟎 factors as:

(𝒙𝒙 − 𝟖𝟖)(𝒙𝒙 + 𝟐𝟐) = 𝟎𝟎 solves as:

𝒙𝒙 = 𝟖𝟖 𝒃𝒃𝒐𝒐 𝒙𝒙 = −𝟐𝟐

𝒙𝒙𝟐𝟐 − 𝟏𝟏𝟔𝟔 = 𝟎𝟎 factors as:

(𝒙𝒙 − 𝟒𝟒)(𝒙𝒙 + 𝟒𝟒) = 𝟎𝟎 solves as:

𝒙𝒙 = 𝟒𝟒 𝒃𝒃𝒐𝒐 𝒙𝒙 = −𝟒𝟒

As you can see, the only thing that changes here is the fact that problem B doesn’t have a middle term. That affects the choice of factors, but nothing else.

1. Copy each problem into your notes. Then solve each problem according to the steps in the bulleted list shown above:

a. 𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟐𝟐 = 𝟎𝟎 b. 𝒙𝒙𝟐𝟐 − 𝟏𝟏𝟔𝟔 = 𝟑𝟑𝟑𝟑

By the end of this lesson, you should be able to: • find the 𝒙𝒙-intercepts of a quadratic function that is factorable • solve a quadratic equation involving a trinomial that is factorable

Example 2 There are two types of problems where factoring out a Greatest Common Factor becomes necessary. Here is an example of each type:

𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟐𝟐𝒙𝒙 − 𝟏𝟏𝟐𝟐 = 𝟎𝟎 factors as:

𝟑𝟑�𝒙𝒙𝟐𝟐 + 𝟒𝟒𝒙𝒙 − 𝟐𝟐� = 𝟎𝟎 and then:

𝟑𝟑(𝒙𝒙 + 𝟐𝟐)(𝒙𝒙 − 𝟒𝟒) = 𝟎𝟎

𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟒𝟒𝒙𝒙 = 𝟎𝟎 factors as:

𝟐𝟐𝒙𝒙(𝒙𝒙+ 𝟕𝟕) = 𝟎𝟎

You can see that in problem C, the greatest common factor was a constant, 2, while in problem D, the constant was a monomial, 𝟐𝟐𝒙𝒙. That is going to affect the way that the two problems continue. Compare the next few steps:

𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟐𝟐𝒙𝒙 − 𝟏𝟏𝟐𝟐 = 𝟎𝟎 𝟑𝟑�𝒙𝒙𝟐𝟐 + 𝟒𝟒𝒙𝒙 − 𝟐𝟐� = 𝟎𝟎 𝟑𝟑(𝒙𝒙 + 𝟐𝟐)(𝒙𝒙 − 𝟒𝟒) = 𝟎𝟎

set each factor equal to zero: 𝟑𝟑 = 𝟎𝟎 𝒙𝒙 + 𝟐𝟐 = 𝟎𝟎 𝒙𝒙 − 𝟒𝟒 = 𝟎𝟎

𝒏𝒏𝒃𝒃 𝒃𝒃𝒃𝒃𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒃𝒃𝒏𝒏 −𝟐𝟐 − 𝟐𝟐 +𝟒𝟒 + 𝟒𝟒 𝒙𝒙 = −𝟐𝟐 𝒙𝒙 = 𝟒𝟒

𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟒𝟒𝒙𝒙 = 𝟎𝟎 𝟐𝟐𝒙𝒙(𝒙𝒙+ 𝟕𝟕) = 𝟎𝟎

set each factor equal to zero:

𝟐𝟐𝒙𝒙 = 𝟎𝟎 𝒙𝒙 + 𝟕𝟕 = 𝟎𝟎 ÷ 𝟐𝟐 ÷ 𝟐𝟐 −𝟕𝟕 − 𝟕𝟕 𝒙𝒙 = 𝟎𝟎 𝒙𝒙 = −𝟕𝟕

Notice that in problem C, where we factored out a constant GCF, no solution came from that constant. But in problem D, the GCF of a coefficient and variable generated a solution. Let’s see that again, with some different numbers:

𝟒𝟒𝒙𝒙𝟐𝟐 + 𝟑𝟑𝟐𝟐𝒙𝒙+ 𝟒𝟒𝟖𝟖 = 𝟎𝟎 factors as:

𝟒𝟒�𝒙𝒙𝟐𝟐 + 𝟖𝟖𝒙𝒙 + 𝟏𝟏𝟐𝟐� = 𝟎𝟎 and then:

𝟒𝟒(𝒙𝒙 + 𝟐𝟐)(𝒙𝒙 + 𝟔𝟔) = 𝟎𝟎 so the next step is:

𝟒𝟒 = 𝟎𝟎 𝒃𝒃𝒐𝒐 𝒙𝒙 + 𝟐𝟐 = 𝟎𝟎 𝒃𝒃𝒐𝒐 𝒙𝒙 + 𝟔𝟔 = 𝟎𝟎 which solves as:

𝒙𝒙 = −𝟐𝟐 𝒃𝒃𝒐𝒐 𝒙𝒙 = −𝟔𝟔 (since 𝟒𝟒 = 𝟎𝟎 doesn’t have a solution,

we can ignore that part)

−𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟐𝟐𝒙𝒙 = 𝟎𝟎 factors as:

−𝟑𝟑𝒙𝒙(𝒙𝒙 − 𝟒𝟒) = 𝟎𝟎 so the next step is:

−𝟑𝟑𝒙𝒙 = 𝟎𝟎 𝒃𝒃𝒐𝒐 𝒙𝒙 − 𝟒𝟒 = 𝟎𝟎 which solves as:

𝒙𝒙 = 𝟎𝟎 𝒃𝒃𝒐𝒐 𝒙𝒙 = 𝟒𝟒

So the bottom line is this: a GCF that’s only a constant will not generate any solutions, but a GCF that

includes a variable will contribute an automatic answer of 𝒙𝒙 = 𝟎𝟎, regardless of its coefficient.

2. Copy each trinomial into your notes. Put all of the terms on the left side. Factor the trinomial. Set each factor equal to zero, then solve.

a. 𝟐𝟐𝒙𝒙𝟐𝟐 = 𝟔𝟔𝒙𝒙 + 𝟖𝟖 b. 𝒙𝒙𝟐𝟐 + 𝟑𝟑𝒙𝒙 − 𝟒𝟒 = −𝟒𝟒 Example 3 In previous lessons, we have dealt with problems whose leading coefficient is not a GCF, but rather is incorporated within one of the two binomial factors. For example, in the trinomial 𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟐𝟐𝒙𝒙 − 𝟐𝟐, the leading coefficient is not a factor of each term. So we are unable to remove it as a GCF. Instead, we break up the 𝟑𝟑𝒙𝒙𝟐𝟐 between the two binomial factors, like this:

(𝟑𝟑𝒙𝒙 )(𝒙𝒙 ) Then we start to guess & check with the factor pairs of 5, or we use a diamond & box process, to fill in the rest of the binomials. Eventually, we end up with this:

(𝟑𝟑𝒙𝒙 + 𝟐𝟐)(𝒙𝒙 − 𝟏𝟏) So, how does that coefficient alter the solution process? The entire thing would look like this:

𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟐𝟐𝒙𝒙 − 𝟐𝟐 = 𝟎𝟎 (𝟑𝟑𝒙𝒙 + 𝟐𝟐)(𝒙𝒙 − 𝟏𝟏) = 𝟎𝟎

𝟑𝟑𝒙𝒙 + 𝟐𝟐 = 𝟎𝟎 𝒃𝒃𝒐𝒐 𝒙𝒙 − 𝟏𝟏 = 𝟎𝟎 −𝟐𝟐 − 𝟐𝟐 +𝟏𝟏 + 𝟏𝟏 𝟑𝟑𝒙𝒙 = −𝟐𝟐 𝒃𝒃𝒐𝒐 𝒙𝒙 = 𝟏𝟏

÷ 𝟑𝟑 ÷ 𝟑𝟑 𝒙𝒙 = − 𝟐𝟐

𝟑𝟑

The bottom line here is that when there is a coefficient inside the binomial, that binomial’s solution will be a fraction. (This will always be true, assuming that any GCF is already removed.)

3. Copy each trinomial into your notes. Put all the terms on one side. Factor and solve. a. 𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟐𝟐 = 𝟏𝟏𝟏𝟏𝒙𝒙 b. 𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟐𝟐𝒙𝒙 + 𝟏𝟏 = 𝟕𝟕𝒙𝒙 − 𝒙𝒙𝟐𝟐 + 𝟐𝟐

Example 4 Sometimes, the leading coefficient is composite, rather than prime (for example, 6 instead of 3). In that case, you might find yourself with coefficients in both binomials, meaning both answers are fractions. It’s also possible that you might remove a GCF and still have a leading coefficient other than one.

Consider these examples:

𝟔𝟔𝒙𝒙𝟐𝟐 − 𝟕𝟕𝒙𝒙 + 𝟐𝟐 = 𝟎𝟎 factors as:

(𝟑𝟑𝒙𝒙 − 𝟐𝟐)(𝟐𝟐𝒙𝒙 − 𝟏𝟏) = 𝟎𝟎 and solves as:

𝒙𝒙 = 𝟐𝟐 𝟑𝟑

𝒃𝒃𝒐𝒐 𝒙𝒙 = 𝟏𝟏 𝟐𝟐

𝟏𝟏𝟎𝟎𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟑𝟑𝒙𝒙 − 𝟑𝟑 = 𝟎𝟎 factors as:

(𝟐𝟐𝒙𝒙 + 𝟑𝟑)(𝟐𝟐𝒙𝒙 − 𝟏𝟏) = 𝟎𝟎 and solves as:

𝒙𝒙 = −𝟑𝟑𝟐𝟐

𝒃𝒃𝒐𝒐 𝒙𝒙 = 𝟏𝟏 𝟐𝟐

𝟒𝟒𝒙𝒙𝟐𝟐 − 𝟐𝟐𝒙𝒙 − 𝟔𝟔 = 𝟎𝟎 factors as:

(𝟒𝟒𝒙𝒙 + 𝟑𝟑)(𝒙𝒙 − 𝟐𝟐) = 𝟎𝟎 and solves as:

𝒙𝒙 = − 𝟑𝟑 𝟒𝟒

𝒃𝒃𝒐𝒐 𝒙𝒙 = 𝟐𝟐

𝟒𝟒𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟎𝟎𝒙𝒙 + 𝟔𝟔 = 𝟎𝟎 factors as:

𝟐𝟐�𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟐𝟐𝒙𝒙 + 𝟔𝟔� = 𝟎𝟎 𝟐𝟐(𝟐𝟐𝒙𝒙 + 𝟑𝟑)(𝒙𝒙 + 𝟏𝟏) = 𝟎𝟎

and solves as: 𝒙𝒙 = − 𝟑𝟑

𝟐𝟐 𝒃𝒃𝒐𝒐 𝒙𝒙 = − 𝟏𝟏

𝟑𝟑

Notice that the GCF of 2 in the lower left does not contribute a solution to the equation. On top of all of these possibilities, there is also the possibility that the problem may be presented with the terms out of order, or with like terms on each side of the equals sign, or some other condition that requires us to rearrange the equation before beginning the solution process. Remember, what we need in order to factor and solve is:

• All the terms on one side, with the 𝒙𝒙𝟐𝟐 term positive • All like terms combined • Powers in order from highest to lowest

For example:

𝟐𝟐𝟎𝟎𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟎𝟎𝒙𝒙 = 𝟑𝟑 + 𝒙𝒙 − 𝟏𝟏𝟎𝟎𝒙𝒙𝟐𝟐 +𝟏𝟏𝟎𝟎𝒙𝒙𝟐𝟐 − 𝒙𝒙 − 𝟑𝟑 − 𝟑𝟑 − 𝒙𝒙 + 𝟏𝟏𝟎𝟎𝒙𝒙𝟐𝟐

𝟑𝟑𝟎𝟎𝒙𝒙𝟐𝟐 + 𝟗𝟗𝒙𝒙 − 𝟑𝟑 = 𝟎𝟎 Now we factor completely: 𝟑𝟑�𝟏𝟏𝟎𝟎𝒙𝒙𝟐𝟐 + 𝟑𝟑𝒙𝒙 − 𝟏𝟏� = 𝟎𝟎

𝟑𝟑(𝟐𝟐𝒙𝒙 + 𝟏𝟏)(𝟐𝟐𝒙𝒙 − 𝟏𝟏) = 𝟎𝟎 Next, set each factor that involves a variable equal to zero, and solve:

𝟐𝟐𝒙𝒙 + 𝟏𝟏 = 𝟎𝟎 𝒃𝒃𝒐𝒐 𝟐𝟐𝒙𝒙 − 𝟏𝟏 = 𝟎𝟎 𝒙𝒙 = −𝟏𝟏

𝟐𝟐 𝒃𝒃𝒐𝒐 𝒙𝒙 = 𝟏𝟏

𝟐𝟐

4. Copy each trinomial into your notes. Solve like the example above.

a. 𝟏𝟏𝟐𝟐𝒃𝒃𝟐𝟐 − 𝟑𝟑𝟑𝟑𝒃𝒃 + 𝟏𝟏𝟐𝟐 = 𝟑𝟑 − 𝟗𝟗𝒃𝒃 b. 𝟏𝟏𝟐𝟐𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟒𝟒 = 𝟐𝟐𝒙𝒙 + 𝟔𝟔

Problem Set Solve the following equations. Show all work. 1. 𝑥𝑥2 − 64 = 0 2. 𝑥𝑥2 − 45 = 4 3. 4𝑥𝑥2 − 4𝑥𝑥 − 24 = 0 4. 2𝑥𝑥2 + 6𝑥𝑥 − 18 = 2 5. 𝑝𝑝2 − 7𝑝𝑝 = 0 6. 5𝑝𝑝2 − 13𝑝𝑝 = 2𝑝𝑝 7. 3𝑎𝑎2 − 75 = 0 8. 2𝑛𝑛2 − 32 = 0 9. 4𝑥𝑥2 + 𝑥𝑥 = 3 10. 5𝑥𝑥2 + 2 = 3𝑥𝑥 11. 3𝑎𝑎2 + 9𝑎𝑎 + 2 = 2𝑎𝑎 12. 4𝑚𝑚2 − 11𝑚𝑚 + 4 = 1 − 3𝑚𝑚 13. 12𝑥𝑥2 − 24 = 2𝑥𝑥 + 6 14. 17𝑏𝑏2 + 4𝑏𝑏 + 10 = 13 − 5𝑏𝑏 − 13𝑏𝑏2 15. 6𝑟𝑟2 − 11𝑟𝑟 = 0 16. 3𝑥𝑥2 + 2𝑥𝑥 = 0 Partial answers: 1) 8 2) 7 3) 3 4) 2 5) 0 6) 3 7) 5 8) 4 9) -1 10) 1 11) −2 12) 1

2 13) 5

3 14) 1

5 15) 0 16) −2

Lesson 8: Solving by Taking the Square Root Last edited on 7/20/17

Lesson 8: Solving by Taking the Square Root

Example 1

Because this module focuses mostly on variables that are squared, it’s best to have a thorough

understanding of what it means to square something.

On a physical level, squaring means it’s possible to create a square array out of the amount. For example:

42 = 16 means

16 dots in

4 rows and

4 columns

In comparison, a number like 15 can only be arranged in a rectangular array:

15 dots in

3 rows and or 5 rows and

5 columns 3 columns

Or, possibly, in a partially complete square array:

15 dots in

a 4 by 4 array,

missing one dot

On a mathematical level, squaring something means that you are multiplying it by itself:

42 𝑚𝑒𝑎𝑛𝑠 4 × 4

This is why 42 and (−4)2 come out to the same result. Both calculations involve exact duplicates:

42 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 ⇒ 4 × 4 and (−4)2

𝑏𝑒𝑐𝑜𝑚𝑒𝑠 ⇒ −4 × −4

On the other hand, −42 represents −1 × 4 × 4.

solve a quadratic equation by taking the square root

identify when a quadratic can be solved by taking the square root and when it can’t

simplify the square root of a perfect square

estimate the value of a square root of a non-perfect square as between two integers

So if that is what squaring means, then what does it mean to “square root” something?

The following pattern shows the mathematical meaning:

if… then…

(5)2 and (−5)2 both equal 25 the square root of 25 is either 5 or −5

Squaring and taking the square root are inverse operations --- one undoes the other. But there is an

issue with the potential sign of the square root result.

When mathematicians want the negative answer to a square root, we write this:

−√25 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 ⇒ − 5

If we want the positive version (sometimes referred to as the principal root), we write it like this:

√9 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 ⇒ 3

And for those situations where we want both, we write it like this:

±√16 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 ⇒ ± 4

This is read out loud as “plus or minus the square root of 16 becomes plus or minus 4”.

Generally speaking, if the square root symbol is already there, then the sign is already chosen. But if

we introduce the square root symbol during a solution process, then we write the plus-or-minus

symbol as well.

The square root symbol is called a radical sign and the number underneath is called the radicand.

Certain radicands are known as perfect squares because their square roots are integers. The first

twelve perfect squares are:

1 4 9 16 25 36 49 64 81 100 121 144

If you were to take the square root of any of these, you would get an integer. For example:

√100 ⇒ 10 − √4 ⇒ −2 ± √49 ⇒ ±7

But if you were to take the square root of any numbers in between the perfect squares, you would get

an irrational number --- something whose decimal equivalent is non-terminating (doesn’t stop) and

non-repeating (doesn’t have a pattern). For example:

√99 = 9.949874371066….

We can estimate the approximate decimal value of an irrational square root by comparing the two

perfect squares on either side. For example:

√28 is between 5 and 6 because 28 is between 25 and 36

− √8 is between −2 and −3 because 8 is between 4 and 9

1. Copy each square root into your notes. Find the integer value, if possible. If not, estimate the

approximate decimal value.

a. √36

b. −√144

c. √10

d. − √60

2. Write the list of the first 20 perfect squares in your notes. Use your calculator to find the

values beyond 122.

Example 2

In the previous lesson, we looked at how to solve equations that were factorable.

Consider this problem, for example: 𝑥2 − 16 = 0

We solved the equation by factoring: (𝑥 − 4)(𝑥 + 4) = 0

And then solving each factor: 𝑥 = 4 𝑜𝑟 𝑥 = −4

Now, consider the quadratic equation: 𝑥2 = 16

See how that’s basically the same equation, but with the 16 on the other side of the equals sign?

We just discussed out that the square root of 16 is either 4 or −4.

So could we solve the equation 𝑥2 = 16 by taking the square root of 16?

Well, sort of… You see, we have to take the square root of both sides, like this:

𝑥2 = 16

±√𝑥2 = ±√16

𝑥 = ±4

Remember, when the radical is already there, the sign is decided, but when we write in the radical, we

also have to write in the plus-or-minus sign. You’ll notice that we don’t bother to keep it on the 𝑥 side.

It’s already a variable, meaning that it can vary. The plus-or-minus is just part of that variability.

The critical part of this method is that each side must be treated as a single item. Square roots can’t

“distribute”. For example:

√𝑥2 − 16 = √0 does NOT become √𝑥2 − √16 = 0 and then 𝑥 − 4 = 0

Here are some examples of equations solved using the square root process, and some equations that

can’t be solved this way. Pay close attention to the steps that can be done and the steps that can’t:

YES NO

𝑥2 = 25

±√𝑥2 = ±√25

𝑥 = ±5

𝑥2 + 25 = 0

±√𝑥2 + 25 = ±√0

can’t finish --- square roots can’t be

“distributed” to multiple terms

𝑥2 − 4 = 0 +𝟒 + 𝟒

𝑥2 = 4

±√𝑥2 = ±√4

𝑥 = ±2

𝑥2 + 25 = 0 −𝟐𝟓 − 𝟐𝟓

𝑥2 = −25

±√𝑥2 = ±√−25

can’t finish --- square roots of negative

radicands can’t be done w/ real numbers

3𝑥2 = 12 ÷ 𝟑 ÷ 𝟑

𝑥2 = 4

𝑥 = ±2

3𝑥2 = 12𝑥 ÷ 𝟑 ÷ 𝟑

𝑥2 = 4𝑥

±√𝑥2 = ±√4𝑥

𝑥 = ±2√𝑥

can’t finish with this method ---

not allowed to divide by a variable

Notice in the last example in the “NO” column that having a variable on each side of the equation

makes the equation unable to be solved with this method. We could solve this equation using the

methods shown in the previous lesson, however. When the equation has both an 𝑥2 and an 𝑥, we

can’t square root to get the answer --- we have to factor.

3. Copy and solve each equation in your notes. Show all steps.

a. 𝒙𝟐 − 𝟔𝟒 = 𝟎 b. 𝟓𝒙𝟐 = 𝟒𝟓

Example 3

As you saw in example 2, we can use the properties of equality (adding/subtracting/dividing, even

multiplying if needed, to both sides) to arrange the equation before the square root is applied.

What we need for this method to work is a single squared item on both sides of the equation. But the

item doesn’t have to be just an 𝒙𝟐. For example:

(𝒙 + 𝟑)𝟐 = 𝟏𝟔 can become ± √(𝒙 + 𝟑)𝟐 = ±√𝟏𝟔

The parentheses around the 𝒙 + 𝟑 make it a single item, which means the square root can be applied.

Here’s how the solution process would start:

(𝒙 + 𝟑)𝟐 = 𝟏𝟔

±√(𝒙 + 𝟑)𝟐 = ±√𝟏𝟔

𝒙 + 𝟑 = ±𝟒 −𝟑 − 𝟑

𝒙 = −𝟑 ± 𝟒

Normally, by the time we have 𝒙 by itself, the problem is done. Here, however, we have the statement

“3 plus or minus 4”. That needs to be finished --- what is 3 plus 4? what is 3 minus 4? --- and those

results are the final answers.

So the last few steps would look like this:

𝒙 = −𝟑 ± 𝟒

𝒙 = −𝟑 + 𝟒 𝒐𝒓 𝒙 = −𝟑 − 𝟒

𝒙 = 𝟏 𝒐𝒓 𝒙 = −𝟕

4. Copy and solve in your notes:

a. (𝑥 + 5)2 = 9 b. (𝑥 − 7)2 = 25

Problem Set

Solve each equation.

1. 𝑥2 = 81

2. 𝑥2 = 144

3. 𝑥2 − 9 = 0

4. 𝑥2 − 1 = 0

5. 2𝑥2 = 98

6. 3𝑥2 = 48

7. (𝑥 + 2)2 = 36

8. (𝑥 − 4)2 = 121

9. (𝑥 + 3)2 = 100

10. (𝑥 − 5)2 = 4

11. Explain why each equation cannot be solved using the square root method from this lesson:

a. 𝑥2 + 4 = 0 b. 𝑥2 + 4𝑥 = 0

NYS COMMON CORE MATHEMATICS CURRICULUM M4Lesson 9ALGEBRA I

Lesson 9: Simplifying Non‐Perfect Square Roots Last edited on 7/19/17

Lesson9:SimplifyingSquareRoots

Example1 Inthelastlesson,wediscussedwhatitmeanstosquareaquantity,withbothaphysicalrepresentationandwithamathematicalrepresentation.Onaphysicallevel,squaringmeansit’spossibletocreateasquarearrayoutoftheamount.Forexample:

4 16means16dotsin4rowsand4columns

Onamathematicallevel,squaringsomethingmeansthatyouaremultiplyingitbyitself:

4 4 4 16Sowhathappenswhenyoutrytotakethesquarerootofsomething?Squaresandsquarerootsareinversefunctions,soone“undoes”theother.Mathematically,itwouldlooklikethis:

3 and 3 bothequal9 …whichmeans…√9equals3

√9equals 3

√9equals 3Physically,itwouldlooklikethis:

9dotssplitupinto3rowsand3columns

Thesquarerootisdescribingthematchinglength&widthofthearray‐‐‐thatbothsidesare3.Sowhathappensifit’snotaperfectsquare,like9or16?Well,welookforaperfectsquareamongitsfactors.Forexample,32isnotaperfectsquare,butitequals16 2.Thenwedealwiththeperfectsquareportionofthenumber,andleavetherestalone.

Bytheendofthislesson,youshouldbeableto: identifywhenasquarerootsimplifiestoanintegerandwhenitdoesn’t

simplifythesquarerootofanon‐perfectsquare solveaquadraticbytakingthesquarerootofanon‐perfectsquare

Here’showitlooksmathematically:

32 16 2 √32 √16 2 √16 √2Now,youmightbethinking,“Hey,Ithoughtsquarerootscouldn’tdistributeovermultipleterms,”andyouwouldbecorrect.Butremember,atermisacollectionofthingsbeingmultiplied,andthat’swhatthe16and2aredoing.Sowecantakeonesquarerootoftwonumbersmultiplied,andrewriteitasonemultiplicationoftwosquarerootsofnumbers.Oncewegettothe“onemultiplicationoftwosquareroots”,whathappensnextisdifferentforthepartthatisaperfectsquarethanforthepartthatisn’t:

√16 √2

4 √2

whichweusuallywriteas: 4√2Whenwetrytorepresentthisprocessusinganarray,itlookssomethinglikethis:

32dotsin2separate4by4arrays

Sointheanswer4√2,the4indicatesthesidelengthofthearray,andthe2isthefactthatweneedtwooftheminordertoholdatotalof32dots.Let’sseethatprocessafewmoretimes,withdifferentnumbers:

75 25 3andso

√75 √25 √3

5 √3

whichwewriteas:5√3

18 9 2andso

√18 √9 √2

3 √2

80 16 5andso

√80 √16 √5

4 √5

600 100 6andso

√600 √100 √6

10 √6

1. Copyandcompletethefollowingsimplifications:

a. 98 49 2 √98 √49 √ ______√_____

b. 48 16 3 √48 √ √3 ______√_____

Example2 So,howdoweknowwhattwofactorstouseinordertoexpressanon‐perfectsquareasaproductofaperfectsquareandsomethingelse?Mostly,byguessingandchecking.

Let’ssaythatyouwantedtosimplify:√128Thelistofthefirst12perfectsquaresis: 1 4 9 16 25 36 49 64 81 100 121 144For128,wecouldstartwiththenextsmallestperfectsquare,andworkourwaydown:

Does121gointo128?NO Does100gointo128?NO Does81?NO Does64?YES‐‐‐64 2

SonowwecancontinuewiththeprocessusedinExample1:

128 64 2 √128 √64 √2

8 √2

whichwewriteas: 8√2Whenusingthisprocess,however,wemustbecarefultostartatthetopofthelistandworkourwaydown.Ifwedon’tusethelargestperfectsquarethatdividesthenumber,thenwe’releavingmoreperfectsquares“trapped”insidethesquareroot.Forexample,128canalsobedividedby16,likethis:

√128 √16 √8

8√8Anditcanalsobedividedby4,likethis:

√128 √4 √32

2√32

But √8 can be divided by

another perfect square: 4

Moralofthestory:findthelargestperfectsquarethatdividesintothenumberyou’retryingtosimplify.Itisalsobestifyouputtheperfectsquarefirst,thentheotherfactor,becauseit’stheperfectsquarethatwillgeneratethecoefficientontheoutsideofthesquarerootsymbol.Ifwemissfindingthelargestperfectsquare,wecan“piggyback”offofwhatwedofind.Takethatlast

versionof√128,forexample:

√128 √4 √32

2√32

2√16 √2

2 4 √2

Forafinalanswerof: 8√2CAUTION:Noteveryradicand(that’sthenumberunderthesquarerootsymbol)canbesimplified.

Forexample,√10isnotsimplifiablebecausenoneofthefactorsof10(2&5)areperfectsquares.Ifnoperfectsquaredividesthenumberevenly,thenthesquarerootstaysasis,nochanges.

2. ThesimplificationsstartedbelowdoNOTusethelargestperfectsquarepossible.Findthelargestperfectsquarethatdividestheradicandandcompletethesimplification.

a. √200 √25 √8 b. √72 √4 √18

3. Eachofthesimplificationsbelowhasbeendoneincorrectly.Describeandfixthemistake.

a. √150 √50 √3 25√3

b. √120 √30 √4 15√2

4. Describewhatiswrongwiththesetwosimplifications:

a. √15 3√5 b. √36 6√6

Example3 Theprevious2examplesusedtheideaofbreakingaradicandintoaperfectsquarefactorandanon‐perfect‐squarefactor.Butthereisanotherwaytovisualizetheideaofsimplifyingasquareroot,anditgoesbacktoamethodyoustudiedinpreviousmathclasses:primefactorization.Theideabehindprimefactorizationisthatinsteadofbreakinganumberintoonlyapairoffactors,weinsteadbreakitdowntoalistoffactorsthatareallprimenumbers.Forexample,insteadofthinkingof18as9 1,wevisualizeitas3 3 2.

Howdoesthisrelatetosimplifyingaperfectsquare?Well,rememberthatsquaringsomethingmeansmultiplyingitbyitself.Sosince18 3 3 2thenwecanwriteitas18 3 2.Andifasquareroot“undoes”asquare,then:

18 3 2 √18 3 √2Thenthesquarerootcancelsthepower,likethis:

√18 3 √2

Leavinguswiththis: √18 3√2Let’sseethatprocessagain,withsomedifferentnumbers:

50 5 10whichisalso5 5 2andthatmeans

√50 5 √2

5 √2

56 8 7whichisalso2 2 2 7

andthatmeans

√56 2 √14

2 √14

ThisprocesscanalsoprovetousthatsomethingisNOTabletobesimplified:

30 6 5whichisalso2 3 5

andsincenoneofthefactorsarerepeated,thereisnowaytosimplifythesquareroot

210 21 10whichisalso3 7 2 5

andsincenoneofthefactorsarerepeated,thereisnowaytosimplifythesquareroot

CAUTION:Thereisadifferencebetweensimplifyingasquarerootandestimatingasquareroot.Tosimplifyasquarerootmeanstoremoveanyfactorsthatrepresentperfectsquares.Toestimateasquarerootmeanstogiveadecimalapproximationofthevalue.

Forexample:√45simplifiesto3√5becauseof45isamultipleof9(specifically,9 5).Buttheestimateofitsvalueis6.708…because45isbetween36(whichis6 )and49(whichis7 ).

5. Usetheprimefactorizationsshowntosimplifythesquareroots:

a. 96 2 2 19andso√96 _____√_____

b. 126 2 3 3 7andso√126 _____√_____

6. Useprimefactorizationstoshowwhythefollowingsquarerootscan’tbesimplified:

a. √66

b. √70

ProblemSetSimplifyeachsquareroot.

1. √24

2. √40

3. √54

4. √90

5. √80

6. √112

7. √98

8. √108

9. √192

10. √28

11. √8

12. √125

13. √216

14. √27

15. √300

16. √288

17. √175

18. √72

19. √44

20. √245

Describe/explainthemistakemadeineachsimplificationshownbelow:

21. √14 2√7

22. √20 5√4

23. √72 3√8

24. √25 √5

Lesson 10: More Solving by Taking the Square Root Last edited on 7/22/17

Lesson 10: More Solving by Taking the Square Root

Example 1

Lesson 8 discussed how to solve equations by taking the square root of both sides. In that lesson, we

only dealt with the square roots of perfect squares. Lesson 9 looked at how to take the square roots of

non-perfect squares. This lesson puts those two ideas together.

Here is how the process looked in Lesson 8:

𝑥2 = 25

±√𝑥2 = ±√25

𝑥 = ±5

𝑥2 − 49 = 0 +𝟒𝟗 + 𝟒𝟗

𝑥2 = 49

±√𝑥2 = ±√49

𝑥 = ±7

3𝑥2 = 12 ÷ 𝟑 ÷ 𝟑

𝑥2 = 4

𝑥 = ±2

2𝑥2 + 10 = 28 −𝟏𝟎 − 𝟏𝟎

2𝑥2 = 18 ÷ 𝟐 ÷ 𝟐

𝑥2 = 9

𝑥 = ±3

In each of these examples, we used the Properties of Equality to isolate the 𝑥2 term and then took the

square root of both sides. If we apply this solution process on equations involving non-perfect

squares, the solution process stays the same but answers look different. Compare these examples:

Perfect Square Non-Perfect Square

𝑥2 = 25

±√𝑥2 = ±√25

𝑥 = ±5

𝑥2 = 24

±√𝑥2 = ±√24

𝑥 = ±√4 × √6

𝑥 = ±2√6

solve a quadratic equation by taking the square root

identify when taking the square root will have integer answers and when it won’t

present non-integer answers as conjugate pairs of rational and irrational numbers

when needed

Perfect Square Non-Perfect Square

𝑥2 − 49 = 0 +𝟒𝟗 + 𝟒𝟗

𝑥2 = 49

±√𝑥2 = ±√49

𝑥 = ±7

𝑥2 − 48 = 0 +𝟒𝟖 + 𝟒𝟖

𝑥2 = 48

±√𝑥2 = ±√48

𝑥 = ±√16 × √3

𝒙 = ±𝟒√𝟑

3𝑥2 = 12 ÷ 𝟑 ÷ 𝟑

𝑥2 = 4

𝑥 = ±2

4𝑥2 = 12 ÷ 4 ÷ 4

𝑥2 = 3

𝒙 = ±√3

2𝑥2 + 10 = 28 −𝟏𝟎 − 𝟏𝟎

2𝑥2 = 18 ÷ 𝟐 ÷ 𝟐

𝑥2 = 9

𝒙 = ±𝟑

2𝑥2 + 10 = 46 −𝟏𝟎 − 𝟏𝟎

2𝑥2 = 36 ÷ 𝟐 ÷ 𝟐

𝑥2 = 18

𝑥 = ±√18 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 ⇒ ± √9 × √2

𝑥 = ±3√2

As you can see, the Properties of Equality are still applied, as usual. But instead of the square roots

providing integer answers, we get instead a radicand that simplifies (but stays irrational).

1. Copy and solve the following in your notes:

a. 5𝑥2 − 40 = 0 b. 2𝑥2 + 5 = 95

(You should get ±2√2 for part a and ±3√5 for part b.)

Example 2

The other type of problem discussed in Lesson 8 was this style:

(𝑥 + 3)2 = 16

±√(𝑥 + 3)2 = ±√16

𝑥 + 3 = ±4 −𝟑 − 𝟑

𝑥 = −3 ± 4

𝑥 = −3 + 4 𝑜𝑟 𝑥 = −3 − 4

𝑥 = 1 𝑜𝑟 𝑥 = −7

What makes this type of problem distinct from the ones in Example 1 is the fact that the quadratic is a

binomial squared instead of a monomial squared (in other words, the fact that it has parentheses).

So, what happens if the binomial squared is not isolated on one side of the equation? Well, our first job

will be to isolate it. And for that, we continue to use the trusty Properties of Equality.

Consider this problem, for example: 2(𝑥 + 3)2 − 40 = 0

We first get rid of the 40: 2(𝑥 + 3)2 = 40

And then get rid of the 2: (𝑥 + 3)2 = 20

At this point, we square root both sides: ±√(𝑥 + 3)2 = ±√20

The root and power cancel: 𝑥 + 3 = ±√20

And we simplify the radical: 𝑥 + 3 = ±√4 × √5

𝑥 + 3 = ±2√5

The next thing is to get rid of the 3: −𝟑 − 𝟑

But at this point, there is a problem. The −3 is rational. The mixed radical 2√5 is irrational ---

a non-repeating, non-terminating decimal. So they are not like terms, and we can’t combine them.

We could add −3√5 with 2√5, or we could add −3 with 2, but we can’t mix a regular number

with a radical. (We could estimate their total approximate decimal value, but we can’t combine them.)

So what does this mean for the final answer? Well, it doesn’t go much farther:

𝑥 = −3 ± 2√5

The “plus-or-minus” indicates that what appears to be a single answer is actually a conjugate pair of

answers, like this:

𝑥 = −3 + 2√5 or 𝑥 = −3 − 2√5

a. 4(𝑥 − 1)2 − 32 = 0 b. 3(𝑥 + 2)2 + 3 = 45

(You should get 1 ± 2√2 for part a and −2 ± √14 for part b.)

Example 3

As you saw in Lesson 8, sometimes a square root reduces to a whole number. It depends on whether

the radicand is a perfect square or not.

When solving an equation with a binomial square, like in Example 2, if the radicand is a perfect square,

then the answers will be two integers. Instead of leaving the answer as a conjugate pair of rational and

irrational numbers mixed together, we will instead continue on for one more step.

Here’s how that would look:

3(𝑥 − 5)2 + 2 = 20 −2 − 2

3(𝑥 − 5)2 = 18 ÷ 3 ÷ 3

(𝑥 − 5)2 = 9

±√(𝑥 − 5)2 = ±√9

𝑥 − 5 = ±3 +5 + 5

𝑥 = 5 ± 3

𝑥 = 5 + 3 𝑜𝑟 𝑥 = 5 − 3

𝑥 = 8 𝑜𝑟 𝑥 = 2

a. 6(𝑥 − 4)2 − 1 = 5 b. −3(𝑥 + 2)2 + 8 = −40

(One of the answers to part a is 5, and one of the answers to part b is −6.)

Problem Set

Solve each equation.

1. 𝑥2 + 1 = 81

2. 12𝑥2 = 144

3. 3𝑥2 − 8 = 10

4. 4𝑥2 − 11 = 49

5. 2(𝑥 − 6)2 + 10 = 98

6. 3(𝑥 + 8)2 − 12 = 48

7. 3(𝑥 + 2)2 − 9 = 36

8. 5(𝑥 − 7)2 + 1 = 126

9. 2(𝑥 + 3)2 + 2 = 100

10. 10(𝑥 − 5)2 + 6 = 46

partial answers: 1) 4√5 3) √6 5) 6 + 2√11 7) −2 + √15 9) −10

Lesson 11: Completing the Square Last edited on 3/5/18

Lesson 11: Completing the Square

Example 1 In Module 3, we spent some time discussing parent functions. One of the functions we studied was 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2. You learned how changes to the function altered the shape and placement of the graph. For example, given the function 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 3)2 − 5, we know that the parent graph has been shifted to the right 3 and down 5. This means that the vertex of the function is the point (3,−5). The format 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − ℎ)2 + 𝑘𝑘 is known as the vertex form of a quadratic because it shows the vertex coordinates. Consider the examples below:

function vertex

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 + 4)2 + 7 (−4, 7)

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 2)2 + 3 (2, 3)

𝑓𝑓(𝑥𝑥) = −(𝑥𝑥 + 5)2 (−5, 0)

𝑓𝑓(𝑥𝑥) = 12

(𝑥𝑥 − 1)2 − 8 (1,−8)

In this lesson, we are going to discuss how to take a function that is presented in standard form, where the terms are in order by highest to lowest power, and transform it into vertex form. This process will take advantage of the pattern that was reviewed in Lesson 4 regarding binomial squares. Here is that pattern again:

(4𝑥𝑥 − 7)2 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 16𝑥𝑥2 − 56𝑥𝑥 + 49 But we are going to focus on problems where the leading coefficient is just a 1. That means the pattern is a little simpler:

(𝑥𝑥 − 7)2 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝑥𝑥2 − 14𝑥𝑥 + 49

By the end of this lesson, you should be able to: • convert a quadratic function from standard form (with a leading coefficient of 1)

into vertex form by completing the square • solve a quadratic function by completing the square

4𝑥𝑥, squared −7, squared 4𝑥𝑥 times −7 times 2

𝑥𝑥, squared −7, squared 𝑥𝑥 times −7 times 2

And then that pattern was able to be worked backwards in order to factor a quadratic back to a binomial square. For example:

Given the problem: 𝑥𝑥2 − 8𝑥𝑥 + 16 This problem factors as:

(𝑥𝑥 − 4)2 What we’re going to be doing is forcing every quadratic we come across to fit that binomial square pattern. This is known as completing the square. Let’s say that we were given the quadratic: 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 8𝑥𝑥 + 17 That’s pretty close to the example at the top of the page: 𝑥𝑥2 − 8𝑥𝑥 + 16 The constant term is just off by 1, at 17 instead of 16. So what if we separated out that extra one? What if we rewrote the problem like this:

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 8𝑥𝑥 + 16��𝑤𝑤𝑤𝑤 𝑘𝑘𝑘𝑘𝑘𝑘𝑤𝑤 𝑡𝑡ℎ𝑎𝑎𝑡𝑡 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖

(𝑥𝑥−4)2

Then we can write the function as: 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 4)2 + 1 which is in vertex form. Let’s see that again:

binomial square: completing the square:

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 10𝑥𝑥 + 25 becomes

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 5)2

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 10𝑥𝑥 + 25��𝑤𝑤𝑤𝑤 𝑘𝑘𝑘𝑘𝑘𝑘𝑤𝑤 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖

(𝑥𝑥−5)2

so the final version is: 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 5)2 + 2

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 14𝑥𝑥 + 49 becomes

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 + 7)2

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 14𝑥𝑥 + 49��𝑤𝑤𝑤𝑤 𝑘𝑘𝑘𝑘𝑘𝑘𝑤𝑤 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖

(𝑥𝑥−5)2

so the final version is: 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 + 7)2 + 5

𝑥𝑥 squared −4 squared 𝑥𝑥 times −4 times 2

𝑥𝑥 squared 7 squared 𝑥𝑥 times 7 times 2

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 3)2

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 6𝑥𝑥 + 9��𝑤𝑤𝑤𝑤 𝑘𝑘𝑘𝑘𝑘𝑘𝑤𝑤 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖

(𝑥𝑥−3)2

so the final version is: 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 3)2 − 1

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 + 10)2

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 20𝑥𝑥 + 100��𝑤𝑤𝑤𝑤 𝑘𝑘𝑘𝑘𝑘𝑘𝑤𝑤 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖

− 10

so the final version is: 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 + 10)2 − 10

1. Copy and complete the table in your notes:

binomial square: completing the square:

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − )2

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 12𝑥𝑥 + _____��𝑤𝑤𝑤𝑤 𝑘𝑘𝑘𝑘𝑘𝑘𝑤𝑤 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖

(𝑥𝑥− )2

+ _____

so the final version is: 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − ______)2 + ______

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 + )2

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + ____ 𝑥𝑥 + _____��𝑤𝑤𝑤𝑤 𝑘𝑘𝑘𝑘𝑘𝑘𝑤𝑤 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖

(𝑥𝑥+ )2

− _____

so the final version is: 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 + ______)2 − ______

Example 2 So far, we’ve relied on being able to recognize a pattern and use it to our advantage. And that’s a pretty good strategy for a lot of problems. But we want to develop this process into a series of steps that will always work, regardless of our ability to recognize an underlying pattern.

𝑥𝑥 squared 10 squared 𝑥𝑥 times 10 times 2

𝑥𝑥 squared _____ squared 𝑥𝑥 times _____ times 2

𝑥𝑥 squared _____ squared 𝑥𝑥 times −6 times 2

Step 1 is to clear the work space a little… having the constant term hovering around, and trying to visualize how it connects to the square you’re trying to make, is distracting. So we will use the Additive Property of Equality to put the constant on the other side of the equals sign, like this:

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 12𝑥𝑥 + 8 −𝟖𝟖 − 𝟖𝟖

𝑓𝑓(𝑥𝑥) − 8 = 𝑥𝑥2 + 12𝑥𝑥 Step 2 is to recognize that the linear term (with the 𝑥𝑥) is always a multiple of 2. We need to see it without that doubling, so we divide the linear term by 2. However, we do NOT divide on both sides. The purpose of dividing is to find what goes in with the 𝑥𝑥 in the binomial square. So it looks like this:

𝑓𝑓(𝑥𝑥) − 8 = 𝑥𝑥2 + 12𝑥𝑥 ÷ 2

𝑓𝑓(𝑥𝑥) − 8 = (𝑥𝑥 + 6) Step 3 is to remember that what we’re trying to create is a binomial square. So the next thing to do is to write a power of 2 on the outside of the binomial, like this:

𝑓𝑓(𝑥𝑥) − 8 = (𝑥𝑥 + 6)2 Once we’ve written in that power, though, there’s an issue --- we’re guaranteeing that the 6 is squared. That adds 36 to the right hand side. (You can’t see the 36 --- it’s hidden as part of the binomial square.) Step 4 is to balance that extra constant by adding the squared constant to the other side:

𝑓𝑓(𝑥𝑥) − 8 = (𝑥𝑥 + 6)2 +𝟑𝟑𝟑𝟑

Now we have this: 𝑓𝑓(𝑥𝑥) + 28 = (𝑥𝑥 + 6)2 Step 5, the last step, is to put the constant back on the right, to serve as part of the vertex:

𝑓𝑓(𝑥𝑥) + 28 = (𝑥𝑥 + 6)2 −𝟐𝟐𝟖𝟖 −𝟐𝟐𝟖𝟖

For a final answer of: 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 + 6)2 − 28 Here’s another example, without the discussion in between each step:

Step 1: move the constant to the left 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 10𝑥𝑥 − 7 +𝟕𝟕 +𝟕𝟕

Step 2: divide linear term by 2; put the # in parentheses with the 𝑥𝑥 Step 3: write a power of 2 on the parentheses

𝑓𝑓(𝑥𝑥) + 7 = 𝑥𝑥2 − 10𝑥𝑥 ÷ 2

𝑓𝑓(𝑥𝑥) + 7 = (𝑥𝑥 − 5)2 Step 4: add the constant squared to the left side +𝟐𝟐𝟐𝟐

𝑓𝑓(𝑥𝑥) + 32 = (𝑥𝑥 − 5)2 Step 5: put the constant back on the right −𝟑𝟑𝟐𝟐 −𝟑𝟑𝟐𝟐

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 5)2 − 28

Note: we do NOT need to add 36 to both sides --- writing in the power already added 36 to the right hand side

And by the way, it wouldn’t matter if the problem presented the quadratic using 𝑓𝑓(𝑥𝑥) = notation, or if it used 𝑦𝑦 = notation. The process would stay the same:

Step 1: move the constant to the left 𝑦𝑦 = 𝑥𝑥2 + 6𝑥𝑥 + 15 −𝟏𝟏𝟐𝟐 −𝟏𝟏𝟐𝟐

Step 2: divide linear term by 2; put the # in parentheses with the 𝑥𝑥 Step 3: write a power of 2 on the parentheses

𝑦𝑦 − 15 = 𝑥𝑥2 + 6𝑥𝑥 ÷ 2

𝑦𝑦 − 15 = (𝑥𝑥 + 3)2 Step 4: add the constant squared to the left side +𝟗𝟗

𝑦𝑦 − 6 = (𝑥𝑥 + 3)2 Step 5: put the constant back on the right +𝟑𝟑 + 𝟑𝟑

𝑦𝑦 = (𝑥𝑥 + 3)2 + 6 2. Copy each equation into your notes. Complete the square to convert them into vertex form. Show

all steps. a. 𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝟐𝟐 − 𝟖𝟖𝒙𝒙 − 𝟐𝟐 b. 𝒚𝒚 = 𝒙𝒙𝟐𝟐 + 𝟐𝟐𝒙𝒙 + 𝟒𝟒

Example 3 The completing the square process can be used to solve quadratic equations as well as to convert quadratic functions into vertex form. The problems wil not be presented in the same way, however --- instead of the left hand side being either 𝒇𝒇(𝒙𝒙) = or 𝒚𝒚 = , the left side will have the quadratic, and the right hand side will have either zero or a constant. Here is an example:

𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟒𝟒𝒙𝒙 − 𝟖𝟖 = −𝟏𝟏𝟐𝟐 The complete the square process will be operating on the part of the equation that has the 𝒙𝒙’s, so we get the 8 to join the −𝟏𝟏𝟐𝟐, and then ignore that subtotal for a little while:

𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟒𝟒𝒙𝒙 − 𝟖𝟖 = −𝟏𝟏𝟐𝟐 +𝟖𝟖 +𝟖𝟖 𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟒𝟒𝒙𝒙 = −𝟒𝟒

÷ 𝟐𝟐 (𝒙𝒙 + 𝟕𝟕)𝟐𝟐 = −𝟒𝟒 +𝟒𝟒𝟗𝟗 (𝒙𝒙 + 𝟕𝟕)𝟐𝟐 = 𝟒𝟒𝟐𝟐

At this point, instead of bringing the 45 back to the same side as the binomial square, we solve the problem by taking the square root of both sides, like we did in Lesson 10:

�(𝒙𝒙 + 𝟕𝟕)𝟐𝟐 = ±√𝟒𝟒𝟐𝟐

And in this case, the radicand is not a perfect square, so we end up simplifying to a mixed radical: �(𝒙𝒙 + 𝟕𝟕)𝟐𝟐 = ±√𝟒𝟒𝟐𝟐

±√𝟗𝟗 × √𝟐𝟐 𝒙𝒙 + 𝟕𝟕 = ±𝟑𝟑√𝟐𝟐

−𝟕𝟕 −𝟕𝟕 𝒙𝒙 = −𝟕𝟕 ± 𝟑𝟑√𝟐𝟐

So the final answers are: 𝒙𝒙 = −𝟕𝟕 + 𝟑𝟑√𝟐𝟐 and 𝒙𝒙 = −𝟕𝟕 − 𝟑𝟑√𝟐𝟐 It is possible for the radicand to be a perfect square, and in that situation, the answers are two integers:

𝒙𝒙𝟐𝟐 − 𝟏𝟏𝟑𝟑𝒙𝒙 + 𝟐𝟐𝟐𝟐 = 𝟏𝟏𝟐𝟐𝟐𝟐 −𝟐𝟐𝟐𝟐 − 𝟐𝟐𝟐𝟐 𝒙𝒙𝟐𝟐 − 𝟏𝟏𝟑𝟑𝒙𝒙 = 𝟖𝟖𝟐𝟐

÷ 𝟐𝟐 (𝒙𝒙 − 𝟖𝟖)𝟐𝟐 = 𝟖𝟖𝟐𝟐 +𝟑𝟑𝟒𝟒 (𝒙𝒙 − 𝟖𝟖)𝟐𝟐 = 𝟏𝟏𝟒𝟒𝟒𝟒

�(𝒙𝒙 − 𝟖𝟖)𝟐𝟐 = ±√𝟏𝟏𝟒𝟒𝟒𝟒 𝒙𝒙 − 𝟖𝟖 = ±𝟏𝟏𝟐𝟐 +𝟖𝟖 + 𝟖𝟖 𝒙𝒙 = 𝟖𝟖 ± 𝟏𝟏𝟐𝟐

𝒙𝒙 = 𝟖𝟖 + 𝟏𝟏𝟐𝟐 𝒐𝒐𝒐𝒐 𝒙𝒙 = 𝟖𝟖 − 𝟏𝟏𝟐𝟐 𝒙𝒙 = 𝟐𝟐𝟐𝟐 𝒐𝒐𝒐𝒐 𝒙𝒙 = −𝟒𝟒

a. 𝑥𝑥2 − 4𝑥𝑥 + 7 = 15 b. 𝑥𝑥2 + 8𝑥𝑥 − 5 = 28 (Part a has a mixed radical/rational answer and part b has two integer answers.)

Problem Set Solve each equation. 1. 𝑥𝑥2 + 2𝑥𝑥 + 9 = 21

2. 𝑥𝑥2 − 4𝑥𝑥 − 7 = 13

3. 𝑥𝑥2 − 8𝑥𝑥 + 20 = 40

4. 𝑥𝑥2 + 10𝑥𝑥 − 4 = 35

Convert each function to vertex form: 5. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 6𝑥𝑥 − 4

6. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 12𝑥𝑥 + 40

7. 𝑦𝑦 = 𝑥𝑥2 + 14𝑥𝑥 + 36

8. 𝑦𝑦 = 𝑥𝑥2 − 16𝑥𝑥 + 50

partial answers: 1) −1 + √13 2) 2 + 2√6 3) −2 4) −13

Lesson 12: Completing the Square with 𝒂𝒂 ≠ 𝟏𝟏 Last edited on 3/5/18

Lesson 12: Completing the Square with 𝒂𝒂 ≠ 𝟏𝟏

Example 1 In the last lesson, we discussed how to use a process known as completing the square. It’s one way to convert a quadratic function into vertex form, and it’s also one way to solve a quadratic equation. You have learned, and will learn in the future, other ways to do both of those things. For example, let’s say you wanted to solve the quadratic equation: 𝑥𝑥2 − 6𝑥𝑥 + 5 = 0 We already know that we can’t just subtract the 5 and divide by 6, for example, nor can we square root to get rid of the power. Whenever there is a mixture of powers, we have to either factor or use the complete the square process. Here is the solution process for the equation, done with both methods:

factoring: completing the square:

𝑥𝑥2 − 6𝑥𝑥 + 5 = 0 think: what multiplies to 5 and adds to −6?

(𝑥𝑥 − 5)(𝑥𝑥 − 1) = 0 𝑥𝑥 − 5 = 0 𝑜𝑜𝑜𝑜 𝑥𝑥 − 1 = 0 +𝟓𝟓 + 𝟓𝟓 +𝟏𝟏 + 𝟏𝟏 𝑥𝑥 = 5 𝑜𝑜𝑜𝑜 𝑥𝑥 = 1

𝒙𝒙𝟐𝟐 − 𝟔𝟔𝒙𝒙 + 𝟓𝟓 = 𝟎𝟎 −𝟓𝟓 − 𝟓𝟓 𝒙𝒙𝟐𝟐 − 𝟔𝟔𝒙𝒙 = −𝟓𝟓

÷ 𝟐𝟐 (𝒙𝒙 − 𝟑𝟑)𝟐𝟐 = −𝟓𝟓 +𝟗𝟗 (𝒙𝒙 − 𝟑𝟑)𝟐𝟐 = 𝟒𝟒

�(𝒙𝒙 − 𝟑𝟑)𝟐𝟐 = ±√𝟒𝟒 𝒙𝒙 − 𝟑𝟑 = ±𝟐𝟐 +𝟑𝟑 + 𝟑𝟑 𝒙𝒙 = 𝟑𝟑 ± 𝟐𝟐

𝒙𝒙 = 𝟑𝟑 + 𝟐𝟐 𝒐𝒐𝒐𝒐 𝒙𝒙 = 𝟑𝟑 − 𝟐𝟐 𝒙𝒙 = 𝟓𝟓 𝒐𝒐𝒐𝒐 𝒙𝒙 = 𝟏𝟏

Most people would agree that the completing the square process looks a little longer, but keep in mind that many of the steps shown here would actually be done using mental math, and not written down. Obviously, if the quadratic is not factorable, then the factoring method won’t work. For example, 𝑥𝑥2 −6𝑥𝑥 + 4 = 0 is not factorable, but it could still be solved using the completing the square process. (The answers would not be integers, however.)

By the end of this lesson, you should be able to: • convert a quadratic function from standard form (with a leading coefficient other

than 1) into vertex form by completing the square • solve a quadratic with a leading coefficient other than 1 by completing the square

Up until now, we have only tried to complete the square with quadratics whose middle term was an even number. But what would happen if it was an odd number? Let’s say that we had the equation: 𝑥𝑥2 − 5𝑥𝑥 + 6 = 0. This factors as: (𝑥𝑥 − 2)(𝑥𝑥 − 3) = 0 so the answers are 2 and 3. Let’s try completing the square and see if we get the same thing:

𝑥𝑥2 − 5𝑥𝑥 + 6 = 0 −6 − 6

𝑥𝑥2 − 5𝑥𝑥 = −6 ÷ 2

�𝑥𝑥 − 52�2

= −6

�𝑥𝑥 − 52�2

��𝑥𝑥 − 52�2

= ±� 1 4

𝑥𝑥 − 52

= ± 12

𝑥𝑥 = 5 2

± 1 2

𝑥𝑥 = 52

𝑜𝑜𝑜𝑜 𝑥𝑥 = 52− 1

𝑥𝑥 = 62

𝑜𝑜𝑜𝑜 𝑥𝑥 = 42

𝑥𝑥 = 3 𝑜𝑜𝑜𝑜 𝑥𝑥 = 2 As we can see from this example, when the problem has an odd middle term, completing the square still works, but we will have to cope with fractions.

1. Use the completing the square process to solve the following quadratic. As you work, explain how the fractions squared, and how you took the square root of them. Also show how you converted to common denominators and how you simplified to lowest terms.

𝑥𝑥2 − 9𝑥𝑥 − 10 = 0 Because this problem is factorable, you should get integer answers --- specifically:

𝑥𝑥 = 10 𝑜𝑜𝑜𝑜 𝑥𝑥 = −1 Be aware that if the problem isn’t factorable, then completing the square will lead to conjugate pairs of mixed fractions & radicals.

When squaring a fraction, the top and the bottom are both squared:

� 5 2�2

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� (5)2

−6 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� −24

and then −24 4

+ 254 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� 1

When taking the square root of a fraction, the root applies to both top & bottom:

±� 1 4 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��± √1

Here is an unfactorable, odd-middle-term quadratic equation, solved by completing the square: 𝑥𝑥2 + 3𝑥𝑥 − 7 = 0

+𝟕𝟕 + 𝟕𝟕 𝑥𝑥2 + 3𝑥𝑥 = 7

�𝑥𝑥 + 32�2

+ 𝟗𝟗𝟒𝟒

�𝑥𝑥 + 32�2

��𝑥𝑥 + 32�2

= ±� 37 4

𝑥𝑥 + 32

= ± √372

−𝟑𝟑𝟐𝟐

− 𝟑𝟑𝟐𝟐

𝑥𝑥 = − 3 2

± √37 2

𝑥𝑥 = −32

+ √372

𝑜𝑜𝑜𝑜 𝑥𝑥 = −32− √37

This particular radicand does not simplify, but even if it could, we wouldn’t be able to combine it with the −3. We could write them as single fractions, like this:

𝑥𝑥 = −3 + √372

𝑜𝑜𝑜𝑜 𝑥𝑥 = −3 − √372

but that doesn’t really make it any nicer, so it’s not strictly necessary.

1. Use the complete the square process to solve these non-factorable quadratics: a. 𝑥𝑥2 − 7𝑥𝑥 + 4 = 0 b. 𝑥𝑥2 + 𝑥𝑥 − 3 = 0

Example 2 Okay, we’ve figured out how to cope when the middle number isn’t the perfect thing to divide by 2. Now it’s time to figure out how to cope when the first number isn’t the perfect thing, either. So far, we’ve focused on quadratics whose leading coefficient is 1, because it’s easier to complete the square with them. That makes learning the process simpler. But now that we’ve seen the process a few times, let’s see what happens when the leading coefficient is something other than 1. The first step of completing the square is to move the constant. It’s a way to clear the decks a little bit and make room to maneuver. When we have a leading coefficient that isn’t 1, then part of that “clearing the decks” process is to get rid of that coefficient.

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±�37 4 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� ± √37

�32�2

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� (3)2

7 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� 28

and then 28 4

+ 94 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� 37

Here’s an example: Step 1: move the constant

AND 3𝑥𝑥2 + 30𝑥𝑥 + 15 = 0

−𝟏𝟏𝟓𝟓 −𝟏𝟏𝟓𝟓 divide away the coefficient

Step 2: divide linear term by 2; put the # in parentheses with the 𝑥𝑥

Step 3: write a power of 2 on the parentheses

3𝑥𝑥2 + 30𝑥𝑥 = −15 ÷ 𝟑𝟑 ÷ 𝟑𝟑 ÷ 𝟑𝟑

𝑥𝑥2 + 10𝑥𝑥 = −5 ÷ 𝟐𝟐

(𝑥𝑥 + 5)2 = −5 Step 4: add the constant squared to the other side +𝟐𝟐𝟓𝟓

(𝑥𝑥 + 5)2 = 20 Step 5: take the square root; simplify as needed Step 6: finish solving for 𝑥𝑥

�(𝑥𝑥 + 5)2 = ±√20 𝑥𝑥 + 5 = ±√4 × √5

𝑥𝑥 + 5 = ±2√5 −𝟓𝟓 −𝟓𝟓

𝑥𝑥 = −5 ± 2√5

2. Copy each equation into your notes. Complete the square to solve. Show all steps. a. 2𝑥𝑥2 − 12𝑥𝑥 − 8 = 0 b. 5𝑥𝑥2 + 10𝑥𝑥 + 40 = 0

Example 3 Examples 1 & 2 used completing the square in order to solve quadratic equations. But the same process can be used to convert quadratic functions into vertex form. Here is an example with an odd middle term:

𝒇𝒇(𝒙𝒙) = 𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟏𝟏𝒙𝒙+ 𝟖𝟖 −𝟖𝟖 −𝟖𝟖 𝒇𝒇(𝒙𝒙) − 𝟖𝟖 = 𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟏𝟏𝒙𝒙

÷ 𝟐𝟐

𝒇𝒇(𝒙𝒙) − 𝟖𝟖 = �𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟐𝟐�𝟐𝟐

+ 𝟏𝟏𝟐𝟐𝟏𝟏𝟒𝟒

𝒇𝒇(𝒙𝒙) − 𝟑𝟑𝟐𝟐𝟒𝟒

+ 𝟏𝟏𝟐𝟐𝟏𝟏𝟒𝟒

= �𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟐𝟐�𝟐𝟐

𝒇𝒇(𝒙𝒙) + 𝟖𝟖𝟗𝟗𝟒𝟒

= �𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟐𝟐�𝟐𝟐

−𝟖𝟖𝟗𝟗𝟒𝟒

𝒇𝒇(𝒙𝒙) = �𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟐𝟐�𝟐𝟐− 𝟖𝟖𝟗𝟗

𝟒𝟒

The quadratic is now in vertex form, and shows the vertex to be the point �− 𝟏𝟏𝟏𝟏𝟐𝟐

,−𝟖𝟖𝟗𝟗𝟒𝟒�.

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Here is an example with a leading coefficient other than 1: 𝒚𝒚 = 𝟒𝟒𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟒𝟒𝒙𝒙 − 𝟐𝟐𝟎𝟎

+𝟐𝟐𝟎𝟎 +𝟐𝟐𝟎𝟎 𝒚𝒚 + 𝟐𝟐𝟎𝟎 = 𝟒𝟒𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟒𝟒𝒙𝒙

÷ 𝟒𝟒 ÷ 𝟒𝟒 ÷ 𝟒𝟒 ÷ 𝟒𝟒

𝒚𝒚 𝟒𝟒

+ 𝟓𝟓 = 𝒙𝒙𝟐𝟐 − 𝟔𝟔𝒙𝒙 ÷ 𝟐𝟐

𝒚𝒚 𝟒𝟒

+ 𝟓𝟓 = (𝒙𝒙 − 𝟑𝟑)𝟐𝟐 +𝟗𝟗

𝒚𝒚 𝟒𝟒

+ 𝟏𝟏𝟒𝟒 = (𝒙𝒙 − 𝟑𝟑)𝟐𝟐 −𝟏𝟏𝟒𝟒 −𝟏𝟏𝟒𝟒

𝒚𝒚 𝟒𝟒

= (𝒙𝒙 − 𝟑𝟑)𝟐𝟐 − 𝟏𝟏𝟒𝟒 At this point, we multiply through by the 4 to get the 𝒚𝒚 by itself:

𝟒𝟒�𝒚𝒚𝟒𝟒� = 𝟒𝟒(𝒙𝒙 − 𝟑𝟑)𝟐𝟐 − 𝟒𝟒(𝟏𝟏𝟒𝟒)

𝒃𝒃𝒃𝒃𝒃𝒃𝒐𝒐𝒃𝒃𝒃𝒃𝒃𝒃 �� 𝒚𝒚 = 𝟒𝟒(𝒙𝒙 − 𝟑𝟑)𝟐𝟐 − 𝟓𝟓𝟔𝟔

Now this quadratic is in vertex form, showing a vertex of (𝟑𝟑,−𝟓𝟓𝟔𝟔) and having a stretch factor of 4.

3. Copy and solve each equation in your notes. Show all steps. a. 𝑦𝑦 = 2𝑥𝑥2 − 4𝑥𝑥 + 7 b. 𝑦𝑦 = 𝑥𝑥2 + 7𝑥𝑥 − 5

It is possible for a quadratic to have both a leading coefficient other than one and an odd middle term:

𝒚𝒚 = 𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟐𝟐𝟏𝟏𝒙𝒙 + 𝟏𝟏𝟖𝟖 −𝟏𝟏𝟖𝟖 −𝟏𝟏𝟖𝟖

𝒚𝒚 − 𝟏𝟏𝟖𝟖 = 𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟐𝟐𝟏𝟏𝒙𝒙 ÷ 𝟑𝟑 ÷ 𝟑𝟑 ÷ 𝟑𝟑 ÷ 𝟑𝟑

𝒚𝒚 𝟑𝟑− 𝟔𝟔 = 𝒙𝒙𝟐𝟐 + 𝟕𝟕𝒙𝒙

÷ 𝟐𝟐 𝒚𝒚 𝟑𝟑− 𝟐𝟐𝟒𝟒

𝟒𝟒= �𝑥𝑥 − 𝟕𝟕

𝟐𝟐�𝟐𝟐

+ 𝟒𝟒𝟗𝟗𝟒𝟒

𝒚𝒚 𝟑𝟑

+ 𝟐𝟐𝟓𝟓𝟒𝟒

= �𝒙𝒙 − 𝟕𝟕𝟐𝟐�𝟐𝟐

−𝟐𝟐𝟓𝟓𝟒𝟒

𝒚𝒚 𝟑𝟑

= �𝑥𝑥 − 𝟕𝟕𝟐𝟐�𝟐𝟐− 𝟐𝟐𝟓𝟓

𝟒𝟒

𝒚𝒚 = 𝟑𝟑�𝒙𝒙 − 𝟕𝟕𝟐𝟐�𝟐𝟐− 𝟕𝟕𝟓𝟓

𝟒𝟒

Notice how dividing affects both items on

the left side

Notice how dividing affects both items on

the left side

The 6 is converted

into 244

in order to add

all terms are multiplied by 3

Problem Set Solve each equation.

1. 𝑥𝑥2 + 5𝑥𝑥 − 9 = 0

2. 𝑥𝑥2 − 9𝑥𝑥 + 7 = 0

3. 5𝑥𝑥2 − 10𝑥𝑥 + 20 = 0

4. 6𝑥𝑥2 + 24𝑥𝑥 − 42 = 0

Convert each function to vertex form:

5. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 3𝑥𝑥 + 4

6. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 9𝑥𝑥 − 1

7. 𝑦𝑦 = 2𝑥𝑥2 + 12𝑥𝑥 + 36

8. 𝑦𝑦 = 3𝑥𝑥2 − 12𝑥𝑥 − 9

Lesson 13: Deriving the Quadratic Formula Last edited on 3/5/18

Lesson 13: Deriving the Quadratic Formula

Example 1 In the last two lessons, we discussed how to complete the square, both to convert a quadratic function into vertex form, and also to solve a quadratic equation. Completing the square can also be used to generate a formula known as the Quadratic Formula. This formula enables us to solve any quadratic equation by substituting the values from the equation into locations in the formula, without needing to complete the square. To generate the formula, we need to apply the completing the square process to a generic equation --- one in which the usual numeric coefficients have been replaced with letters:

𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 We already know how to complete the square on an equation with numeric coefficients. We will use that to help us understand the generic equation by putting the two kinds side-by-side:

with numeric coefficients: with generic coefficients:

first, we “clear the decks” because we want the first term to be just 𝒙𝒙𝟐𝟐 :

3𝑥𝑥2 + 11𝑥𝑥 + 5 = 0 −𝟓𝟓 −𝟓𝟓

3𝑥𝑥2 + 11𝑥𝑥 = −5 ÷ 𝟑𝟑 ÷ 𝟑𝟑 ÷ 𝟑𝟑

𝑥𝑥2 + 113𝑥𝑥 = −5

𝒂𝒂𝒙𝒙𝟐𝟐 + 𝒃𝒃𝒙𝒙 + 𝒄𝒄 = 𝟎𝟎 −𝒄𝒄 − 𝒄𝒄 𝒂𝒂𝒙𝒙𝟐𝟐 + 𝒃𝒃𝒙𝒙 = −𝒄𝒄

÷ 𝒂𝒂 ÷ 𝒂𝒂 ÷ 𝒂𝒂

𝒙𝒙𝟐𝟐 + 𝒃𝒃𝒂𝒂𝒙𝒙 = − 𝒄𝒄

𝒂𝒂

because the terms do not divide evenly by the leading coefficient, we end up with fractions

next, we divide the middle term by 2, but because it doesn’t divide evenly, and with the way that fractions work, the denominator gets larger:

𝑥𝑥2 + 113𝑥𝑥 = −5

÷ 𝟐𝟐

�𝑥𝑥 + 116�2

𝒙𝒙𝟐𝟐 + 𝒃𝒃𝒂𝒂𝒙𝒙 = − 𝒄𝒄

𝒂𝒂

÷ 𝟐𝟐

�𝒙𝒙 + 𝒃𝒃𝟐𝟐𝒂𝒂�𝟐𝟐

By the end of this lesson, you should be able to: • apply the completing the square process to a generic quadratic equation • connect the completing the square process to the Quadratic Formula • use the Quadratic Formula on simple quadratic equations

When we write the power on the outside of the parentheses, we balance that constant squared by adding the it to the other side as well:

�𝑥𝑥 + 116�2

= − 5 3

+𝟏𝟏𝟐𝟐𝟏𝟏𝟑𝟑𝟑𝟑

= − 𝒄𝒄𝒂𝒂

+𝒃𝒃𝟐𝟐

𝟒𝟒𝒂𝒂𝟐𝟐

*remember, (𝟐𝟐𝒂𝒂)𝟐𝟐

𝒃𝒃𝒃𝒃𝒄𝒄𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 �� 𝟒𝟒𝒂𝒂𝟐𝟐

but now we need to convert to a common denominator in order to combine the fractions on the right side:

�𝑥𝑥 + 116�2

= − 5 3 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� −60

+𝟏𝟏𝟐𝟐𝟏𝟏𝟑𝟑𝟑𝟑

= − 𝒄𝒄𝒂𝒂 𝒃𝒃𝒃𝒃𝒄𝒄𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 �� −𝟒𝟒𝒂𝒂𝒄𝒄

𝟒𝟒𝒂𝒂𝟐𝟐

+𝒃𝒃𝟐𝟐

𝟒𝟒𝒂𝒂𝟐𝟐

�𝒃𝒃𝟐𝟐−𝟒𝟒𝒂𝒂𝒄𝒄�𝟒𝟒𝒂𝒂𝟐𝟐

*because the letters are not like terms, the numerator does not simplify

the next step is to take the square root of both sides, remembering that with fractions, the root applies to both the numerator and denominator:

� �𝑥𝑥 + 116�2

= ± √61√36

𝑥𝑥 + 116

= ± √616

��𝒙𝒙 + 𝒃𝒃𝟐𝟐𝒂𝒂�𝟐𝟐

= ± �(𝒃𝒃𝟐𝟐−𝟒𝟒𝒂𝒂𝒄𝒄)�𝟒𝟒𝒂𝒂𝟐𝟐

𝒙𝒙 + 𝒃𝒃𝟐𝟐𝒂𝒂

= �(𝒃𝒃𝟐𝟐−𝟒𝟒𝒂𝒂𝒄𝒄)𝟐𝟐𝒂𝒂

last but not least, we isolate the variable using the Additive Property of Equality:

x + 116

= ± √616

−𝟏𝟏𝟏𝟏𝟑𝟑

− 𝟏𝟏𝟏𝟏𝟑𝟑

𝑥𝑥 =−11 ± √61

𝒙𝒙 + 𝒃𝒃𝟐𝟐𝒂𝒂

= �(𝒃𝒃𝟐𝟐−𝟒𝟒𝒂𝒂𝒄𝒄)𝟐𝟐𝒂𝒂

− 𝒃𝒃𝟐𝟐𝒂𝒂

𝑥𝑥 =−𝑏𝑏 ± √𝑏𝑏2 − 4𝑎𝑎𝑐𝑐

2𝑎𝑎

And that final result is the Quadratic Formula:

2𝑎𝑎

You can see that we usually drop the parentheses around the portion that is underneath the square root symbol.

Example 2 Okay, so we derived the Quadratic Formula. Now, how do we use it to solve a quadratic equation? Let’s start with problems whose answers will not involve simplifying a square root. Here’s an example:

5𝑥𝑥2 + 7𝑥𝑥 − 2 = 0

Step 1: identify 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐

5𝑥𝑥2 + 7𝑥𝑥 − 2 = 0

𝒂𝒂 = 𝟓𝟓 𝒃𝒃 = 𝟕𝟕 𝒄𝒄 = −𝟐𝟐

Step 2: calculate 𝑏𝑏2 and 4𝑎𝑎𝑐𝑐

𝑏𝑏2 = 49

4𝑎𝑎𝑐𝑐

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��4(5)(−2)

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� −40

Step 3: write out the Quadratic Formula 𝑥𝑥 =−𝑏𝑏 ± √𝑏𝑏2 − 4𝑎𝑎𝑐𝑐

2𝑎𝑎

Step 4: substitute the values from steps 1 & 2 into the formula

𝑥𝑥 =−7 ± √49 −−40

Step 5: combine the #’s under the square root symbol AND under the fraction bar 𝑥𝑥 =

−7 ± √89 10

You can see that because the radicand is prime, the square root can’t be simplified. This is the end of the active calculation phase of the problem. We could write the two answers as:

𝑥𝑥 =−7 + √89

10 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 =

−7 − √8910

but that would be as far as we would go in this situation.

1. Copy the equation into your notes. Then copy and complete the Quadratic Formula set-up shown underneath.

a. 𝟕𝟕𝒙𝒙𝟐𝟐 + 𝟑𝟑𝒙𝒙 − 𝟖𝟖 = 𝟎𝟎

𝑏𝑏2 = 4𝑎𝑎𝑐𝑐 =

𝑥𝑥 =−3 ± √ −

2( ) 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��

b. 𝟐𝟐𝒙𝒙𝟐𝟐 − 𝟗𝟗𝒙𝒙 + 𝟑𝟑 = 𝟎𝟎

𝑏𝑏2 = 4𝑎𝑎𝑐𝑐 =

𝑥𝑥 =9 ± √ −

2( ) 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��

Page 59 Page 59 Page 66

Example 3 There are a few unusual things to watch out for when using the Quadratic Formula. One thing to realize is that if the middle term (the “b” value) is negative, the Quadratic Formula will turn it back into a positive, like this:

2𝑥𝑥2 − 5𝑥𝑥 + 1 = 0 In this problem, = −5 , which would mean that 𝑏𝑏2 = +25 (not negative 25). When we apply the Quadratic Formula, it would look like this:

𝑥𝑥 =+5 ± √25 − 8

And the problem would proceed as usual. Another thing to realize is that if the first term (the “a”value) is negative, the Quadratic Formula will have a negative in the denominator, like this:

−3𝑥𝑥2 + 9𝑥𝑥 + 5 = 0 In this problem, 𝑎𝑎 = −3, which will mean that 4𝑎𝑎𝑐𝑐 = 4(−3)(5)

𝑤𝑤ℎ𝑖𝑖𝑏𝑏ℎ 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� −60

When we apply the Quadratic Formula, it would look like this:

𝑥𝑥 =−9 ± √81 −−60

2(−3)

As the formula calculations proceed, we end up here:

𝑥𝑥 =−9 ± √141

And technically speaking, we’re not supposed to leave a negative number in a denominator. (Remember, the denominator of a fraction is supposed to show how many pieces you cut a whole unit into, and it’s not possible to cut something into a negative number of pieces.) So we multiply both the top and bottom by −1:

𝑥𝑥 =�−9 ± √141 � × −1

(−6) × −1

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��

9 ± √1416

You can see that the sign change does not affect the radical --- it was already “plus-or-minus”. One last thing to watch for is to make sure that the quadratic is set up with all the like terms brought onto the same side and combined:

4𝑥𝑥2 + 10𝑥𝑥 = 3𝑥𝑥 − 2 −𝟑𝟑𝒙𝒙 + 𝟐𝟐 −𝟑𝟑𝒙𝒙 + 𝟐𝟐

4𝑥𝑥2 + 7𝑥𝑥 + 2 = 0 And then the Quadratic Formula can be applied:

𝑥𝑥 =−7 ± √49 − 32

2(4) 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��

−7 ± √178

Page 67

Problem Set Use the Quadratic Formula:

2𝑎𝑎 with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0

to solve each equation.

1. 3𝑥𝑥2 + 5𝑥𝑥 − 1 = 0

2. 2𝑥𝑥2 − 11𝑥𝑥 + 7 = 0

3. 5𝑥𝑥2 − 𝑥𝑥 − 3 = 0

4. 𝑥𝑥2 + 3𝑥𝑥 − 2 = 0

5. 4𝑥𝑥2 + 7𝑥𝑥 + 2 = 0

6. −7𝑥𝑥2 − 9𝑥𝑥 + 1 = 0

Page 68

Lesson 14: Using the Quadratic Formula Last edited on 3/5/18

Lesson 14: Using the Quadratic Formula

Example 1 In the last lesson, we introduced how to use the Quadratic Formula, but we limited ourselves to equations whose answers involved non-reducible radicals. In this lesson, we’ll expand the use of the Quadratic Formula to equations whose answers involve radicands that simplify. For example:

5𝑥𝑥2 + 9𝑥𝑥 + 2 = 0 will have a radical that does not simplify

5𝑥𝑥2 + 8𝑥𝑥 + 2 = 0 will have a radical that simplifies

We already know how to use the Quadratic Formula. Let’s see the differences between these two equations:

𝟓𝟓𝒙𝒙𝟐𝟐 + 𝟕𝟕𝒙𝒙 + 𝟐𝟐 = 𝟎𝟎 𝟓𝟓𝒙𝒙𝟐𝟐 + 𝟖𝟖𝒙𝒙 + 𝟐𝟐 = 𝟎𝟎

first, we identify a, b, and c, then calculate 𝒃𝒃𝟐𝟐 and 𝟒𝟒𝟒𝟒𝟒𝟒 :

𝑎𝑎 = 5, 𝑏𝑏 = 9, 𝑐𝑐 = 2

𝑏𝑏2 = 81, 4𝑎𝑎𝑐𝑐 = 4(5)(2) = 40

𝟒𝟒 = 𝟓𝟓, 𝒃𝒃 = 𝟖𝟖, 𝟒𝟒 = 𝟐𝟐

𝒃𝒃𝟐𝟐 = 𝟔𝟔𝟒𝟒, 𝟒𝟒𝟒𝟒𝟒𝟒 = 𝟒𝟒(𝟓𝟓)(𝟐𝟐) = 𝟒𝟒𝟎𝟎

next, we plug into the Quadratic Formula:

𝑥𝑥 =−9 ± √81 − 40

2(5) 𝑥𝑥 =−𝟖𝟖 ± √𝟔𝟔𝟒𝟒 − 40

now we combine inside the radical, and in the denominator:

𝑥𝑥 =−9 ± √41

10 𝑥𝑥 =

−8 ± √44 10

in the left hand column, the radicand is prime, and does not simplify. in the right hand column, the radicand is a multiple of 4, and so it will simplify:

𝑥𝑥 =−9 ± √41

10 𝑥𝑥 =

−8 ± √4 × √1110 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏

��−8 ± 2√11

This is where the calculations end --- we can’t combine integers with mixed radicals. The only other

By the end of this lesson, you should be able to: • use the Quadratic Formula on any quadratic equation • use the discriminant to predict the number of solutions for a quadratic equation

thing we could do to change the appearance of the answers at this point would be to write them as separate conjugate pairs of fractions, like this:

𝑥𝑥 =−9 ± √41

10 𝑥𝑥 =

−8 ± √4 × √1110 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏

��−8 ± 2√11

𝑥𝑥 = − 910

+ √4110

and 𝑥𝑥 = − 910− √41

10 𝑥𝑥 = − 8

10+ 2√11

10 and 𝑥𝑥 = − 8

10− 2√11

which reduces to

𝑥𝑥 = −45

+ √1110

and 𝑥𝑥 = −45− √11

In the left hand column, we could have done the reducing without separating into two fractions:

𝑥𝑥 =−8 ± √4 × √11

10 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��

−8 ± 2√1110

𝑤𝑤ℎ𝑖𝑖𝑏𝑏ℎ 𝑟𝑟𝑏𝑏𝑟𝑟𝑟𝑟𝑏𝑏𝑏𝑏𝑏𝑏 𝑡𝑡𝑏𝑏 ��

−4 ± √1110

But it is important to realize that all three integers --- the −8, the 2, and the 10 --- must ALL have a common factor in order to do this reduction. If not, then it’s better to split into two fractions first, and then reduce to lowest terms

1. Copy each equation into your notes, then apply the Quadratic Formula to solve for 𝑥𝑥: a. 3𝑥𝑥2 − 5𝑥𝑥 − 1 = 0 b. 3𝑥𝑥2 − 6𝑥𝑥 − 1 = 0

* part A should not simplify, but part B should

Example 2 Whether a quadratic equation will have an answer where the radical simplifies, or whether it will not, depends on how the values of 𝑏𝑏2 and 4𝑎𝑎𝑐𝑐 combine. Because 4𝑎𝑎𝑐𝑐 is already a multiple of 4, then if 𝑏𝑏2 is also a multiple of 4, then we are guaranteed to have something that is reducible. In example 1, we contrasted 5𝑥𝑥2 + 9𝑥𝑥 + 2 = 0 with 5𝑥𝑥2 + 8𝑥𝑥 + 2 = 0. In the second equation, because the middle term is even, 𝑏𝑏2 is a multiple of 4, and the calculations underneath the radical symbol ended up as another multiple of 4, and therefore could be reduced. That’s not the only way that you can end up with an answer involving a mixed radical, but it is the most likely way. But it’s also possible that the radicand simplifies to an integer, instead of to a mixed radical like 2√11. Here’s an example:

5𝑥𝑥2 + 9𝑥𝑥 − 2 = 0

Page 59 Page 59 Page 65 Page 70

identify 𝑎𝑎, 𝑏𝑏, and 𝑐𝑐

5𝑥𝑥2 + 9𝑥𝑥 − 2 = 0

𝑎𝑎 = 5 𝑏𝑏 = 9 𝑐𝑐 = −2

calculate 𝑏𝑏2 and 4𝑎𝑎𝑐𝑐

𝑏𝑏2 = 81

4𝑎𝑎𝑐𝑐 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ��4(5)(−2)

write out the Quadratic Formula (this is optional) 𝑥𝑥 =

−𝑏𝑏 ± √𝑏𝑏2 − 4𝑎𝑎𝑐𝑐2𝑎𝑎

substitute into the formula 𝑥𝑥 =

−9 ± √81 −−40 2(5)

combine the #’s under the square root symbol AND under the fraction bar 𝑥𝑥 =

−9 ± √121 10

This time around, the radicand ended up as a perfect square. The square root of 121 is just 11 --- not a radical, but an integer. So the problem now involves no square root symbols:

𝑥𝑥 =−9 ± √121

−9 ± 1110

This means that the answers will end up as either a rational number, like this:

𝑥𝑥 =−9 + 11

𝑤𝑤ℎ𝑖𝑖𝑏𝑏ℎ 𝑟𝑟𝑏𝑏𝑟𝑟𝑟𝑟𝑏𝑏𝑏𝑏𝑏𝑏 𝑡𝑡𝑏𝑏 ��

Or an integer, like this:

𝑥𝑥 =−9− 11

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� −

𝑤𝑤ℎ𝑖𝑖𝑏𝑏ℎ 𝑟𝑟𝑏𝑏𝑟𝑟𝑟𝑟𝑏𝑏𝑏𝑏𝑏𝑏 𝑡𝑡𝑏𝑏 �� −2

2. Copy the equation into your notes, then use the Quadratic Formula to solve the equation. a. 4𝑥𝑥2 + 4𝑥𝑥 − 3 = 0 b. 6𝑥𝑥2 − 7𝑥𝑥 − 10 = 0

Example 3 It can be helpful to predict the type of answers that a quadratic equation will have. Knowing that the answers will NOT involve radicals is particularly important, because it means that the equation is factorable. Solving a quadratic from its factors is usually faster than any other method. The equation in example 2 was factorable, like this:

Page 71

5𝑥𝑥2 + 9𝑥𝑥 − 2 = 0 𝑓𝑓𝑓𝑓𝑏𝑏𝑡𝑡𝑏𝑏𝑟𝑟𝑏𝑏 𝑓𝑓𝑏𝑏 �� (5𝑥𝑥 − 1)(𝑥𝑥 + 2)

And from the factors we can see why the two answers are 15

and − 2:

5𝑥𝑥 − 1 = 0 𝑤𝑤ℎ𝑒𝑒𝑒𝑒 𝑥𝑥 = 15

𝑥𝑥 + 2 = 0 𝑤𝑤ℎ𝑒𝑒𝑒𝑒 𝑥𝑥 = −2 Being able to predict what type of answers a quadratic equation will have depends on finding the value that will be underneath the square root symbol. This radicand decides the style of answer because it determines whether the square root will simplify to a mixed radical or to an integer. The radicand is based on the value we get when calculating 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐. We call this portion of the Quadratic Formula the discriminant because it discriminates (which in this case means to separate into categories) between several types of answers. if the discriminant is… then the answers are… a perfect square ---

like 25 or 49 or 121 two rational numbers ---

two fractions, two integers, or one of each

positive, but not a perfect square--- like 24 or 50 or 120

two mixtures of rational and irrational numbers ---

like 4 5

+ √3 5

or −5±√172

zero one rational number --- fraction or integer

negative not possible as real numbers --- there are no answers

Here is an example of each type for you to try:

3. Copy the equation, then calculate the discriminant. When done, list the number and type of solutions that the quadratic equation will have.

a. 10𝑥𝑥2 − 7𝑥𝑥 + 1 = 0

b. 9𝑥𝑥2 + 12𝑥𝑥 + 4 = 0

c. 6𝑥𝑥2 − 5𝑥𝑥 − 3 = 0

d. 2𝑥𝑥2 − 𝑥𝑥 + 7 = 0 Partial answers: a) rational b) one answer only c) mixtures of rational/irrational d) zero answers

Problem Set Use the Quadratic Formula:

2𝑎𝑎 with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0

to solve each equation.

1. 3𝑥𝑥2 + 8𝑥𝑥 − 1 = 0

2. 2𝑥𝑥2 − 10𝑥𝑥 + 7 = 0

3. 5𝑥𝑥2 − 2𝑥𝑥 − 3 = 0

4. 𝑥𝑥2 + 4𝑥𝑥 − 2 = 0

5. 4𝑥𝑥2 + 6𝑥𝑥 + 2 = 0

6. 7𝑥𝑥2 − 8𝑥𝑥 + 1 = 0

Lesson 15: Graphing from Standard Form Last edited on 3/5/18

Lesson 15: Graphing from Standard Form

Example 1 The last few lessons focused on solving a quadratic equation using different methods. But what are we finding when we do that? The difference between a function and an equation is important. A function is a way to describe all of the possible values generated by a particular set of arithmetic operations. An equation takes that function and assigns it to one specific value. For example, consider this linear function and linear equation:

𝒚𝒚 = 𝟐𝟐𝟐𝟐 − 𝟓𝟓 𝟐𝟐𝟐𝟐 − 𝟓𝟓 = 𝟏𝟏𝟏𝟏 this is a function this is an equation

that describes a line that says the height of the line is 11 you use it to generate ordered pairs you use it to find the x-coordinate

like (2,−1) and (7, 9) of just that point: ( _____ , 11) The same sort of thing is true for quadratic functions and quadratic equations:

𝒇𝒇(𝟐𝟐) = 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 − 𝟖𝟖 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐 − 𝟖𝟖 = 𝟎𝟎 this is a function this is an equation

it describes a quadratic it says the quadratic has a height of zero you use it to generate ordered pairs you use it to find the 𝑥𝑥-intercepts

like (1,−5) and (−3,−5) which are (−4, 0) and (2, 0) How do we know that these points are the 𝑥𝑥-intercepts? Take a look:

𝑥𝑥2 + 2𝑥𝑥 − 8 = 0 Factors as: (𝑥𝑥 + 4)(𝑥𝑥 − 2) = 0 Which means that: 𝑥𝑥 + 4 = 0 or 𝑥𝑥 − 2 = 0 And solves as: 𝑥𝑥 = −4 or 𝑥𝑥 = 2 The underlying function is: 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 2𝑥𝑥 − 8 Watch what happens if we plug in −4 for 𝑥𝑥:

𝑓𝑓(−4) = (−4)2 + 2(−4) − 8 = 16 − 8 − 8 = 0 So when 𝑥𝑥 = −4, the function’s height is zero. That only happens when it’s crossing the 𝑥𝑥-axis.

By the end of this lesson, you should be able to: • graph a quadratic function by finding the vertex from the 𝟐𝟐-intercepts • graph a quadratic function by finding the vertex by completing the square

The same is true when = 2 : 𝑓𝑓(2) = (2)2 + 2(2) − 8

= 4 + 4 − 8 = 0

1. Each quadratic equation is factored. Use the factors to find the 𝑥𝑥-intercepts of the function: a. 𝑥𝑥2 − 5𝑥𝑥 − 24 = 0

(𝑥𝑥 − 8)(𝑥𝑥 + 3) = 0 For 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 54 − 24 The 𝑥𝑥-intercepts are:

b. 𝑥𝑥2 + 6𝑥𝑥 + 5 = 0 (𝑥𝑥 + 5)(𝑥𝑥 + 1) = 0 For 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 = 6𝑥𝑥 + 5 The 𝑥𝑥-intercepts are:

Example 2 So what can you do with the 𝑥𝑥-intercepts? Well, you know from Module 3 that quadratic functions are symmetric --- the right hand half is a mirror image of the right hand half. And the line that goes through the middle of the two halves will also go through the vertex. When you graphed quadratic functions in Module 3, Topic C, you used the vertex form. For example, the function 𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 3)2 − 4 has its vertex at the point (3,−4) and opens up, like this: As you can see from the graph, its 𝑥𝑥-intercepts are the points (1,0) and (5,0). And halfway between 𝑥𝑥 = 1 and 𝑥𝑥 = 5 is the value 3. The vertex’s 𝑥𝑥-coordinate is the midpoint between the 𝑥𝑥-intercepts:

1 + 52 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏

�� 6 2

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� 3

What this all means is that one way to find the vertex of a quadratic equation is to find the 𝑥𝑥-intercepts. Once we average the intercepts, we will know the first half of the vertex’s ordered pair. To find the second half, the 𝑦𝑦-coordinate, we can then plug the 𝑥𝑥-coordinate into the function.

Page 59 Page 59 Page 75

Here’s what the whole process would look like: 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 6𝑥𝑥 − 7

Set the function equal to zero: 𝑥𝑥2 + 6𝑥𝑥 − 7 = 0 Factors as: (𝑥𝑥 + 7)(𝑥𝑥 − 1) = 0 Solves as: 𝑥𝑥 = −7 𝑜𝑜𝑜𝑜 𝑥𝑥 = 1 So the 𝑥𝑥-intercepts are: (−7, 0) and (1,0) And the 𝑥𝑥-coordinate of the vertex is: −7+1

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� − 6

Which means the 𝑦𝑦-coordinate of the vertex is: (−3)2 + 6(−3) − 7 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� 9 − 18 − 7

𝑏𝑏𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑏𝑏�� −16

And now we plot the vertex of (−3,−16) as well as the 𝑥𝑥-intercepts: To find more points, we can use the quadratic pattern off of the vertex, either as a table of values or as a series of movements, which we then connect into a smooth curve:

2. Copy the equation into your notes. Factor it, then identify the 𝑥𝑥-intercepts. Use the intercepts to find the vertex.

a. 𝑥𝑥2 + 4𝑥𝑥 − 5 = 0 b. 𝑥𝑥2 − 6𝑥𝑥 + 8 = 0

Example 3 Not every quadratic function will be factorable. We have seen that in previous lessons in this module. So how else can we find the vertex and graph the shape? Another option besides factoring is to complete the square. You saw this process back in Lesson 11. Here is an example:

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 8𝑥𝑥 + 10 This is not factorable --- there are no factor pairs for 10 that would add to −8. But we can complete the square fairly easily, since the middle number is an even number:

𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 8𝑥𝑥 + 10 −10 − 10

𝑓𝑓(𝑥𝑥) − 10 = 𝑥𝑥2 − 8𝑥𝑥 ÷ 2

𝑓𝑓(𝑥𝑥) − 10 = (𝑥𝑥 − 4)2

+16 𝑓𝑓(𝑥𝑥) + 6 = (𝑥𝑥 − 4)2

𝑓𝑓(𝑥𝑥) = (𝑥𝑥 − 4)2 − 6 Now we know that this quadratic has a vertex of (4,−6) and opens up: Here is an example for you to try in your notes:

3. Complete the square, then graph the quadratic: 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 6𝑥𝑥 + 11 = 0

Did you get a vertex of (−3,−2)? Congratulations!

Example 4 Of course, not every quadratic lends itself to an easy complete-the-square process, either. For example, the middle term might not divide easily by 2, or there might be a coefficient other than 1 on the leading term. In those situations, we turn instead to using part of the quadratic formula. The Quadratic Formula looks like this:

𝑥𝑥 =−𝑏𝑏 ± √𝑏𝑏2 − 4𝑎𝑎𝑎𝑎

2𝑎𝑎 with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑎𝑎 = 0

and the first portion of it is also the formula for finding the 𝑥𝑥-coordinate of the vertex of the quadratic. We usually refer to the vertex as (ℎ,𝑘𝑘), so the formula for finding the 𝑥𝑥-coordinate is:

ℎ = − 𝑏𝑏2𝑒𝑒

And then, to find the 𝑘𝑘 value, we plug the ℎ value back into the function. Here’s an example of how the process would look:

𝑓𝑓(𝑥𝑥) = 2𝑥𝑥2 − 8𝑥𝑥 + 7 𝑎𝑎 = 2, 𝑏𝑏 = −8, 𝑎𝑎 = 7 ℎ = − 𝑏𝑏

2𝑒𝑒 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� +8

2(2) 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� 8

𝑏𝑏𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑏𝑏 �� 2

𝑘𝑘 = 𝑓𝑓(2) 𝑤𝑤ℎ𝑖𝑖𝑏𝑏ℎ 𝑖𝑖𝑏𝑏 �� 2(2)2 − 8(2) + 7

𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 �� 2(4) − 16 + 7

𝑏𝑏𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑏𝑏�� −1

We now know the vertex is the point (2,−1) and that the quadratic has been stretched by a factor of 2: Here are two practice problems for you to try in your notes:

4. Use the formula ℎ = − 𝑏𝑏2𝑒𝑒

to find the 𝑥𝑥-coordinate of the vertex. Then plug that value into the function to find the 𝑦𝑦-coordinate. You do not need to graph the function.

a. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 5𝑥𝑥 − 4 = 0 (the vertex will have fraction coordinates)

b. 𝑓𝑓(𝑥𝑥) = −𝑥𝑥2 − 6𝑥𝑥 + 1

Problem Set Find the vertex of each quadratic function: 1. 𝑓𝑓(𝑥𝑥) = 3𝑥𝑥2 + 6𝑥𝑥 − 1

2. 𝑓𝑓(𝑥𝑥) = −𝑥𝑥2 − 10𝑥𝑥 − 17

3. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 2𝑥𝑥 − 3

4. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 4𝑥𝑥 − 2

5. 𝑓𝑓(𝑥𝑥) = −𝑥𝑥2 + 6𝑥𝑥 + 2

6. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 8𝑥𝑥 + 1

Find the vertex and graph the function: 7. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 − 4𝑥𝑥 + 3

Page 79

8. 𝑓𝑓(𝑥𝑥) = −𝑥𝑥2 + 6𝑥𝑥 − 2

9. 𝑓𝑓(𝑥𝑥) = 𝑥𝑥2 + 2𝑥𝑥 − 6

10. 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥2 + 12𝑥𝑥 + 9

pbworksmrshicklin.pbworks.com/w/file/fetch/124298196/a1 - m4 a & b - quadratics.pdfm4 nys common...

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