## Engage NY Eureka Math Algebra 1 Module 4 Lesson 5 Answer Key

### Eureka Math Algebra 1 Module 4 Lesson 5 Example Answer Key

Example 1.

For each of the related questions below, use what you know about the zero product property to find the answers.

a. The area of a rectangle can be represented by the expression x^{2} + 2x – 3. If the dimensions of the rectangle are known to be the linear factors of the expression, write each dimension of this rectangle as a binomial. Write the area in terms of the product of the two binomials.

Answer:

The factors of x^{2} + 2x – 3 are (x + 3)(x – 1), so the dimensions of the rectangle are (x – 1) on the shorter side (width) and (x + 3) on the longer side (length).

b. Draw and label a diagram that represents the rectangleâ€™s area.

The diagram below models the problem in a way that avoids negative areas.

c. Suppose the rectangleâ€™s area is 21 square units. Can you find the dimensions of the rectangle?

Answer:

Yes. By setting the area expression equal to 21, I can solve for x and from there find the dimensions.

d. Rewrite the equation so that it is equal to zero and solve.

Answer:

x^{2} + 2x – 3 – 21 = 0 becomes x^{2} + 2x – 24 = 0

(x + 6)(x – 4) = 0, which leads to x = – 6 or 4

e. What are the actual dimensions of the rectangle?

Answer:

The dimensions of the rectangle are (x – 1) by (x + 3), so the dimensions would be (4 – 1) = 3 by (4 + 3) = 7.

f. A smaller rectangle can fit inside the first rectangle, and it has an area that can be represented by the expression x^{2} – 4x – 5. If the dimensions of the rectangle are known to be the linear factors of the expression, what are the dimensions of the smaller rectangle in terms of x?

Answer:

Factoring the quadratic expression, we get (x – 5)(x + 1).

g. What value for x would make the smaller rectangle have an area of 1/3 that of the larger?

Answer:

\(\frac{1}{3}\) of the larger area is \(\frac{1}{3}\) times 21, which is 7 square units. x^{2} – 4x – 5 = 7 is easier to solve if we subtract 7 from each side to get x^{2} – 4x – 12 = 0. Factoring the left side gives us (x – 6)(x + 2) = 0. So, x = 6 or – 2. Discuss which answer is correct based on this context.

### Eureka Math Algebra 1 Module 4 Lesson 5 Exercise Answer Key

Opening Exercise

Consider the equation aâˆ™bâˆ™câˆ™d = 0. What values of a, b, c, and d would make the equation true?

Answer:

Any set of values where one of the four variables was equal to zero, and also values where two, three, or even all four of the variables were equal to zero.

Exercises 1â€“4

Find values of c and d that satisfy each of the following equations. (There may be more than one correct answer.)

Exercise 1.

cd = 0

Answer:

Either c or d must be zero, but the other can be any number, including zero (i.e., both c and d MIGHT be equal to zero at the same time).

Exercise 2.

(c – 5)d = 2

Answer:

There are an infinite number of correct combinations of c and d, but each choice of c leads to only one choice for d and vice versa. For example, if d = 2, then c must be 6, and if c = 4, then d must be – 2.

Exercise 3.

(c – 5)d = 0

Since the product must be zero, there are only two possible solution scenarios that make the equation true, c = 5 (and d can be anything) or d = 0 (and c can be anything); specifically, one solution would be c = 5 and d = 0.

Exercise 4.

(c – 5)(d + 3) = 0

Answer:

c = 5 or d = – 3. Either makes the product equal zero; they could both be true, but both do not have to be true. However, at least one must be true.

Exercises 5â€“8

Solve. Show your work.

Exercise 5.

x^{2} – 11x + 19 = – 5

Answer:

x^{2} – 11x + 19 = – 5

x^{2} – 11x + 24 = 0

(x – 3)(x – 8) = 0

x = 3 or 8

Exercise 6.

7x^{2} + x = 0

Answer:

7x^{2} + x = 0

x(7x + 1) = 0

x = 0 or – \(\frac{1}{7}\)

(This problem has two points to remember: All terms have a factor of 1, and sometimes the solution is a fraction.)

Exercise 7.

7r^{2} – 14r = – 7

Answer:

7r^{2} – 14r = – 7

7r^{2} – 14r + 7 = 0

7(r^{2} – 2r + 1) = 0

7(r – 1)(r – 1) = 0 or 7(r – 1)^{2} = 0

r = 1

(There is only one solution; rather, both solutions are 1 in this case.)

Exercise 8.

2d^{2} + 5d – 12 = 0

Answer:

2d^{2} + 5d – 12 = 0

Two numbers for which the product is – 24 and the sum is + 5: – 3 and + 8.

So, we split the linear term: 2d^{2} – 3d + 8d – 12 = 0.

And group by pairs: d(2d – 3) + 4(2d – 3) = 0.

Then factor: (2d – 3)(d + 4) = 0.

So, d = – 4 or \(\frac{3}{2}\).

### Eureka Math Algebra 1 Module 4 Lesson 5 Problem Set Answer Key

Solve the following equations.

Question 1.

(2x – 1)(x + 3) = 0

Answer:

x = \(\frac{1}{2}\) or – 3

Question 2.

(t – 4)(3t + 1)(t + 2) = 0

Answer:

t = 4 or – \(\frac{1}{3}\) or – 2

Question 3.

x^{2} – 9 = 0

Answer:

(x + 3)(x – 3) = 0

x = – 3 or 3

Question 4.

(x^{2} – 9)(x^{2} – 100) = 0

Answer:

(x + 3)(x – 3)(x + 10)(x – 10) = 0

x = – 3 or 3 or – 10 or 10

Question 5.

x^{2} – 9 = (x – 3)(x – 5)

Answer:

(x + 3)(x – 3) – (x – 3)(x – 5) = 0

(x – 3)(x + 3 – (x – 5)) = 0

(x – 3)(8) = 0

x = 3

Question 6.

x^{2} + x – 30 = 0

Answer:

(x + 6)(x – 5) = 0

x = – 6 or 5

Question 7.

p^{2} – 7p = 0

Answer:

p(p – 7) = 0

p = 0 or 7

Question 8.

p^{2} – 7p = 8

Answer:

p^{2} – 7p – 8 = 0

(p – 8)(p + 1) = 0

p = 8 or – 1

Question 9.

3x^{2} + 6x + 3 = 0

Answer:

3(x + 1)(x + 1) = 0

x = – 1

Question 10.

2x^{2} – 9x + 10 = 0

Answer:

(2x – 5)(x – 2) = 0

x = \(\frac{5}{2}\) or 2

Question 11.

x^{2} + 15x + 40 = 4

Answer:

x^{2} + 15x + 36 = 0

(x + 12)(x + 3) = 0

x = – 12 or – 3

Question 12.

7x^{2} + 2x = 0

Answer:

x(7x + 2) = 0

x = 0 or – \(\frac{2}{7}\)

Question 13.

7x^{2} + 2x – 5 = 0

Answer:

(7x – 5)(x + 1) = 0

x = \(\frac{5}{7}\) or – 1

Question 14.

b^{2} + 5b – 35 = 3b

Answer:

b^{2} + 2b – 35 = 0

(b + 7)(b – 5) = 0

b = – 7 or 5

Question 15.

6r^{2} – 12r = 18

Answer:

6r^{2} – 12r – 18 = 0

6(r – 3)(r + 1) = 0

r = 3 or – 1

Question 16.

2x^{2} + 11x = x^{2} – x – 32

Answer:

x^{2} + 12x + 32 = 0

(x + 8)(x + 4) = 0

x = – 8 or – 4

Question 17.

Write an equation (in factored form) that has solutions of x = 2 or x = 3.

Answer:

c(x – 2)(x – 3) = 0 where c is any constant

Question 18.

Write an equation (in factored form) that has solutions of a = 0 or a = – 1.

Answer:

ca(a + 1) = 0 where c is any constant

Question 19.

Quinn looks at the equation (x – 5)(x – 6) = 2 and says that since the equation is in factored form it can be solved as follows:

(x – 5)(x – 6) = 2

x – 5 = 2 or x – 6 = 2

x = 7 or x = 8 .

Explain to Quinn why this is incorrect. Show her the correct way to solve the equation.

Answer:

This method is incorrect because the factored expression is equal to 2 and not 0. Only zero has the special property that if the product of two numbers equals zero then one (or both) of the numbers must equal zero. In order to solve this equation, we need to rewrite it as a factored expression equal to zero.

x^{2} – 11x + 30 = 2

x^{2} – 11x + 28 = 0

(x – 7)(x – 4) = 0

x = 7 or 4

### Eureka Math Algebra 1 Module 4 Lesson 5 Exit Ticket Answer Key

Question 1.

Factor completely: 3d^{2} + d – 10.

Answer:

(3d – 5)(d + 2)

Question 2.

Solve for d: 3d^{2} + d – 10 = 0.

Answer:

(3d – 5)(d + 2) = 0, so d = \(\frac{5}{3}\) or – 2.

Question 3.

In what ways are Problems 1 and 2 similar? In what ways are they different?

Answer:

Both involve the same quadratic expression. The first is just an expression (no equal sign), so it cannot be solved, but there are two factors that are irreducible over the integers. The second is an equation in d and has two solutions that are related to those factors.