2. Creating and Solving Quadratic Equations in One Variable Eureka Math Algebra 1 Module 3 Lesson 7

3. Objectives Students interpret word problems to create equations in one variable and solve them (i.e., determine the solution set) using factoring and the zero product property. Objectives A.SSE.B.3a A.CED.A.A MP.2

6. (h)(w) = A (4x)(3x) = 192 12x ² - 192 = 0 12 (x ² - 16) = 0 12 ( x – 4 ) (x + 4) = 0 x = 4 or -4 Since -4 would give negative dimensions, the only viable answer is x = 4 for dimensions of 16 in. and 12 in.

11. Both are possible answers for the length. So if the length is 10 ft., the width is 24 ft. Likewise, if the length is 24 ft., the width is 10 ft.

13. We’ll share our answers in class and see who has the dimensions with the biggest area.