Grade: 9 Subject: Math (Algebra I) Unit: Introduction to Quadratics Lesson: 4 of 6 SAT: AdvancedMath ACT: Math

Word Problems

Learning Objectives

In this lesson, you will learn to:

Solve these 10 word problems. Click each question to reveal the answer.

Question 1: The product of two consecutive positive integers is 72. Find the integers.

Answer: 8 and 9

Solution: Let n = first integer. n(n+1) = 72. n^2 + n - 72 = 0. Factor: (n+9)(n-8) = 0. n = 8 (positive), so integers are 8 and 9.

Question 2: A rectangle has length 3 cm more than its width. Its area is 70 cm^2. Find the dimensions.

Answer: Width = 7 cm, Length = 10 cm

Solution: Let w = width. w(w+3) = 70. w^2 + 3w - 70 = 0. Factor: (w+10)(w-7) = 0. w = 7 (positive). Length = 10.

Question 3: A ball is thrown upward with h = -16t^2 + 48t + 4 (height in feet, time in seconds). When does it hit the ground?

Answer: t = 3.08 seconds (approximately)

Solution: Set h = 0: -16t^2 + 48t + 4 = 0. Divide by -4: 4t^2 - 12t - 1 = 0. Use quadratic formula: t = (12 + sqrt(160))/8 = 3.08 seconds.

Question 4: The sum of a number and its square is 56. Find the number.

Answer: 7 or -8

Solution: Let n = number. n + n^2 = 56. n^2 + n - 56 = 0. Factor: (n+8)(n-7) = 0. n = 7 or n = -8.

Question 5: A square garden has a 2-meter walkway around it. The total area (garden + walkway) is 196 m^2. Find the garden's side length.

Answer: 10 meters

Solution: Let s = garden side. Total side = s + 4. (s+4)^2 = 196. s + 4 = 14 (positive root). s = 10 meters.

Question 6: A rocket's height is h = -5t^2 + 30t. When is it at 40 meters?

Answer: t = 2 seconds and t = 4 seconds

Solution: Set h = 40: -5t^2 + 30t = 40. -5t^2 + 30t - 40 = 0. Divide by -5: t^2 - 6t + 8 = 0. Factor: (t-2)(t-4) = 0.

Question 7: The difference of two numbers is 5, and their product is 84. Find the numbers.

Answer: 12 and 7 (or -7 and -12)

Solution: Let x = larger, y = smaller. x - y = 5, xy = 84. x = y + 5. (y+5)y = 84. y^2 + 5y - 84 = 0. (y+12)(y-7) = 0. y = 7 or -12.

Question 8: A triangular plot has base 4 ft longer than height. Area is 48 ft^2. Find the base.

Answer: Base = 12 ft

Solution: Let h = height, b = h + 4. Area: (1/2)bh = 48. (1/2)(h+4)h = 48. h^2 + 4h = 96. h^2 + 4h - 96 = 0. (h+12)(h-8) = 0. h = 8, base = 12.

Question 9: A picture frame (uniform width) is added to an 8x10 photo. Total area is 168 in^2. Find the frame width.

Answer: 2 inches

Solution: Let w = frame width. (8+2w)(10+2w) = 168. 80 + 36w + 4w^2 = 168. 4w^2 + 36w - 88 = 0. w^2 + 9w - 22 = 0. (w+11)(w-2) = 0. w = 2.

Question 10: A number minus twice its reciprocal equals 1. Find the number.

Answer: 2 or -1

Solution: Let x = number. x - 2/x = 1. Multiply by x: x^2 - 2 = x. x^2 - x - 2 = 0. (x-2)(x+1) = 0. x = 2 or x = -1.