Check out these four great word problems involving quadratic equations in this lesson.
Problem #1: A rectangular garden has an area of 14 m2 and a perimeter of 18 meters. Find the dimensions of the rectangular garden. The figure above shows how to set up the problem.
Let w = width and l = length
2l + 2w = 18
lw = 14
Divide both sides of 2l + 2w = 18 by 2 to get l + w = 9.
Using l + w = 9, solve for l. We get l = 9 - w.
Substitute 9 - w for l in lw = 14
( 9 - w)w = 14
9w - w2 = 14
9w - w2 - 14 = 0
- w2 + 9w - 14 = 0
w2 - 9w + 14 = 0
(w - 7)(w - 2) = 0
w = 7 and w = 2
Problem #2: The sum of two numbers is 12 and their product is 35. What are the two numbers?
Let n and m be the two numbers.
n + m = 12 (1)
n × m = 35 (2)
Using (1), n = 12 - m
(12 - m) × m = 3512m - m2 = 35
Problem #3: The quadratic equation for the cost in dollars of producing automobile tires is given below where x is the number of tires the company produces. Find the number of tires that will minimize the cost.C = 0.00002x2 - 0.04x + 38
The standard form of a quadratic equation is ax² + bx + c. To solve this problem, we just need 2 important concepts about quadratic equations. First, when we are trying to maximize or minimize, we need to use the formula below that will help us find the x-coordinate of the vertex. Second, if a > 0, the vertex is a minimum. if a < 0, the vertex is a maximum.
Problem #4: You want to frame a collage of pictures with a 9-ft strip of wood. What dimensions will help you maximize the area?
First, we need to find the quadratic equation.
Area = l × w Perimeter = 2l + 2w
9 = 2l + 2w. Solve for l and replace l in the formula for the area.
9 - 2w = 2l
A = -w2 + 4.5w
A = -w2 + 4.5w + 0
If you found the word problems involving quadratic equations on this lesson difficult to understand, review the lesson about factoring trinomials