# Word problems involving quadratic equations

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 below shows how to set up the problem.

Solution:

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?

Solution:

Let n and m be the two numbers.

n + m = 12  (1)
n × m = 35  (2)

Using (1), n = 12 - m

(12 - m) × m = 35

12m - m2 = 35

- m2 + 12m - 35 = 0

m2 + -12m + 35 = 0

(m - 5)(m - 7) = 0

m = 5 and m = 7. The two numbers are 5 and 7

## Interesting word problems involving quadratic equations.

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

Solution:

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.

x =
-b / 2a

Since a = 0.00002 and 0.00002 is bigger than 0, the quadratic equation will give a minimum.

x =
- -0.04 / 2 × 0.00002

x =
0.04 / 0.00004
=1000

To minimize the cost, the company should produce 1000 tires.

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?

Solution:

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

l =
9 - 2w / 2

A =
9 - 2w / 2
× w

A =
9w - 2w2 / 2

A = -w2 + 4.5w

A = -w2 + 4.5w + 0

Since a = -1 and -1 is smaller than 0, the quadratic equation will give a maximum.

x =
-b / 2a

w =
-4.5 / -2
= 2.25

l =
9 - 2 × 2.25 / 2

l =
9 - 4.5 / 2
= 2.25

To maximize the area, l = w = 2.25

If you found the word problems involving quadratic equations on this lesson difficult to understand, review the lesson about factoring trinomials

100 Tough Algebra Word Problems.

If you can solve these problems with no help, you must be a genius!