# 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

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