Quadratic Applications

Question

The width of a rectangle is 10 ft less than the length. The area is 10 ft² . Find the length and width.

Answer 1

Hello Natalie.  I will show you the procedure to find your length as well as your width.

 

 

Let's define variables to represent the quantities we want to find.

 

Let length = x

Let width = x - 10

Area = 10 ft²

 

 

We know the area of a rectangle is length times width.  We can write an equation that represents this using the variables we just defined.

 

x(x - 10) = 10

 

 

Distribute the x to get rid of parenthesis.

 

x² - 10x = 10

 

 

Subtract 10 on both sides of equation.

 

x² - 10x - 10 = 0

 

 

We have a quadratic equation.  We can use the quadratic formula to solve for x.  Note that when we solve for x, we will also find the length because the variable x represents length.

 

x = (-b ± √(b² - 4ac)) / 2a

 

where:

a = 1

b = -10

c = -10

 

Plug in these values into the formula.

 

x = (10 ± √(100 - 4(-10))) / 2

x = (10 ± √(140)) / 2

x = (10 ± 11.83) / 2

 

The plus/minus sign indicates that we will have two values of x.

 

x = (10 + 11.83) / 2           and            x = (10 - 11.83) / 2

 

x = 10.915                       and             x = -0.915

 

 

We have two values of x.  Next, is to determine which value to accept.  Since the variable x represents length, and we cannot have a negative value for any dimension, we accept the positive value of x.

 

x = 10.915

 

Substitute this value of x into the width.

 

Width = x - 10 = 10.915 - 10 = 0.915 ft

 

Length = 10.915 ft

Width = 0.915 ft

 

If we multiply these values, we get a result of 10 ft², which is the area.

 

 

Answer 2

L = length

W = width

 

Area = L*W

 

 

Write an equation for W in terms of the length (refer to the question about how width relates to length).

 

 

W = ______

 

A = L* (______)

 

Substitute what you found for W into the area equation, substitute the given area, then simplify and solve for L. Then use the length to solve for width.