McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
1. Solving Quadratic Equations by Factoring
Continue to next subchapter

Exercise 46 Page 174

To find the area of a rectangle, we multiply its length by its width.

x=10
Dimensions: 8ft by 12ft

Practice makes perfect

To find the value of x and the dimensions of the given rectangle, recall that the area of a rectangle is found by multiplying its length by its width. We see in the diagram that the area is A=96ft^2, the length is l = x-2ft, and the width is w=x+2ft.

Solving for x

We can create an equation to solve for x by substituting the given expressions into the formula for the area of a rectangle. A=l * w ⇒ 96=(x-2)(x+2) Let's simplify this equation. Notice that on the right-hand side we have difference of two squares.
96=(x-2)(x+2)
â–Ľ
(a+b)(a-b)=a^2-b^2
96=x^2-2^2
96=x^2-4
100=x^2
x^2=100
Now, we will use inverse operations to solve the equation.
x^2=100
sqrt(x^2)=sqrt(100)

sqrt(a^2)=± a

x=±sqrt(100)
x=±10
The solutions of this equation are x=10 and x=- 10.

Finding the Dimensions

We need to determine which of the solutions that we found will satisfy the given conditions of our rectangle. To do this, let's substitute these values into the expressions for the length and the width of the rectangle. Then we can evaluate the reasonableness of each measurement.

Length (l) Width (w)
x= 10 10-2=8 10+2=12
x= - 10 - 10-2=- 12 - 10+2=- 8

If x=- 10, the length and the width are both negative. This does not make sense, because a rectangle cannot have negative dimensions. Therefore, x=10 and the dimensions of the rectangle are l = 8ft and w=12ft.