McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
1. Solving Quadratic Equations by Factoring
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Exercise 48 Page 174

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

x=12
Dimensions: 32ft by 14ft

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=448ft^2, the length is l = 3x-4ft, 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 ⇒ 448=(3x-4)(x+2)Let's simplify the this equation. We will start by distributing the parentheses on the right-hand side.
448=(3x-4)(x+2)
â–Ľ
Multiply parentheses
448=3x(x+2)-4(x+2)
448=3x^2+6x-4(x+2)
448=3x^2+6x-4x-8
448=3x^2+2x-8
0=3x^2+2x-456
3x^2+2x-456=0
Now we can factor the quadratic trinomial on the left-hand side of the equation. Here you can see a step-by-step guide on how to factor quadratic trinomials.
3x^2+2x-456=0
3x^2-36x+38x-456=0
â–Ľ
Factor out 3x & 38
3x(x-12)+38x-456=0
3x(x-12)+38(x-12)=0
(x-12)(3x+38)=0
Next, we will use the Zero Product Property to solve the equation.
(x-12)(3x+38)=0
lcx-12=0 & (I) 3x+38=0 & (II)
â–Ľ
(I), (II): Solve for x
lx=12 3x+38=0
lx=12 3x=- 38
lx=12 x=- 383
The solutions of this equation are x=12 and x=- 383.

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= 12 3( 12)-4=
32
12+2=
14
x= - 38/3 3( - 383 ) -4=
- 42
- 383+2=
- 32/3

If x=- 383, the length and the width are both negative. This does not make sense, because a rectangle cannot have negative dimensions. Therefore, x=12 and the dimensions of the rectangle are l = 32ft and w=14ft.

Checking Our Solution

We can check this solution by solving and seeing that the area is 448ft^2 when the dimensions are multiplied.
A=l * w
488? =32* 14
488=488