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=432in^2, the length is l = x-2in, and the width is w=x+4in.
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 ⇒ 432=(x-2)(x+4)Let's simplify the this equation. We will start by distributing the parentheses on the right-hand side.
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.
The solutions of this equation are x=20 and x=- 22.
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= 20
20-2=18
20+4=24
x= - 22
- 22-2=- 24
- 22+4=- 18
If x=- 22, the length and the width are both negative. This does not make sense, because a rectangle cannot have negative dimensions. Therefore, x=20 and the dimensions of the rectangle are l = 18in and w=24in.
