13 MA11001 13A Quadratics Applications

Quadratic Applications KEY

Part 1: Geometry

Since area is a square unit, often quadratic equations must be used to solve problems involving area. Draw a picture to model each problem. Solve each using any of the following methods: factoring, graphing, or tables. Show all work.

1. The length of a rectangle is 7 meters more than the width. The area is 60 square meters. Find the length and width.
Let x=width and x+7=width

A=lw
60=(x+7)(x)
60=x²+7x x²+7x-60=0 (x+12)(x-5) x= -12 x=5

x must equal 5, since the -12 solution doesn’t make sense. A side cannot have negative length. The width is 5 meters and length is 12 meters.

2. A picture frame is 18 inches by 14 inches. If 221 square inches of picture shows, find the width of the frame, assuming the distance is equal all the way around.

Let x=width of frame
A=lw
221=(18-2x)(14-2x)
4x²-64x+31=0
x=15.5 x=0.5

15.5 is not possible. That is larger than 14, one of the sides of the frame. The frame has a width of 0.5 inches.

Quadratic Applications KEY

3. A building architect is modifying the visual space of a building with mirrors. One mirror is to be in the shape of a triangle whose height is 6 feet less than twice the base of the mirror. If the mirror has an area of 108 square feet, what are the base and height of the mirror?

Let x=base and 2x-6=height
A=bh
108=(x)(2x-6)
216=2x²-6x
2x²-6x-216=0 x=12 x= -9
The base of the mirror must be 12 feet. Since -9 does not make sense, the height is 18 feet.

4. The length of the side of a large square is 1 cm less than twice the length of the side of a smaller square. The area of the large square is 33 cm2 more than the area of the small square. Find the length of the sides of the two squares.

Let x=side of small square
2x-1=side of large square
AL=AS+33
(2x-1)²=x²+33
4x²-4x+1=x²+33