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General Quadratic Word Problems (page 2 of 3)

Sections: Projectile motion, General word problems, Max/min problems

Most quadratic word problems should seem very familiar, as they are built from the linear problems that you're used to doing.

• A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?

The height is 4/3 the width, so h = (4/3)w. Then the area is A = hw = (4/3)w² = 192. I have to solve this to find the width, and then back-solve to find the height.

(4/3)w² = 192
w² = 144
w = ± 12

Since I can't have a negative width, I can ignore "w = –12". Then the width is 12 and the height is h = (4/3)(12) = 16.

The enlargement will be 12 inches by 16 inches.

• The product of two consecutive negative integers is 1122. What are the numbers?

Remember that consecutive integers are one unit apart, so my numbers are n and n + 1. Multiplying to get the product, I get:

n(n + 1) = 1122
n² + n = 1122
n² + n – 1122 = 0
(n + 34)(n – 33) = 0   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

The solutions are n = –34 and n = 33. I need a negative value, so I'll ignore "n = 33" and
take n = –34. Then the other number is n + 1 = (–34) + 1 = –33.

The two numbers are –33 and –34.

Note that the second value could have been gotten by changing the sign on the extraneous solution. Many students get in the very bad habit of arbitrarily changing signs to get the answers they need, but this does not always work, and will very likely get them in trouble later on. Take the extra half a second to find the right answer the right way.

• A garden measuring 12 meters by 16 meters is to have a pedestrian pathway installed all around it, increasing the total area to 285 square meters. What will be the width of the pathway?
    

The first thing I need to do is draw a picture. Since I don't know how wide the path will be, I'll label this as "x".    Looking at my picture, this says that the total width will be 12 + x + x = 12 + 2x, and the total length will be 16 + x + x = 16 + 2x.    Then the new area is given by:

(12 + 2x)(16 + 2x) = 285
192 + 56x + 4x² = 285
4x² + 56x – 93 = 0

This quadratic is messy enough that I won't bother with trying to use factoring to solve; I'll just go straight to the Quadratic Formula:

Obviously the negative value won't work in this context, so I'll ignore it.

The width of the pathway will be 1.5 meters.

• You have to make a square-bottomed, unlidded box with a height of three inches and a volume of approximately 42 cubic inches. You will be taking a piece of cardboard, cutting three-inch squares from each corner, scoring between the corners, and folding up the edges. What should be the dimensions of the cardboard, to the nearest quarter inch?
   

When dealing with geometric sorts of word problems, it is usually helpful to draw a picture. Since I'll be cutting equal-sized squares out of all of the corners, and since the box will have a square bottom, I know I'll be starting with a square piece of cardboard. I don't know how big the cardboard will be yet, so I'll label the sides as having length "w".
Since I know I'll be cutting out three-by-three squares to get sides that are three inches high, I can mark that on my drawing. The dashed lines show where I'll be scoring the cardboard and folding up the sides.
Since I'll be losing three inches on either end of the cardboard when I fold up the sides, the final width of the bottom will be the original "w" inches, less three on the one side and another three on the other side. That is, the width of the bottom will be w – 3 – 3 = w – 6. Then the volume of the box, from the drawing, is: (w – 6)(w – 6)(3) = 42 (w – 6)(w – 6) = 14

This is the quadratic I need to solve. Simplifying and applying the Quadratic Formula, I get two solutions:

That is, the width of the original cardboard is either about 2.26 inches or else about 9.74 inches. How do I know which one is right? By checking in the original word problem. If the cardboard is only 2.26 inches wide, how on earth would I be able to fold up three-inch-deep sides? But if the cardboard is 9.74 inches, then I can fold up three inches on either side, and still be left with 3.74 inches in the middle. Checking:

(3.74)(3.74)(3) = 41.9628

So the second solution works, giving me the required volume:

The cardboard should measure 9.75 inches on a side.

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