General Quadratic Word Problems (page 2 of 3)
Most quadratic word problems should seem very familiar, as they are built from the linear problems that you've done in the past.
The height is defined in terms of the width, so I'll pick a variable for "width", and then create an expression for the height.
Let "w" stand for the width of the picture. The height h is 4/3 the width, so h = (4/3)w. Then the area is A = hw = [(4/3)w][w] = (4/3)w2 = 192. I need to solve this "area" equation for the value of the width, and then back-solve to find the value of the height.
Since I can't have a negative width, I can ignore the "w = –12" solution. Then the width must be 12 and the height is h = (4/3)(12) = 16.
The enlargement will be 12 inches by 16 inches.
Remember that consecutive integers are one unit apart, so my numbers are n and n + 1. Multiplying to get the product, I get:
+ 1) = 1122
The solutions are n
= –34 and n
= 33. I need a negative
value, so I'll ignore "n
= 33" and
The two numbers are –33 and –34.
Note that the second value could have been gotten by changing the sign on the extraneous solution. Warning: 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.
(12 + 2x)(16
+ 2x) = 285
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. Checking the original exercise to verify what I'm being asked to find, I notice that I need to have units on my answer:
The width of the pathway will be 1.5 meters.
This is the quadratic I need to solve. I can take the square root of either side, and then add the to the right-hand side:
...or I can multiply out the square and apply the Quadratic Formula:
Either way, I get two solutions which, when expressed in practical decimal terms, tell me that the width of the original cardboard is either about 2.26 inches or else about 9.74 inches.
How do I know which solution value for the width is right? By checking each value in the original word problem. If the cardboard is only 2.26 inches wide, then 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 of cardboard on either side, and still be left with 3.74 inches in the middle. Checking:
(3.74)(3.74)(3) = 41.9628
This isn't exactly 42, but, taking round-off error into account, it's close enough that I can trust that I have the correct value:
The cardboard should measure 9.75 inches on a side.
In this last exercise above, you should notice that each solution method gave the same final answer for the cardboard's width. But the Quadratic Formula took longer and provided me with more opportunities to make mistakes. Warning: Don't get stuck in the rut of always using the Quadratic Formula!