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Using Quadratic Equations to Solve Word Problems
SOLVED EXAMPLES

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 1999-2009 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.

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 the
width as "x".
Looking at my picture, I see that the total width will
be
x + 12 + x = 12 + 2x, and the total length will be
x + 16 + 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:
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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.

You have a 500-foot roll of fencing and a large field. You want to
construct a rectangular playground area. What are the dimensions of the
largest such yard? What is the largest area?
The fencing-length information gives me perimeter. If the length of the
enclosed area is L and the width is w, then the perimeter is 2L + 2w = 500, so
L = 250 – w. By solving the perimeter equation for one of the variables, I can
substitute into the area formula and get an equation with only one variable:
A = Lw = (250 – w)w = 250w – w2 = –w2 + 250w
To find the maximum, I have to find the vertex (h, k).
h = –b/2a = –(250)/2(–1) = –250/–2 = 125
In my area equation, I plug in "width" values and get out "area" values. So the
h-value in the vertex is the maximizing width, and the k-value will be the
maximal area:
k = A(125) = –(125)2 + 250(125) = –15 625 + 31 250 = 15 625
The problem didn't ask me "what is the value of the variable w?", but "what are
the dimensions?" I have w = 125. Then the length is L = 250 – w = 250 – 125
= 125.
The largest area will have dimensions of 125' by 125',
for a total area of 15 625 square feet.
Note that the largest rectangular area was a square. This is always true: for a given
perimeter, the largest rectangular area will be that of a square.

The sum of the digits of a two-digit number is 7. When the digits are
reversed, the number is increased by 27. Find the number.
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I'll use "t" for the "tens" digit of the original number and "u" for the "units" (or
"ones") digit. Keeping in mind that the tens digit stands for "ten times this
value", I then have:
t+u=7
Just as "26" is "10 times 2, plus 6 times 1", so also the two-digit number will be
ten times the "tens" digit, plus one times the "units" digit.
Original number: 10t + 1u
The new number will have the values of the digits (represented by the
variables) in reverse order:
New number: 10u + 1t
And this new number is twenty-seven more than the original number:
(New number) is (Old number) increased by (twenty-seven)
10u + 1t = 10t + 1u + 27
Now I have a system that I can solve:
t+u=7
10u + t = 10t + u + 27
First I'll simplify the second equation:
10u + t = 10t + u + 27
9u – 9t = 27
u–t=3
After reordering the variables in the first equation, I now have:
u+t=7
u–t=3
Adding down, I get 2u = 10, so u = 5. Then t = 2. Checking, this means that
the original number was 25 and the new number (gotten by switching the
digits) is 52. Since 52 – 25 = 27, this solution checks out. Copyright ©
Elizabes Reserved
The number is 25.

A landscaping company placed two orders with a nursery. The first order
was for 13 bushes and 4 trees, and totalled $487. The second order was
for 6 bushes and 2 trees, and totalled $232. The bills do not list the peritem price. What were the costs of one bush and of one tree?
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I could try to add the bushes and trees, to get 19 bushes and 6 trees, but this
wouldn't get me anywhere, because I don't have subtotals for the bushes and
trees. So I'll pick variables ("b" for the number of bushes and "t" for the number
of trees) and set up a system of equations:
first order: 13b + 4t = 487
second order: 6b + 2t = 232
Multiplying the second row by –2, I get:
13b + 4t = 487
–12b – 4t = –464
This says that b = 23. Back-solving, I get that t = 47. Of course, the exercise
didn't ask for the values of the two variables. Translating back into English, the
solution is:
Bushes cost $23 each; trees cost $47 each.
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