8.3 QUADRATIC EQUATION WORD PROBLEMS
APPLICATIONS – Sometimes the negative answers will not make
sense in the context of the problem.
Example 6
Ronald’s living room is 2 feet longer than it is wide, and its area is
168 square feet. What are the dimensions of the room?
What are we asked to find? The dimensions – length and width
l = length
w = width
Given: Living room is 2 feet longer than it is wide.
l=w+2
Area = LW = 168
(w + 2)w = 168
Put in standard form, then factor and solve.
w2 + 2w – 168 = 0
(w - 12)(w + 14) = 0
w = 12, or w = -14 w = -14 does not make sense for width.
w = 12
l = 12 + 2 = 14
SOLUTION: Width is 12 feet and length is 14 feet.
Example 7
Shirley used 14 meters of fencing to enclose a rectangular region.
To be sure that the region was a rectangle, she measured the
diagonals and found that they were 5 meters each (since the
diagonals were equal then it is a rectangle). What are the length
and width of the rectangle?
5
w
Asked to find: length (l) and width (w)
l
Given: She used 14 meters of fencing. (Perimeter or Area?)
2l + 2w = 14
Use this equation to put w in terms of l.
2l + 2w = 14
2w = 14 -2l
w=7–l
Diagonal = 5 meters Use Pythagorean Theorem
l 2 + w 2 = 52
l2 + (7 – l)2 = 25
l2 + 49 – 14l +l2= 25
2l2 – 14l + 24 = 0
first
Factor. Don’t forget to factor out the GCF
2(l2 – 7l + 12)
2(l – 3)(l – 4) = 0
l = 3,
l=4
w = 7 -3 = 4
w=7–4=3
Length should be greater than width.
Length is 4 metersWidth is 3 meters
Example : A community garden sits on a corner lot and is in the shape of a right triangle. One side is 10 ft
longer than the shortest side, while the longest side is 20 ft longer than the shortest side. Find the lengths of the
sides of the garden.
Hint:
Step 2: Notice that the longest side is always the hypotenuse. Let variables represent the unknown quantities.
Use given information to put other uknowns in terms of one.
Step 3: Use given information and your variables to form an equation.
Because it is a right triangle, you can use the Pythagorean theorem.
x2 + (x+10)2 = (x + 20)2
Use distributive property and combine like terms. If it is a degree 2 equation, but it in
standard form by putting 0 on one side.
x2 + x2 + 20x + 100 = x2 + 40x + 400
2 x2 + 20x + 100 = x2 + 40x + 400
x2 - 20x -300 = 0
Step 4: Solve the equation.
Now factor this polynomial and use the zero-product-rule.
(x - 30)(x + 10) = 0
x = 30 or x = -10
The length of a side of a triangle cannot be negative, so we cannot use x = -10 as an answer.
Therefore x = 30 feet = length of shortest side
x + 30 = 30 + 10 = 40 feet = length of second side
x + 20 = 30 + 20 = 50 feet = length of the longest side.
Step 5: Check your answer and state conclusion. Does 302 + 402 = 502? 900 + 1600 = 2500 ? Yes
CONCLUSION: The lengths of the sides of the garden are 30 ft, 40 ft and 50 ft.
Projectile Motion Example:
After the semester is over, Herman discovers that the math
department has changed textbooks (again) so the bookstore won't
buy back his nearly-new book. Herman goes to the roof of the math
building, which is 160 feet high, and chucks his book straight down
at 48 feet per second. How many seconds does it take his book to
strike the ground?
Use the formula h(t) = –16t2 – 48t + 160
I need to find the time for the book to reach a height of zero ("zero"
being "ground level"), so:
0 = –16t2 – 48t + 160
Factor out -16 so it is in standard form.
0 = -16(t2 + 3t – 10)
You can now divide both sides by -16 to simplify.
t2 + 3t – 10 = 0
(t + 5)(t – 2) = 0
t = –5 or t = 2
t = -5 does not make sense because time can’t be negative, so
Correct answer is t=2 seconds.
CONCLUSION: It takes 2 seconds for the book to strike the ground.