Download Problem Solving: Roots of a Quadratic Equation

Problem Solving: Roots of a Quadratic Equation
Example Problem 1: In a conservation park, a lifeguard has used 620 m of
marker buoys to rope off a safe rectangular swimming enclosure. If one side
of the enclosure uses the beach as a border, find the dimensions of the
rectangular enclosure given that the area is 16 800 m2 .
Solution:
beach
Let w = width
Let 620 – 2w = length
w
620 -2w
Develop a quadratic equation for this problem
Area = Length x width
16 800 = (w)(620 – 2w)
16 800 = -2w2 + 620w
0 = -2w2 + 620w – 16800
Remember: A quadratic equation
is an equation of the form
ax2 + bx + c = 0
Now you need to solve for w. Remember that finding the solution to a quadratic equation is the same as
finding the roots of the function. Therefore, choose a method to find roots and use it. Shown below is an
example solution shown using the method of completing the square to find the roots. (Refer to previous
notes or p.43 in your text for elaboration on the different methods)
Completing the Square
0 = -2w2 + 620w -16800
16800
0
 2 2 620
=
w +
w2
2 2
2
2
0 = w – 310w + 8400
-8400 = w2 -310w
-8400 + 24025 = w2 – 310w + 24025
15625 = (w – 155)2
15625 = (w  155)2
Divide though by the a-value
Isolate the variable terms
Complete the square and balance
Factor the trinomial
Solve for w
±125 = w – 155
w = 155 ± 125
The two roots are 30 and 280
Note: verify both
Verify root of 30
W = 30m
L = 620 – 2(30) = 560m
A = (560)(30) = 16800 verified
P = 2w + L = 60 +560 = 620
Verify root of 280
W = 280m
L = 620 -2(280) =60m
A = (60)(280) = 16800
P = 2w + L = 560 + 60 = 620
Therefore, the possible dimensions of the swimming enclosure are 560m x 30m
and 60 m x 280m.
Example Problem 2: The sum of the squares of two consecutive even
numbers is 452. What are the two numbers?
Solution:
Let x = the 1st number
Let x+2= the 2nd number
x2 + (x+2)2 = 452
x2 + x2 + 4x + 4 = 452
2x2 + 4x - 448 = 0
set up the equation
expand the brackets (FOIL)
collect like terms and set to quadratic equation form
Now find roots using one of a number of methods
Example:
Method: Factoring by inspection
2x2 + 4x -448 = 0
x2 + 2x -224 = 0
divide through by value of leading coefficient
(x-14)(x+16) = 0
Therefore, the roots are 14 and -16
If x = 14, then x+2 = 16. 142 + 162 = 452. Works!
If x = -16, then x+2 = -14. (-16)2 + (-14)2 = 452. Works!
Note: There is no inadmissible root in this problem
Example Problem 3: A rectangular supermarket with dimensions 90m by 60m is built in
the centre of a rectangular city lot with an area of 9000 m2. The area of the lot that
surrounds the supermarket will be a uniform band of pavement that will be used for
parking. What is the width of the uniform band of pavement?
90 + 2x
90m
supermarket
90 m
60m
mmmmmm
Ff
Let x = the width of the uniform pavement
Let 90+2x = Length of lot
Let 60 + 2x = Width of lot
60+2x