Module 4 quadratic functions

Module 4
Quadratic Functions
What this module is about
This module is about the application of quadratic equations and functions
in everyday situations. As you go over the different problems you will apply your
knowledge and skills related to quadratic equations and functions in solving
problems. Frequently, only one solution of the equation is relevant to the problem
at hand. The root which does not satisfy the conditions of the problem must be
rejected. It is always a good practice, then, to check the solutions to determine if
one or both may be used.
What you are expected to learn
This module is designed for you to:
1. recall the different steps in solving word problems
2. translate verbal statements into symbols
3. apply knowledge and skills related to quadratic equations and functions in
solving problems.
How much do you know
A. Write the letter of the correct answer.
1. The roots of a quadratic equation are 8 and -8. What is the equation?
a. x2
+ 64 = 0 b. x2
– 64 = 0 c. x2
– 8 = 0 d. x2
+ 8 = 0
2. What are the values of a, b, and c in the equation 6x = 2x2
+ 1
a. a = -2 b = -6 c = 1

2
b. a = 2 b = -6
c. a = -2 b = 6 c. = -1
d. a = 2 b = -6 c. = 1
3. Which of the following is a solution of ( x – 3 ) ( x – 4 ) = 20
a = -5 b = 4 c = 8 d = 20
4. Which of the following has roots of -3 and 7?
a. x2
+ 4x – 21 = 0 b. x2
– 4x – 21 = 0
c. x2
– 4x + 21 = 0 d. x2
+ 4x + 21 = 0
5. Find the product of the roots of 2x2
= x + 3
a.
2
1
b.
2
3
c. -1 d.
2
3−
B. Write an equation to show the functional relationship between the quantities
involved in the problems using the indicated variables.
1. The area (A) of a square of side s is s2
.
2. The perimeter of a rectangle is equal to twice the Area.
3. 12x plus seven is equal to 3x reduced by 4
4. The circumference (C) of a circle is twice the product of π and the radius
(r).
5. The total distance (d) covered is equal to the product of the rate (r) and
the time (t).
C. Solve the following problems.
1. The product of two consecutive odd integers is 143. Find the integers.
2. One side of a rectangle is 10 cm longer than its width. The area of the
rectangle is 96 sq. cm. Find the dimensions of the rectangle.
3. The sum of two numbers is 22 and the sum of their squares is 250. Find
the numbers.

3
4. The sum of two numbers is 24. What is the maximum possible product of
the two numbers?
What you will do
Lesson 1
Number Problems
Many students could not solve this type of problem easily because they
fail to translate correctly the different expressions given by the word problem into
a correct equation. In the following examples, you will see that before you solve a
word problem, you must first translate the word problem into an equation.
Steps in Solving word problems involving quadratic equation and function:
1. Identify your unknowns.
2. Write your equation.
3. Operate on the numbers.
4. Convert the equation into the standard form. Remember to change signs.
5. Factor the equation into its two corresponding linear equations. By
factoring, you reduce the quadratic equation into two equivalent linear
equations.
Examples:
1. Find two consecutive odd integers whose product is 195.
Solution:
1st
step: Identify your unknowns.
Let x = one odd number,
so x + 2 = next consecutive odd number.

4
2nd
step: Write your equation.
x ( x + 2 ) = 195
one odd consecutive their
number odd number product
3rd
step: Operate on the numbers. Multiply the two quantities.
Thus:
x2
+ 2x = 195
4th
step: Bring all terms to the left side of the equal sign and equate to zero.
Remember to change signs.
So x2
+ 2x - 195 = 0
5th
step: Factor the equation.
Thus: (x + 15) (x – 13) = 0
6th
step: Solve for x.
x + 15 = 0 or (x – 13) = 0
x = -15 or x = 13
The consecutive numbers are :
x = 13 and x + 2 = 15, 13 and 15
or
x = -15 and x +2 = -15 + 2 = -13 , -13 and -15
Check:
x ( x + 2) = 195 or x ( x + 2) = 195
(13)(15) = 195 (-13)(-15) = 195
195 = 195 195 = 195

5
2. The sum of two numbers is 19 and their product is 60. Find the numbers.
Solution:
1st
step: Identify the unknown.
Let x = one number
So 19 - x = the other number
And x(19 – x) = their product.
2nd
step: Write the equation.
x ( 19 – x ) = 60
3rd
step: Operate the numbers to remove the parentheses grouping symbol.
19x – x2
= 60
4th
step: Convert the equation to its standard form.
Thus:
19x - x2
- 60 = 0 or x2
- 19x + 60 = 0
5th
( x – 15 ) ( x - 4 ) = 0
6th
step: Solve for x.
x – 15 = 0 and x – 4 = 0
x = 15 x = 4
19 – x 19 – 15 = 4 19 – x 19 –4 = 15
The numbers are therefore 15 and 4
Check:
Their sum: 15 + 4 = 19
19 = 19 check.
Their product: (15) (4) = 60
60 = 60 check.

6
the numbers.
Solution:
1st
step: Identify your unknown
Let x = one of the two numbers
22 – x = the other number
2nd
step: Write the equation
x2
+ ( 22 – x )2
= 250
3rd
step: Operate on the numbers to remove the parentheses grouping
symbol.
x2
+ ( 22 – x )2
= 250
x2
+ 484 - 44x + x2
= 250
2x2
- 44x + 484 = 250
4th
2x2
- 44x + 484 - 250 = 0
2x2
- 44x + 234 = 0 divide the equation by 2
x2
- 22x + 117 = 0
5th
step: Factor the quadratic equation to obtain the two equivalents linear
equations.
( x - 13 ) ( x - 9 ) = 0
6th
step: Solve for x.
x - 13 = 0 and x - 9 = 0
x = 13 and x = 9
If x = 13, 22 – 13 = 9.

7
If x = 9, 22 - 9 = 13.
The numbers are 9 and 13.
Check:
9 + 13 = 22;
92
+ 132
= 81 + 169 = 250.
Try this out
Analyze and solve.
1. Find 2 consecutive integers such that three times the square of the first is
equal to seven more than five times the second.
2. Find 2 consecutive even integers whose product is 224.
3. Find an integer such that the square of the integer is eighty one less than
eighteen times the integer.
4. One number is three times another and their product is equal to their sum
increased by fifty five. Find the numbers.
the numbers.
6. A number is
15
16
less than its multiplicative inverse. Find the number.
7. Find two numbers whose difference is 5 and the difference of their
squares is 65.
8. One number is three more than a second number. The sum of their
squares is 37 more than the product of the numbers. Find the two
numbers.
9. The product of two consecutive integers is 47 more than the next
consecutive integer. Find the two numbers.
10. The tens digit of a certain number is 7 less than the units digit. The sum of
the squares of the two digits is 85. Find the numbers.

8
Lesson 2
Geometry Problems
Many government and classroom examinations deals with problems
involving measurements of different geometric figures. These problems are also
called geometry problems.
Examples:
1. The altitude of a triangle is 3 cm. less than the base. The area of the
triangle is 35 square centimeters. What are its dimensions?
Solution:
1st
Let x cm = length of the base
So (x – 3) cm = length of the altitude
Area of a triangle =
2
))(( altitudebase
2nd
35 =
2
)3( −xx
3rd
x ( x – 3 ) = 70
x2
- 3x = 70
4th
x2
- 3x - 70 = 0
5th
(x – 10)(x + 7) = 0

9
6th
step: Solve for x.
x – 10 = 0 and x + 7 = 0
x = 10 x = -7
We can not accept -7 as a value for the base of the triangle. Therefore the
base of the triangle is 10 cm while the altitude is x - 7 = 7 cm.
2. The perimeter of a rectangle is 138 meters and the area is 1080 square
meters. Find the length and the width of the rectangle.
Solution:
1st
Let x cm = length of the rectangle
138 cm = Perimeter
P = 2l + 2w
138 = 2x + 2w express perimeter in terms of x.
138 = 2(x + w)
69 = x + w
w = 69 – x the width
Area of a rectangle = length x width
2nd
1080 = x ( 69 – x )
3rd
x ( 69 - x ) = 1080
69x - x2
= 1080
4th
x2
- 69x + 1080 = 0
5th
( x – 45) ( x - 24 ) = 0

10
6th
step: Solve for x.
x – 45 = 0 and x - 24 = 0
x = 45 x = 24
x = 45 m is the length of the rectangle
69 – 45 = 24 m is the width
Perimeter = 2( 45 ) + 2(24) = 138 m
Area = 45( 24 ) = 1080 square meters
3. If the hypotenuse of a right triangle measures 13 cm and one leg is 7 cm
more than the other, what are the lengths of the two legs?
Solution:
1st
Let x cm = length of shorter leg
(x + 7 ) cm = length of other leg
Formula: a2
+ b2
= c2
2nd
x2
+ ( x + 7 )2
= 133
3rd
x2
+ x2
+ 14x + 49 = 169
2x2
+ 14x + 49 = 169
4th
2x2
+ 14x + 49 - 169 = 0
2x2
+ 14x - 120 = 0
simplify by dividing by 2
x2
+ 7x - 60 = 0
5th
( x + 12 ) ( x - 5 ) = 0

11
6th
step: Solve for x.
x + 12 = 0 and x - 5 = 0
x = -12 x = 5
Disregard -12 as a solution because it has no real meaning. You cannot
have a length of negative 12.
x = 5 m is the length of of shorter leg
x + 7 = 12 m is the length of the other leg
Check: a2
+ b2
= c2
52
+ (5 + 7)2
= 132
25 + 144 = 169
169 = 169
Try this out
Solve:
1. The length of a rectangular floor is twice the width. The area of the floor is
32 m2
. What are the dimensions of the room?
2. If the perimeter of a rectangular garden is 76m and the area is 360 square
meters, what are the dimensions of the garden?
3. The base of a triangle is 7 cm more than its height. If the area is 294
square centimeters, find the base and the height of the triangle.
4. A square pool was surrounded by 3 meters wide Bermuda grass. If the
total area of the sidewalk and the square pool is 196 sq. m , how long is
each side of the pool? What is the area of the Bermuda grass sidewalk?
5. A rectangular flower garden with dimensions 3m by 7m is surrounded by
the walk of uniform width. If the area of the walk is 11 sq. m, what is the
width in meters?

12
Lesson 3
Motion Problems
Motion problems deal with three quantities. They are:
Distance
Rate or speed
Time
All uniform motion problems are tied-up with the formula:
Distance = Time x Rate
or D = rt
Examples:
1. A car travels 10 kilometers per hour faster than a truck. The car goes 600
kilometers in 5 hours less time than it takes the truck to travel the same
distance. Find the rate of each vehicle in kilometers per hour.
Solution:
1st
Rate Time Distance
Truck x
x
600
600
Car x + 10
10
600
+x
600
2nd
x
600
=
10
600
+x
+ 5
3rd
Multiply both sides by the least common denominator which is
x( x + 10 )
600( x + 10 ) = 600x + 5( x2
+ 10x )

13
600x + 6000 = 600x + 5x2
+ 50x
4th
Thus:
5x2
+ 600x - 600x + 50x - 6000 = 0
Simplify by dividing the equation by 5
x2
+ 10x - 1200 = 0
5th
(x + 40) (x - 30) = 0
6th
step: Solve for x.
x + 40 = 0 and x – 30 = 0
x = -40 x = 30
We must disregard the negative value. Thus, the rate of the truck is 30
kilometers per hour. The rate of the car is x + 10 = 40 kilometers per hour.
2. A man drives 500 km to a business convention. On the return trip, he
increases his speed by 25 km per hour and saves 1 hour of driving time.
How fast did he go in each direction?
Solution:
1st
Let r = speed in going to the convention
r + 25 = speed of the return trip
r
500
= length of time to the convention
25
500
+r
= length of time in returning from the convention

14
(length of time in going) = (length of time in returning) + 1
2nd
r
500
=
25
500
+r
+ 1
3rd
r( r + 25 )
500( r+ 25 ) = 500r + r( r + 25 )
500r + 12, 500 = 500r + r2
+ 25r
500r + 12,500 = 525r + r2
4th
Thus:
0 = r2
+ 525r - 500r - 12,500
5th
0 = ( r + 125 ) (r - 100 )
6th
step: Solve for x.
r + 125 = 0 or r – 100 = 0
r = -125 r = 100
We must disregard the negative value. Thus, the rate of the man is 100
kilometers per hour in going to the convention and returns at 100 + 25 = 125
kilometers per hour.
2. A man can row 8 kilometers downstream and back in 6 hours. If the rate of
the stream is 1 kilometer per hour, what is the rate of the man rowing in
still water?

15
Solution:
1st
Let x = rate of the man in still water
D r t
Downstream 8 x + 1
1
8
+x
Upstream 8 x - 1
1
8
−x
2nd
1
8
+x
+
1
8
−x
= 6
3rd
( x + 1) (x - 1)
8(x – 1) + 8(x + 1) = 6(x + 1(x – 1)
8x – 8 + 8x + 8 = 6(x2
– 1)
16x = 6x2
– 6
4th
Thus:
0 = 6x2
- 16x - 6
Simplify by dividing the equation by 2
0 = 3x2
- 8x - 3
5th
0 = (3x + 1)( x - 3)

16
6th
step: Solve for x.
3x + 1 = 0 or x – 3 = 0
x =
3
1−
x = 3
We must disregard the negative value. Thus, the rate of the man is 3
kilometers per hour.
Try this out
Answer the following problems:
1. Two cyclist A and B traveled the same distance of 120 kilometers. A
traveled 4 kilometers per hour faster than B and covered the distance in
one hour less than B. Find the rate of each.
2. If Mr. Cruz will increase his usual rate by 10 kilometers per hour, it will take
him one hour shorter to cover a distance of 200 kilometers. What is his
usual speed in driving?
3. If you can row upstream to a landing 5 kilometers away and then row back
to your starting point all in three hours and 20 minutes and if the river has
an average current of 2 kilometers per hour, at what rate are you able to
row in still water?
4. A horse travels 30 kilometers per hour faster than a mule. The horse goes
360 kilometers in two hours less time than the mule goes 360 kilometers.
Find the speed of each animals?
5. An airplane flies 900 miles against a headwind of 25 miles per hour. The
plane took 15 minutes longer for this flight than with a tailwind of 25 miles
per hour. How fast could the plane fly in still air?

17
Lesson 4
Work Problems
Work problems are important not only to students but also to industrialists,
engineers, production executives and managers as well. This is because
everyone often meets work problems in real life.
Common work problems have the following characteristics
1. They ask you how long individual can finish a job alone, when working with
others when his rate of working is given; and
2. They ask you how long a group of workers can finish a job, if rate of work
of each member of the group is given.
Examples:
1. Mario takes 5 days more to do a job than Jose. Together they can do it in
6 days. How long does it take each alone to do the job?
Solution:
1st
No. of days to do job
Amount of work done in 1
day
Jose x
x
1
Mario x + 5
5
1
+x
Together 6
6
1
2nd
x
1
+
5
1
+x
=
6
1

18
3rd
Multiply both sides by the least common denominator, 6x(x + 5).
6( x + 5 ) + 6x = x( x + 5 )
6x + 30 + 6x = x2
+ 5x
4th
Thus:
0 = x2
+ 5x - 12x - 30
0 = x2
- 7x - 30
5th
0 = ( x + 3 ) ( x - 10 )
6th
step: Solve for x.
x + 3 = 0 or x – 10 = 0
x = -3 x = 10
The solutions of the fractional equation are 10 and -3.
However, the solution of the problem is x = 10 and x stands for the
number of days it takes Jose to do the work alone x + 5 = 15 is the number of
days it takes Mario to do the work alone.
2. tank has a supply pipe and an exhaust pipe. The exhaust pipe takes 5
minutes longer to empty the tank than for the supply pipe to fill it. If both
are open, it takes the supply pipe 30 minutes to fill the tank. Find how long
it takes the supply pipe to fill the tank when the exhaust pipe is closed.
Solution:
1st
Let x = no. of minutes to fill in the tank
x
1
= part of the tank filled in one minute

19
x + 5 = no. of minutes to empty the tank
5
1
+x
= part of the tank emptied in one minute
It takes 30 minutes to fill the tank with both pipes open
2nd
x
30
-
5
30
+x
= 1
3rd
Multiply both sides by the least common denominator which is x(x + 5).
30( x+ 5 ) - 30r = x( x + 5 )
30x + 150 - 30x = x2
+ 5x
4th
Thus:
0 = x2
+ 5x - 150
5th
0 = ( x + 15 ) ( x - 10 )
6th
step: Solve for x.
x + 15 = 0 or x – 10 = 0
x = -15 x = 10
It takes the supply pipe 10 minutes to fill the tank when the exhaust pipe is
closed.

20
Try this out
1. Charlie can construct 100 identical boxes in two days less than Cholo.
If they work together they can finish the work in 2
12
11
days. How long
will it take each of them to finish 100 boxes?
2. It takes Peter 6 hours longer than Luis to do a certain job. Together
they can do it in 4 hours. How long would it take each working alone to
do the job?
3. Mang Tano and his son Mario can finish planting their crops in 4 days if
they work together. Working alone, Mario will need 2
3
1
days more than
the number of days that it will take his father to finish the work alone.
How many days will each of the father and son need to finish the work if
each will work alone?
4. Mr. Cruz can paint a house in 2 days less than his son. When they work
together they can do the job in 4
9
4
days. How long would it take each
working alone to do job?
Lesson 5
Problems Involving Quadratic Function
(Maximum or Minimum Point)
Quadratic functions have practical applications. Many applications of
quadratic functions are in simple maximization and minimization problems.
The technique for solving such problems is based on the fact that a
quadratic function defined by f(x) = ax2
+ bx + c attains its maximum value

21
( for a 〈 0 ) or minimum value ( for a 〉 0 ) at x = h = -
a
b
2
and the maximum /
minimum value is k = f(h) =
a
bac
4
4 2
−
.
Examples:
1. The sum of two numbers is 24. Find the maximum product and the value
of the two numbers.
Solution:
Let x = the first number
24 – x = the second number
f(x) = the product of the two numbers
y = f(x) = x(24 – x)
= 24x – x2
or y = -x2
+ 24x attains its maximum value at
x =
a
b
2
−
=
)1(2
24
−
−
= 12
The maximum product is 12(24 – 12) = 144
2. A rectangular garden is x meters long and (18 – x) meters wide. If y is the
area, find the maximum area that the garden could have. What could be
its dimensions?
Solution:
Let x = the length
18 – x = width
y(area) = x(18 – x)

22
= 18x – x2
or y = -x2
+ 18x attains its maximum value
at
x =
a
b
2
−
=
2
18
−
−
= 9
The maximum area is 9(18 – 9) = 18 square meters and the dimensions
are length = 9 m and the width = 9 m.
3. A rectangular lot is bounded on the side by a river
and on the other three sides by a total of 80 m fencing.
Find the possible dimensions of the lot.
Solution:
Step 1. The quantity to be maximized is the area of the lot.
Step 2. Let x = be the width
y = be the length
If A is the area, then A = xy
Step 3. Since the total length of fencing is 80 meters,
2x + y = 80 or y = 80 – 2x
Substitute 80 – 2x for y in A = xy
A = x(80 – 2x) = -2x2
+ 80x
Step 4. x =
a
b
2
−
=
)2(2
80
−
−
=
4
80
−
−
= 20
Area = -2x2
+ 80x
= -2(20)2
+ 80(20)
= -800 + 1600
= 800
Then maximum value of A is 800 and occurs when x = 20.
Step 5. Since x = 20 and 2x + y = 80, y = 40. The dimensions of the
largest possible rectangular lot are 40 meters by 20 meters.

23
4. A company charges P200 for each leather bag on order of 150 or less.
The cost of each bag is reduced by P1 for each order in excess of 150.
How many bags on order would result in a maximum revenue? what is the
maximum revenue?
Solution:
Step 1. The quantity to be maximized is total revenue (R) in pesos from the
leather bag to be sold.
Step 2. Let x = be the number of bags ordered
x - 150 = be number of bag ordered in excess of 150
200 – 1(x – 150) = 200 – x + 150
= 350 – x is the cost of each bag
R = (cost per bag) (number of bags sold )
= (350 - x )x
= 350x - x2
or -x2
+ 350x
Step 3. x =
a
b
2
−
=
)1(2
350
−
−
=
2
350
−
−
= 175
R = -x2
+ 350x
= -( 175 )2
+ 350( 175 )
= 30,625
The maximum value of R is 30,625 and it occurs when x = 175

24
Step 5. A purchase of 175 bags maximizes the company’s revenue.
The maximum revenue is P30,625
Try this out
1. What are the dimensions of the largest rectangular field that can be
enclosed with 60 meters of wire?
2. The sum of two numbers is 40. If one is x, what is the other? What is the
maximum product that the 2 numbers could have?
3. A rectangular lot is bordered on one side by a string and on the other
three sides by 600 meters of fencing. Find the dimensions of the lot if its
area is a maximum.
4. A theater seats 2000 people and charges P10 for a ticket. At this price, all
the tickets can be sold. A survey indicates that if the ticket price is
increased by a peso, the number of tickets sold will decrease by 100.
What ticket price result in the greatest revenue?
Let’s summarize
Steps in Solving word problems involving quadratic equation and
function:
1. Identify your unknowns.
2. Write your equation.
3. Operate on the numbers.
4. Convert the equation into the standard form. Remember to change signs.
5. Factor the equation into its two corresponding linear equations. By
factoring, you reduce the quadratic equation into two equivalent linear
equations.

25
What have you learned
A. Define a variable and write a quadratic function to describe each of the
following.
1. The product of two numbers whose sum is 40.
2. The product of two numbers whose difference is 25.
3. The area of a rectangle whose perimeter is 20 centimeter.
4. The product of two consecutive integers is nine less than the square
of the second.
5. Write a quadratic function to describe the area of a circle in terms of
its radius.
B. Answer each of the following problems completely.
1. Find two consecutive even integers whose product is 168.
2. The sum of two numbers is 11. The difference of their squares is 11.
What are the numbers?
3. The perimeter of a garden is 66 meters and its area is 270 sq.
meters. Find the dimensions of the garden.
4. What two integers having a sum equal to 66 will have a maximum
product?
5. Two bikers started at the same corner, one going east, the other
going north. One biker is traveling at 3 kph faster than the other.
After one hour, the two bikers are 15 kilometers apart. Find the rate
of each.

26
Answer Key
How much do you know
A. 1. b B. 1. A = s2
C. 1. 11 and 13 or -13 and -11
2. d 2. P = 2A 2. l = 16 cm. w = 6 cm
3. c 3. 12x + 7 = 3x – 4 3. 9 and 13
4. b 4. c = 2π 4. 144
5. d 5. d = rt
Try this out
Lesson 1
1. x = 3 and x + 1 = 4 6.
3
5−
and
5
3
2. 14 and 16 7. 4 and 9
3. n = 9 8. -7 and -4 or 4 and 7
4. 5 and 15 9. 7 and 8
5. 5 and 11 10. n = 29
Lesson 2
1. width = 4 meters and length = 8 meters
2. length = 20meters and width = 18 meters
3. base = 28 meters height = 21 meters
4. The length of each side of the pool = 8 meters
area of the Bermuda grass sidewalk = 132 square meters
5.
2
1
m is the width of the walk

27
Lesson 3
1. B = 20kph, 2. The speed of Mr Cruz is 40 kph
A = 24 kph
3. The rate in still water 4. Rate of mule = 30 kph,
= 4 km/hr Rate of horse = 60 kph
5. r = 425 kph
Lesson 4
1. Charlie = 5 days, 2. Luis = 6 hours
Cholo = 7 days Peter = 12 hours
3. Mario = 9
3
1
days 4. son = 10 days
Mang Tano = 7 days Mr. Cruz = 8 days
Lesson 5
1. width = 15 meters length = 15 meters
2. 20 is the other number; 400 is the maximum product
3. length = 300m width = 150 meters
4. P25 ticket price would result to P22,500 greatest revenue.
What have you learned
A. 1. y = x( 40 – x ) 2. y = x( x – 25 )
3. A = l( 10 – l ) 4. P = x( x + 1 ) = ( x + 1 )2
- 9 5. A = πr2
B. 1. -14 and -12 or 12 and 14
2. 6 and 5
3. length = 18 meters and width = 15 meters
4. the maximum product is 1089, the numbers are 33 and 33.
5. rate of slower biker = 9 kph
rate of faster biker = 12 kph

Module 4 quadratic functions

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Module 4 quadratic functions