# REAL LIFE APPLICATION QUADRATIC FUNCTIONS - SOLVED EXAMPLES AND EXERCISE

if you have not already done so.

That knowledge is a prerequisite here.

Here is a collection of proven tips,
tools and techniques to turn you into
a super-achiever - even if you've never
thought of yourself as a "gifted" student.

and remember large chunks of information
with the least amount of effort.

If you apply what you read from the above
collection, you can achieve best grades without
giving up your fun, such as TV, surfing the net,
playing video games or going out with friends!

Know more about the Speed Study System.

### Example 1 :

Solve the Following Math Word Problem.

A cyclist covers a distance of 60 km in a given time. If he increaseshis speed by 2 kmph, he will cover the distance one hour before.Find the original speed of the cyclist.

Solution to Example 8 of Math Word Problems :

Let the original speed of the cyclist be x kmph.
Then, the time cyclist takes to cover a distance of 60 km = 60⁄x
If he increases his speed by 2 kmph, the time taken = 60⁄(x + 2)
By data, the second time is less than the first by 1 hour.
∴ 60⁄(x + 2) = 60⁄x - 1
Multiplying both sides with (x + 2)x, we get
60x = 60(x + 2) - 1(x + 2)x = 60x + 120 - x2 - 2x
x2 + 2x - 120 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = 2 and c = -120
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {-b ± √(b2 - 4ac)}⁄2a
= [-2 ± √{ (2)2 - 4(1)( -120)}]⁄2(1)= [-2 ± √{ 4 + 4(1)(120)}]⁄2 = [-2 ± √{4(1 + 120)}]⁄2
= [-2 ± √{4(121)}]⁄2 = [-2 ± 2(11)}]⁄2 = -1±11 = -1+11 or -1-11 = 10 or -12But x can not be negative. ∴ x = 10.
∴ The original speed of the cyclist = x kmph. = 10 kmph. Ans.

Thus the problem on
is solved.

### Example 2 of Real Life Application Quadratic Functions

Solve the Following Problem from

Two years ago, a man's age was three times the square of his son's age.In three years time, his age will be four times his son's age.Find their present ages.

Solution to Example 2 of Real Life Application Quadratic Functions :

Let the present age of the son be x years.
Two years ago, the son's age = x - 2
By data, Two years ago, a man's age was three times the square of his son's age.
⇒ Two years ago, the man's age = 3(x - 2)2
⇒ Present age of man = 3(x - 2)2 + 2
In three years time, man's age = {3(x - 2)2 + 2} + 3
In three years time, son's age = x + 3
By data, In three years time, man's age will be four times his son's age.
⇒ {3(x - 2)2 + 2} + 3 = 4(x + 3) ⇒ 3(x2 - 4x + 4) + 2 + 3 = 4x + 12
⇒ 3x2 - 12x + 12 + 5 = 4x + 12 ⇒ 3x2 - 16x + 5 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 3, b = -16 and c = 5
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {-b ± √(b2 - 4ac)}⁄2a
= [-(-16) ± √{ (-16)2 - 4(3)(5)}]⁄2(3)= [16 ± √{ 256 - 60}]⁄6 = [16 ± √{196}]⁄6
= [16 ± 14]⁄6 = (8 ± 7)⁄3
= (8+7)⁄3, (8-7)⁄3 = 15⁄3, 1⁄3 = 5, 1⁄3
But x can not be 1⁄3. ∴ x = 5.
Present age of son = 5 years. Ans.
Present age of man = 3(x - 2)2 + 2 = 3(5 - 2)2 + 2 = 3 x 9 + 2 = 29 years. Ans.

Thus the problem on
is solved.

Great Deals on School & Homeschool Curriculum Books

## Research-based personalized Math Help tutoring program : Real Life Application Quadratic Functions

Here is a resource for Solid Foundation in
Math Fundamentals from Middle thru High School.
You can check your self by the

### FREE TRIAL.

Are you spending lot of money for math tutors to your
child and still not satisfied with his/her grades ?

Do you feel that more time from the tutor and
more personalized Math Help to identify and fix
the problems faced by your child will help ?

Here is a fool proof solution I strongly recommend
and that too With a minuscule fraction of the amount
you spent on tutors with unconditional 100% money
back Guarantee, if you are not satisfied.

### SUBSCRIBE, TEST, IF NOT SATISFIED, RETURN FOR FULL REFUND

It is like having an unlimited time from an excellent Tutor.

It is an Internet-based math tutoring software program
that identifies exactly where your child needs help and
then creates a personal instruction plan tailored to your
child’s specific needs.

If your child can use a computer and access
the Internet, he or she can use the program.
And your child can access the program anytime
from any computer with Internet access.

### Unique program to help improve math skills quickly and painlessly.

There is an exclusive, Parent Information Page provides YOU
with detailed reports of your child’s progress so you can
monitor your child’s success and give them encouragement.
These Reports include

• Time spent using the program
• Assessment results
• Personalized remediation curriculum designed for your child
• Details the areas of weakness where your child needs additional help
• Provides the REASONS WHY your child missed a concept
• List of modules accessed and amount of time spent in each module
• Quiz results
• Creates reports that can be printed and used to discuss issues with your child’s teachers
These reports are created and stored in a secure section
of the program, available exclusively to you, the parent.
The section is accessed by a password that YOU create and use.
No unauthorized users can access this information.

### Personalized remediation curriculum designed for your child

Thus The features of this excellent Tutoring program are

• Using detailed testing techniques
• Identifing exactly where a student needs help.
• Its unique, smart system pinpointing precise problem areas -
• slowly and methodically guiding the student
• raising to the necessary levels to fix the problem.

### Not a “one-size-fits-all” approach!

Its research-based results have proven that
it really works for all students! in improving
math skills and a TWO LETTER GRADE INCREASE in
math test scores!,if they invest time in using
the program.

Proven for More than 10,000 U.S. public school
students who increased their math scores.

### Example 3 of Real Life Application Quadratic Functions

Solve the Following Problem from

A number consists of two digits whose product is 18. If 27 is added to the number, the digits interchange their places.Find the number.

Solution to Example 3 of Real Life Application Quadratic Functions :

Let the unit's digit of the original number be x.
By data, the product of the digits 18 ∴ The ten's digit of the original number = 18⁄xOriginal number's value = 1x unit's digit + 10 x ten's digit = 1 x x + 10 x 18⁄x = x + 180⁄xDigits reversed number value = 1 x 18⁄x + 10 x x = 18⁄x + 10x
By data, Original number + 27 = Digits reversed number
x + 180⁄x + 27 = 18⁄x + 10x
Multiplying both sides with x, we get
x2 + 180 + 27x = 18 + 10x2 ⇒ -9x2 + 27x + 162 = 0
Dividing by -9, we get
x2 - 3x - 18 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -3 and c = -18
We know by Quadratic Formula, x = {-b ± √(b2 - 4ac)}⁄2a
Applying this Quadratic Formula here, we get
x = {-b ± √(b2 - 4ac)}⁄2a
= [-(-3) ± √{ (-3)2 - 4(1)(-18)}]⁄2(1)= [3 ± √{ 9 + 72}]⁄2 = [3 ± √{81}]⁄2
= [3 ± 9]⁄2 = (3 + 9)⁄2, (3 - 9)⁄2 = 6, -3
But x can not be negative. ∴ x = 6.
unit's digit of the original number = x = 6
The ten's digit of the original number = 18⁄x = 18⁄6 = 3
∴ the required number is 36. Ans.

### Exercise on Real Life Application Quadratic Functions

Solve the following problems from

1. A rectangle is 100 m by 60 m. It has grass for certain width all around iton the outside. Area of grass is 3⁄5 of the area of the rectangle.Find the width of the grass.
2. The sum of the numerator and denominator of a certain fraction is 10.If 1 is subtracted from both the numerator and denominator, the fractionis decreased by 2⁄21. Find the fraction.
3. A man walks a distance of 48 kms. in a given time. If he walks 2 kmph faster, he will perform the journey 4 hours before. Find hisnormal rate of walking.

For Answers on these problems of
See at the bottom of the Page.

## Progressive Learning of Math : Real Life Application Quadratic Functions

Recently, I have found a series of math curricula
(Both Hard Copy and Digital Copy) developed by a Lady Teacher
who taught everyone from Pre-K students to doctoral students
and who is a Ph.D. in Mathematics Education.

This series is very different and advantageous
over many of the traditional books available.
These give students tools that other books do not.
Other books just give practice.
These teach students “tricks” and new ways to think.

These build a student’s new knowledge of concepts
from their existing knowledge.
These provide many pages of practice that gradually
increases in difficulty and provide constant review.

These also provide teachers and parents with lessons
on how to work with the child on the concepts.

The series is low to reasonably priced and include

Elementary Math curriculum

and

Algebra Curriculum.