REAL LIFE APPLICATION QUADRATIC FUNCTIONS - SOLVED EXAMPLES AND EXERCISE

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Quadratic Formula before Real Life Application Quadratic Functions

if you have not already done so.

That knowledge is a prerequisite here.



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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
Real Life Application Quadratic Functions
is solved.





Example 2 of Real Life Application Quadratic Functions

Solve the Following Problem from
Real Life Application Quadratic Functions


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
Real Life Application Quadratic Functions
is solved.

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Example 3 of Real Life Application Quadratic Functions

Solve the Following Problem from
Real Life Application Quadratic Functions


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.

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Exercise on Real Life Application Quadratic Functions

Solve the following problems from
Real Life Application Quadratic Functions.

  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
Real Life Application Quadratic Functions
See at the bottom of the Page.





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Anwers to Exercise on Real Life Application Quadratic Functions

Answers to the problems from
Real Life Application Quadratic Functions.

  1. 10 m
  2. 3⁄7
  3. 4 kmph.