SOLVING WORD PROBLEMS INVOLVING QUADRATIC EQUATIONS

Problem 1 :

If the difference between a number and its reciprocal is ²⁴⁄₅, find the number.

Solution :

Let x be the required number. Then its reciprocal is ¹⁄ₓ be its reciprocal.

5(x² - 1) = 24x

5x² - 5 = 24x

5x² - 24x - 5 = 0

5x² - 25x + 1x - 5 = 0

5x(x - 5) + 1(x - 5) = 0

(5x + 1)(x - 5) = 0

5x + 1 = 0  or  x - 5 = 0

x = -⅕  or  x = 5

So, the required number is -⅕ or -5.

Problem 2 :

A garden measuring 12m by 16m is to have a pedestrian pathway that is w meters wide installed all the way around so that it increases the total area to 285 m². What is the width of the pathway?

Solution :

From the picture above, length of the garden including pathway is (12 + 2w) and width is (16 + 2w).

Total Area = 285 m²

Length ⋅ Width = 285

(12 + 2w)(16 + 2w) = 285

  192 + 24w + 32w + 4w² = 285

 4w² + 56w + 192 - 285 = 0

4w² + 56w - 93 = 0

Solve the above quadratic equation using quadratic formula.

w = -15.5  or  w = 1.5 

Since w is the width of the pathway, it can not be negative. So, w = 1.5.

Therefore , the width of the pathway is 1.5 m.

Problem 3 :

A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the bus.

Solution :

Distance covered = 90 km

Let x be the original speed  of the bus.

Increased speed = x + 15

Time taken by the bus in original speed :

= ⁹⁰⁄ₓ

Time taken by the bus in increased speed :

= ⁹⁰⁄₍ₓ ₊ ₁₅₎

From the given information, the difference between and ⁹⁰⁄ₓ and ⁹⁰⁄₍ₓ ₊ ₁₅₎ is 30 minutes or ½ hour.

2700 = x² + 15x 

x² + 15x - 2700 = 0

x² + 60x - 45x - 2700 = 0

x(x + 60) - 45(x + 60) = 0

(x - 45)(x + 60) = 0

x = 45  or  x = -60

Since x represents the original speed of the bus, it can never be negative. So, x = 45.

Therefore, the original speed of the bus is 45 km per hour.

Problem 4 :

John and Jivanti together had 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they have now is 124. Find the number marbles each one them had initially.

Solution :

Let x be the number of marbles that john has.

Then, the number of marbles that Jivanthi has

= 45 - x

After losing 5 marbles, 

number of marbles with John = x - 5

number of marbles with Jivanti = 45 - x - 5 = 40 - x

The product of number of marbles = 124

(x - 5)(40 - x) = 124

40x - x² - 200 + 5x = 124

45x - x² - 200 = 124

x² - 45x + 124 + 200 = 0

x² - 45x + 324 = 0

x²- 36x - 9x + 324 = 0

x(x - 36) - 9(x - 36) = 0

(x - 9)(x - 36) = 0

x - 9 = 0  or  x - 36 = 0

x = 9  or  x = 36

If x = 9, 

= 45 - x

= 45 - 9

= 36

If x = 36, 

= 45 - x

= 45 - 36

= 9

So, John had 36 marbles, when Jivanti had 9 marbles or Jivanti had 36 marbles, when John had 9 marbles. 

Problem 5 :

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in dollars) was found to be 55 minus the number of toys produced in a day, the total cost of production was $750. Find the number of toys produced on that day.

Solution :

Let x be the number of toys produced in a particular day.

Cost of production  of one toy (in dollars) :

= 55 - x

Total cost =  Number of toys ⋅ Cost of one toy

750 = x (55 - x)

750 = 55x - x²

x² - 55x + 750 = 0

x² - 30x - 25x + 750 = 0

x(x - 30) - 25(x - 30) = 0

(x - 30)(x - 25) = 0

  x - 30 = 0  or  x - 25 = 0

   x = 30  or  x = 25

So, the number of toys produced on that particular day is 30 or 25.

Problem 6 :

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Solution :

Let x be the length of shorter side of the rectangle

Then,

length of diagonal = x + 60

length of longer side = x + 30

Using Pythagorean Theorem in the right triangle ABC,

(x + 60)² = x² + (x + 30)²

x² + 60² + 2(x)(60) = x² + x² + 2(x)(30) + 30²

x² + 3600 + 120x = 2x² + 60x + 900

2x² - x² + 60x - 120x + 900 - 3600 = 0

x² - 60x - 2700 = 0

x² - 90x + 30x - 2700 = 0

x(x - 90) + 30(x - 90) = 0

 (x + 30)(x - 90) = 0

  x + 30 = 0  or  x - 90 = 0

   x = -30  or  x = 90

Because x represents breadth of the rectangle, it can never be negative. So, x = 90.

Therefore, 

length of the shorter side = 90 m

length of the longer side = 90 + 30 = 120 m

Problem 7 :

The difference of squares of two positive numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution :

Let x be the larger number and y be the smaller number.

The square of the smaller number is 8 times the larger number.

y² = 8x -----(1)

The difference of squares of two numbers is 180.

 x² - y² = 180

Substitute y² = 8x and solve for x.

  x² - 8x = 180

x² - 8x - 180 = 0

  x² - 18x + 10x - 180 = 0

x(x - 18) + 10(x - 18) = 0

 (x - 18)(x + 10) = 0 

 x - 18 = 0  or  x + 10 = 0

 x = 18  or  x = -10

Because the numbers are positive, x = -10 can not be accepted. So, x = 18.

Substitute x = 18 into (1).

y² = 8(18)

y² = 144

 y = √144

 y = 12

Therefore, the larger number is 18 and the smaller number is 12.

Problem 8 :

A train travels 360 miles at a uniform speed. If the speed had been 5 miles/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution :

Distance covered = 360 miles

Let x be the original speed of the train.

If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey.

Increased speed = x + 5

Time taken by the train in original speed :

= ³⁶⁰⁄ₓ

Time taken by the train in increased speed :

= ³⁶⁰⁄₍ₓ ₊ ₅₎

From the given information, the difference between and ³⁶⁰⁄ₓ and ³⁶⁰⁄₍ₓ ₊ ₅₎ is 1 hour.

1800 = x² + 5x

x² + 5x - 1800 = 0

x² - 40x + 45x - 1800 = 0

x(x - 40) + 45(x - 40) = 0

(x - 40)(x + 45) = 0

 x - 40 = 0  or  x + 45 = 0

 x = 40  or  x = -45

Since x represents the original speed of the train, it can never be negative. So, x = 40.

Therefore, the original speed of the train is 40 miles/hr.

Problem 9 :

Find two consecutive positive even integers whose squares have the sum 340.

Solution :

Let x and (x + 2) be the two positive even integers.

Sum of their squares = 340

x² + (x + 2)² = 340

x² + (x + 2)(x + 2) = 340

x² + x² + 2x + 2x + 4 = 340

2x² + 4x + 4 = 340

2x² + 4x - 336 = 0

Divide both sides by 2.

x² + 2x - 168 = 0

Solve by factoring.

x² - 12x + 14x - 168 = 0

x(x - 12) + 14(x - 12) = 0

(x - 12)(x + 14) = 0

x - 12 = 0  or  x + 14 = 0

x = 12  or  x = -14

Since the integers are positive, x can not be negative.

So,

x = 12

x + 2 = 14

Therefore, the two positive even integers are 12 and 14.

Problem 10 :

The sum of squares of three consecutive natural numbers is 194. Determine the numbers.

Solution :

Let x, x + 1 and x + 2 be three consecutive natural numbers.

Sum of their squares = 194

x² + (x + 1)² + (x + 2)² = 194

x² + (x + 1)(x + 1) + (x + 2)(x + 2) = 194

x² + x² + x + x + 1 + x² + 2x + 2x + 4 = 194

3x² + 6x + 5 = 194

 3x² + 6x - 189 = 0

Divide both sides by 3.

x² + 2x - 63 = 0

Solve by factoring.

x² - 7x + 9x - 63 = 0

x(x - 7) + 9(x - 7) = 0

(x - 7)(x + 9) = 0

x - 7 = 0  or  x + 9 = 0

x = 7  or  x = -9

Since the numbers are natural numbers, x can not be negative.

So,

x = 7

x + 1 = 8

x + 2 = 9

Therefore, the three consecutive natural numbers are 7, 8 and 9.

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