WORD PROBLEMS ON QUADRATIC EQUATIONS

Problem 1 :

The area of a rectangular field is 2000 m2 and its perimeter is 180 m. Find the length and width of the field.

Solution :

Let x and y be the length and width of the rectangular field  respectively.

Given : Perimeter of the rectangle is 180 m.

2x + 2y = 180

Divide both sides by 2.

x + y = 90

Subtract x from both sides.

y = 90 - x

Given : Area of the rectangle is 2000 m2.

length ⋅ width = 2000

⋅ y = 2000

Substitute y = 90 - x.

x(90 - x) = 2000

90x - x2 = 2000

x2 - 90x + 2000 = 0

(x - 50)(x - 40) = 0

x = 50  or  x = 40

If x = 50,

y = 90 - 50

y = 40

If x = 40,

y = 90 - 40

y = 50

(x, y) = (50, 40) or (40, 50)

So, the length and width of the rectangle are 40 m and 50 m respectively.

Problem 2 :

A distributor of apple juice has 5000 bottles in the store that it wishes to distribute in a month. From experience, it is known that demand D is given by D = -2000p2 + 2000p + 17000. Find the price per bottle that will result zero inventory.

Solution :

Stock in the store is 5000 bottles.

If the inventory is zero, the demand has to be 5000 bottles.

To know the price for zero inventory, we have to substitute D = 5000 in the quadratic equation

D = -2000p2 + 2000p + 17000

5000 = -2000p2 + 2000p + 17000

2000p2 - 2000p - 12000 = 0

Divide both sides by 2000.

p2 - p - 6 = 0

(p - 3)(p + 2) = 0

p - 3 = 0  or  p + 2 = 0

p = 3  or  p = -2

p = -2 can not be accepted. Because, price can not be negative.

So, price per bottle that will result zero inventory is $3.

Problem 3 :

Two squares have sides p cm and (p + 5) cm. The sum of their squares is 625 sq.cm. Find the sides of the squares.

Solution :

According to the question, we have

p2 + (p + 5)2 = 625

p2 + p2 + 10p + 25 = 625

2p210p - 600 = 0

p2 + 5p - 300 = 0

(p - 15)(p + 20) = 0

p = 15  or  p = -20

p = -20 can not be accepted. Because, the side of the square can not be negative.

If p = 15,

p + 5 = 15 + 5

p + 5 = 20

So, the sides of the square are 15 cm and 20 cm.

Problem 4 :

A ball is thrown upwards from a rooftop which is above from the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball h from the ground at time t seconds is given by, h = -16t2 + 64t + 80. How long will the ball take to hit the ground?

Solution :

When the ball hits the ground, height h = 0.

0 = -16t2 + 64t + 80

16t2 - 64t - 80 = 0

t2 - 4t - 5 = 0

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

t = 5  or  t = -1

t = -1 can not be accepted. Because time can never be a negative value.

So, the ball will take 5 seconds to hit the ground.

Problem 5 :

A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?

Solution :

After enlargement, let x be the width of picture.

Then, the height of the picture is

= 4x/3

(From the given information, it is very clear that the picture is in the shape of rectangle. So, the given height can be considered as length of the rectangle) 

Given : Area after enlargement is 192 sq. inches.

length ⋅ width = 192

(4x/3) ⋅ x = 192

4x2/3 = 192

Multiply both sides by 3/4. 

x= 144

Take square root on both sides. 

x = 12

So, the width is 12 inches.

Then, the height is

= 4(12)/3

= 16 inches

So, the dimensions of the picture are 12 inches and 16 inches.

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