# WORD PROBLEMS ON QUADRATIC EQUATIONS 1

Word Problems on Quadratic Equations 1 :

This is the continuity of our content on word problems on quadratic equations.

In this section, we will learn how to solve quadratic equation word problems step by step.

## Word Problems on Quadratic Equations 1

Problem 1 :

The area of a rectangular field is 2000 sq.m 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.

Then, we have

2x + 2y  =  180

Divide both sides by 2.

x + y  =  90

Subtract x from both sides.

y  =  90 - x

Formula to find the area of a rectangle is

Given : Area of the rectangle is 2000 sq.m

That is,

length ⋅ width  =  2000

⋅ y  =  2000

Plug y  =  90 - 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

Hence, 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 be zero, the demand has to be 5000 bottles.

To know the price for zero inventory, we have to plug 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.

Hence, 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

2p² + 10p - 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

Hence, 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.

So, we have

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.

Hence, 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.

That is,

length ⋅ width  =  192

(4x / 3) ⋅ x  =  192

4x² / 3  =  192

Multiply both sides by 3/4.

x²  =  144

Take radical on both sides.

x  =  12

So, the width is 12 inches.

Then, the height is

=  4(12) / 3

=  16 inches

Hence, the dimensions of the picture are 12 inches and 16 inches. Apart from the problems given above, if you need more word problems on quadratic equations

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