**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.

**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

x ⋅ y = 2000

Plug 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

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 = -16t**²** + 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 = -16t**²** + 64t + 80

16t**²** - 64t - 80 = 0

t**²** - 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

4x**² **/ 3 = 192

Multiply both sides by 3/4.

x**²** = 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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