WORD PROBLEMS ON QUADRATIC EQUATIONS

Problem 1 :

The sum of a positive number and its square is 240. Find the number.

Solution :

Let x be the positive number.

x + x² = 240

x² + x - 240 = 0

Solve by factoring.

x² - 15x + 16x - 240 = 0

x(x - 15) + 16(x - 15) = 0

(x - 15)(x + 16) = 0

x - 15 = 0  or  x + 16 = 0

x = 15  or  x = -16

Since x is a positive number, it can not be negative .

So, 

x = 15

Therefore, the number is 15.

Problem 2 :

Find two numbers whose sum is 27 and product is 182.

Solution :

Let x be one of the two numbers.

Then the other number is (27 - x).

Product of those numbers = 182

x(27 - x) = 182

27x - x² = 182

x² - 27x + 182 = 0

Solve by factoring.

x² - 14x - 13x + 182 = 0

x(x - 14) - 13(x - 14) = 0

(x - 14)(x - 13) = 0

x - 14 = 0  or  x - 13 = 0

x = 14  or  x = 13

If x = 14,

= 27 - x

= 27 - 14 

= 13

If x = 13,

= 27 - x

= 27 - 13

= 14

Therefore, the two numbers are 13 and 14.

Problem 3 :

Find two consecutive positive integers, sum of whose squares is 365.

Solution :

Let x and x + 1 be the two consecutive integers.

Sum of squares  of two consecutive integers = 365

x² + (x + 1)² = 365

x² + (x + 1)(x + 1) = 365

x² + x² + x + x + 1 = 365

2x² + 2x + 1 = 365

2x² + 2x - 364 = 0

Divide both sides by 2.

x² + x - 182 = 0

Solve by factoring.

x² - 13x + 14x - 182 = 0

x(x - 13) + 14(x - 13) = 0

(x - 13)(x + 14) = 0

x - 13 = 0  or  x + 14 = 0

 x = 13  or  x = -14

Since x is a positive integer, it can not be negative .

So, 

x = 13

x + 1 = 14

Therefore, the two consecutive integers are 13 and 14.

Problem 4 :

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

x ⋅ 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 5 :

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 6 :

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

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

Problem 7 :

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

Multiply both sides by 16.

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 8 :

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

Multiply both sides by 3.

4x² = 576

Divide both sides by 4. 

x² = 144

x² = 12²

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.

Problem 9 :

The sum of the number and its positive square root is ⁷⁄₃₆. Find the number.

Solution :

Let x be the required number

The sum of the number and its positive square root is ⁷⁄₃₆.

x + √x = ⁷⁄₃₆

√x = ⁷⁄₃₆ - x

By taking squares on both sides, we get

x = (⁷⁄₃₆ - x)²

x = (⁷⁄₃₆ - x)(⁷⁄₃₆ - x)

x = (⁷⁄₃₆)² - (⁷⁄₃₆)x - (⁷⁄₃₆)x + x²

x = (⁴⁹⁄₁₂₉₆) - 2(⁷⁄₃₆)x + x²

x = (⁴⁹⁄₁₂₉₆) - (⁷⁄₁₈)x + x²

Multiply both sides by 1296.

1296x = 49 - 504x + 1296x²

1296x² - 504x - 1296x + 49 = 0

1296x² - 1800x + 49 = 0

Use quadratic formula and solve the above quadraticc equation. 

When x = ⁴⁹⁄₃₆,

The given condition is not satisfied.

When x = ¹⁄₃₆,

The given condition is satisfied.

Therefore, the required number is ¹⁄₃₆.

Problem 10 :

The sum of two numbers is 15 and sum of their reciprocals is ³⁄₁₀. Find the numbers.

Solution :

Let x and y be the two numbers.

Given : The sum of two numbers is 15.

x + y = 15

y = 15 - x ----(1)

Given : The sum of their reciprocals is ³⁄₁₀.

By cross-multiplication,

 10(x + y) = 3xy

 10x + 10y = 3xy

Substitute y = 15 - x.

10x + 10(15 - x) = 3x(15 - x)

10x + 150 - 10x = 45x - 3x²

3x² - 45x + 150 = 0

Divide both sides by 3.

x² - 15x + 50 = 0

Solve by factoring.

x² - 10x - 5x + 50 = 0

x(x - 10) - 5(x - 10) = 0

(x - 10)(x - 5) = 0

x - 10 = 0  or  x - 5 = 0

x = 10  or  x = 5

x = 10  and  y = 5

or

x = 5  and  y = 10

Therefore, the two numbers are 5 and 10.

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