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.
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
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.
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.
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?
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?
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.
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.
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