A polynomial expression with the degree as two is called a quadratic equation. In the previous articles, we have studied the methods of solving quadratic equations. Now in this article, we are going to solve word problems using the quadratic formula. The formula to solve the quadratic equation is x = [-b ± √(b² – 4ac)]/2a. Here you can know the simple techniques to solve quadratic equations.
Word Problems Involving Quadratic Formula with Answers
Before solving the quadratic equation we have to write the expression in the standard form i.e., ax² + bx + c = 0. Learn how to solve the word problems related to quadratic equations using the quadratic formula from this page.
Example 1.
The length of a rectangle is 3 meters more than the width. The area of the rectangle is 70 square meters. Find the dimensions of the rectangle.
Solution:
Given that,
The length of a rectangle is 3 meters more than the width.
The area of the rectangle is 70 square meters.
Let the width be x meters
Let the length be (x + 3) meters
We know that,
Area of the rectangle = length × width
70 sq. meters = x × (x + 3)
x² + 3x = 70
x² + 3x – 70 = 0
x² + 10x – 7x – 70 = 0
x(x + 10) – 7(x + 10) = 0
(x + 10)(x – 7) = 0
x + 10 = 0
x = -10
x – 7 = 0
x = 7
In Quadratic Formula Method:
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 1, b = 3, c = -70
x = [-3 ± √((-3)² – 4(1)(-70))]/2(1)
x = [-3 ± √(9 + 280)]/2
x = [-3 ± √(289)]/2
x = [-3 ± 17]/2
x = [-3 + 17]/2 = 14/2 = 7
x = 7
x = [-3 – 17]/2
x = [-20]/2
x = -10
Measurement cannot be a negative number thus the width is 7 meters
length = x + 3 = 7 + 3 = 10 meters
Example 2.
A plot of land for sale has a width of x meters., and a length that is 8 meters less than its width. A farmer will only purchase the land if it measures 240 square meters. What value of x will cause the farmer to purchase the land?
Solution:
Given,
A plot of land for sale has a width of x meters and a length that is 8 meters less than its width.
A farmer will only purchase the land if it measures 240 square meters.
x(x – 8) = 240
x² -8x = 240
x² – 8x – 240 = 0
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 1, b = -8, c = -240
x = [8 ± √((-8)² – 4(1)(-240))]/2(1)
x = [8 ± √(64 + 960)]/2
x = [8 ± √(1024)]/2
x = [8 ± 32]/2
x = [8 + 32]/2 = 40/2 = 20
x = 20
x = [8 – 32]/2
x = [-24]/2
x = -12
Measurement cannot be a negative number thus the width is 20 meters.
x – 8 = 20 – 8 = 12 meters.
Example 3.
The product of two consecutive odd numbers is 99. Find the unknown number.
Solution:
Given,
The product of two consecutive odd numbers is 99.
Let the two consecutive odd numbers be x, x + 2
x(x + 2) = 99
x² + 2x – 99 = 0
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 1, b = 2, c = -99
x = [-2 ± √((2)² – 4(1)(-99))]/2(1)
x = [-2 ± √(4 + 396)]/2
x = [-2 ± √(400)]/2
x = [-2 ± 20]/2
x = [-2 + 20]/2 = 18/2 = 9
x = 9
x = [-2 – 20]/2
x = [-22]/2
x = -11
Thus the unknown numbers are 9 and -11.
Example 4.
The square of a number exceeds the number by 72. Find the number.
Solution:
4(x² – x – 72) = 0
4x² – 4x – 288 = 0
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 4, b = -4, c = -288
x = [-(-4) ± √((-4)² – 4(4)(-288))]/2(4)
x = [4 ± √(16 + 4608)]/8
x = [4 ± √(4624)]/8
x = [4 ± 68]/8
x = [72]/8 = 9
x = 9
x = [4 – 68]/8
x = [-64]/8
x = -8
Thus the unknown numbers are 9 and -8
Example 5.
The length of the rectangle is 4cm more than the width. The area is 96cm². Find the dimensions of the rectangle.
Solution:
Given,
The length of the rectangle is 4cm more than the width.
The area is 96cm².
Let the width of the rectangle be x
Length = x + 4
We know that,
Area of the rectangle = length × width
96 sq. cm = x(x + 4)
x² + 4x = 96
x² + 4x – 96 = 0
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 1, b = 4, c = -96
x = [-4 ± √((4)² – 4(1)(-96))]/2(1)
x = [-4 ± √(16 + 384)]/2
x = [-4 ± √(400)]/2
x = [-4 ± 20]/2
x = [-4 + 20]/2 = 16/2 = 8
x = 8
x = [-4 – 20]/2
x = [-24]/2
x = -12
Measurement cannot be a negative number thus the width is 8 centimeters.
Length = x + 4 = 8 + 4 = 12 centimeters
Example 6.
If the sides of the square are increased by 3 cm, the area of the square becomes 64 sq. cm. Find the length of the original square.
Solution:
Given that,
If the sides of the square are increased by 3 cm.
the area of the square becomes 64 sq. cm.
Let the unknown side be x.
(x + 3)² = 64
x² + 6x + 9 = 64
x² + 6x + 9 – 64
x² + 6x – 55 = 0
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 1, b = 6, c = -55
x = [-6 ± √((6)² – 4(1)(-55))]/2(1)
x = [-6 ± √(36 + 220)]/2
x = [-6 ± √(256)]/2
x = [-6 ± 16]/2
x = [-6 + 16]/2 = 10/2 = 5
x = 5
x = [-6 – 16]/2
x = [-22]/2
x = -11
Measurement cannot be a negative number.
Thus the length of the original square is 5 + 3 = 8 centimeters.
Example 7.
The height of the triangle is 5 less than its base. The area of the triangle is 42 in². Find the height and base of the triangle.
Solution:
Given,
The height of the triangle is 5 less than its base.
The area of the triangle is 42 in².
Let the b be base
height b – 5
The area of the triangle is 1/2 × b × h
42 sq. in = 1/2 × b × (b – 5)
84 sq. in = b² – 5b
b² – 5b = 84
b² – 5b – 84 = 0
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 1, b = -5, c = -84
x = [5 ± √((-5)² – 4(1)(-84))]/2(1)
x = [5 ± √(25 + 336)]/2
x = [5 ± √(361)]/2
x = [5 ± 19]/2
x = [5 + 19]/2 = 24/2 = 12
x = 12
x = [5 – 19]/2
x = [-14]/2
x = -7
b = 12 in
h = b – 5 = 12 – 5 = 7
Thus the base and height of the triangle are 12 in and 5 in.
Example 8.
Solve the quadratic equation x² + 8x + 16 = 0 by using quadratic formula.
Solution:
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 1, b = 8, c = 16
x = [-8 ± √((8)² – 4(1)(16))]/2(1)
x = [-8 ± √(64 – 64)]/2
x = [-8 ± √(0)]/2
x = [-8]/2
x = -4
x = -4
Example 9.
The product of two consecutive odd integers is 24. Find the integer.
Solution:
Given,
The product of two consecutive odd integers is 24.
Let the two consecutive odd numbers be x, x + 2
x(x + 2) = 24
x² + 2x = 24
x² + 2x – 24 = 0
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 1, b = 2, c = -24
x = [-2 ± √((2)² – 4(1)(-24))]/2(1)
x = [-2 ± √(4 + 96)]/2
x = [-2 ± √(100)]/2
x = [-2 ± 10]/2
x = [-2 + 10]/2 = 8/2 = 4
x = 4
x = [-2 – 10]/2
x = [-12]/2
x = -6
Example 10.
The length of the rectangular field is 10 feet more than its width. The area of the rectangle is 144 square feet. Find the dimensions of the rectangular field.
Solution:
Given,
The length of the rectangular field is 10 feet more than its width.
The area of the rectangle is 144 square feet.
Length = x + 10
width = x
We know that,
Area of the rectangle = length × width
x(x + 10) = 144
x² + 10x = 144
x² + 10x – 144 = 0
The formula to solve the quadratic equation is,
x = [-b ± √(b² – 4ac)]/2a
a = 1, b = 10, c = -144
x = [-10 ± √((10)² – 4(1)(-144))]/2(1)
x = [-10 ± √(100 + 576)]/2
x = [-10 ± √(676)]/2
x = [-10 ± 26]/2
x = [-10 + 26]/2 = 16/2 = 8
x = 8
x = [-10 – 26]/2
x = [-36]/2
x = -18
The dimensions cannot be negative so the width is 8 feet
Length = x + 10
width = x
Length = 8 + 10 = 18 feet
Width = 8 feet
Thus the dimensions of the rectangle are 18 feet and 8 feet.