A polynomial expression with the highest degree as two is known as the quadratic equation. A quadratic equation has only one variable x. Satisfying the equation by finding the unknown value x is called zeros of the equation. The roots of the equation are represented by α and β. Practice the questions from Worksheet on Formation of Quadratic Equation in One Variable to understand the concept deeply. This helps you to secure better grades in the exams and also in the real word quadratic problems.
Do Read:
- Formation of Quadratic Equation in One Variable
- Roots of a Quadratic Equation
- Examples on Quadratic Equations
Formation of Quadratic Equation in One Variable Worksheet with Answers
Learn the fundamental math concepts of quadratic equations with the help of the below examples.
Example 1.
The product of two consecutive positive odd integers is 290. And form the equation?
Solution:
Given that
The product of two consecutive positive odd integers is 290.
The two consecutive odd integers are x, x + 2.
x(x + 2) = 290
x² + 2x = 290
x² + 2x – 290 = 0
Example 2.
The area of the rectangular playground is 32 meters. The length of the rectangular playground is 8 meters more than its width?
Solution:
Given that,
The area of the rectangular playground is 32 square meters.
The length of the rectangular playground is 4 meters more than its breadth.
The width of the rectangular playground is x.
Length = x + 4
We know that,
Area of the rectangle = length × breadth
x + x + 4 = 32
x(x + 4) = 32
x² + 4x – 32 = 0
x² + 8x – 4x – 32 = 0
x(x + 8) – 4(x + 8)
(x – 4) (x + 8)
x – 4 = 0
x = 4
x – 8 = 0
x = -8
The length cannot be in the negative so the length x = 4
Breath = 4 meters
Length = x + 4 = 4 + 4 = 8 meters
Thus the length of the rectangular playground is 12 meters.
Example 3.
The product of the two consecutive odd integers is 675 from the equation?
Solution:
Given that,
The product of the two consecutive odd integers is 675
Let the two consecutive odd integers be x, x + 2.
x(x + 2) = 675
x² + 2x = 675
x² + 2x – 675 = 0
Therefore, the required quadratic equation is x² + 2x – 675 = 0
Example 4.
The length of the diagonal of a rectangular area is 25 m. And the length exceeds its breadth by 5 m.
Solution:
Let the breadth of the rectangle is x m.
Therefore the length = (x – 5) m.
Given length of diagonal = 25 m
We know that the diagonal of rectangle is √ length ² + breadth ²
√(x + 5 )² + x² = 25
(x + 5 )² + x² = 625
2x² + 25 + x² – 625 = 0
2x² – 600 = 0
The required quadratic equation is 2x² – 600 = 0
Example 5.
One person brought some kgs of rice for 60 rupees. If he would get 3 kg more rice with that money, then the price of kg of rice would be less by rupee 1.
Solution:
Price quantity of sugar in kg × rate per kg.
Let the quantity be x.
Price of x kg sugar = Rs. 60
Rate = price/quantity = 60/3.
The same price quantity has increased by 3 kgs and rate has decreased by 1.
Price = quality of sugar in kg × rate per kg.
(x + 3) [(60/x) – 1] = 60
(x + 3) (60 – x)/x = 60x
x² + 30x – 180 = 0
Hence the required quadratic equation is x² + 30x – 180 = 0
Example 6.
The product of two consecutive positive odd integers is 483. And from the equation?
Solution:
Given that
The product of two consecutive positive odd integers is 483.
The two consecutive odd integers are x, x + 2.
x(x + 2) = 483
x² + 2x = 483
x² + 2x – 483 = 0
Thus the quadratic equation is x² + 2x – 483 = 0
Example 7.
The product of the two consecutive odd integers is 323 from the equation?
Solution:
Given that,
The product of the two consecutive odd integers is 323
Let the two consecutive odd integers be x, x + 2.
x(x + 2) = 323
x² + 2x = 323
x² + 2x – 323 = 0
Thus the quadratic equation is x² + 2x – 323 = 0
Example 8.
A bag seller sold a bag by purchasing it at rupees 30. The amount of his profit percentage is as much as the amount with which he bought the bag?
Solution:
Given that
The selling price = 30
The cost price = x
Profit = cost price
Cost price = (selling price – cost price/ cost price) × 100
(30 – x/x) x 100 = x
x² + 100x – 3000 = 0
Therefore the required quadratic equation is x² + 100x – 3000 = 0
Example 9.
The unit digits of a two digits number exceed its ten’s digit by 4 and the product of two digits is less by 6.
Solution:
The ten’s digit be x
Units digit = x + 3
x(x + 3) + 6 = 10x + x + 3
x² + 3x + 6 = 11x + 3
x² – 8x + 3 = 0
Therefore the required quadratic equation is x² – 8x + 3 = 0
Example 10.
The unit digits of a two digits number exceed its ten’s digit by 8 and the product of two digits is less by 16.
Solution:
The ten’s digit be x
Units digit = x + 8
x(x + 8) + 12 = 10x + x + 3
x² + 8x + 12 = 11x + 3
x² – 3x + 9 = 0
Therefore the required quadratic equation is x² – 3x + 9 = 0