# Quadratic Equations Class 10 PDF

## Quadratic Equations Class 10 - Description

Greetings to all, Today we are going to upload the Quadratic Equations class 10  PDF to assist all the students as well as tutors. NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations containing all the solutions to the problems provided in Class 10 Maths NCERT textbook for second term exam preparations. The questions from every section are framed and solved accurately by the subject experts. NCERT Solutions for Class 10 is a detailed and step-by-step guide to all the queries of the students. The exercises present in the chapter should be dealt with with utmost sincerity if one wants to score well in the examinations. Maths is a subject that requires a good understanding and a lot of practice. The tips and tricks to solve the problems easily are also provided here. A quadratic equation in the variable x is an equation of the form ax2+ bx + c = 0, where a, b, c are real numbers, a ≠ 0. That is, ax2+ bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.
Quadratic equations arise in several situations around us. Therefore, students should give special concentration to learning the concepts related to this chapter of the latest CBSE Syllabus for 2021-22 thoroughly to excel in Class 10 Math second term examinations. NCERT Solutions help the students in learning these concepts as well as in evaluating themselves. Practicing these solutions repeatedly is bound to help the students in overcoming their shortcomings. Maths has either a correct answer or a wrong one. Therefore, it is imperative to concentrate while solving the questions to score full marks.

### Detailed Table of Chapter 4 Notes – Quadratic Equations Class 9 Notes PDF

 1 Board CBSE 2 Textbook NCERT 3 Class Class 10 4 Subject Notes 5 Chapter Chapter 4 6 Chapter Name Quadratic Equations 7 Category CBSE Revision Notes

## Quadratic Equations class 10 PDF

Exercise 4.1 Page: 7
Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x – 3)
(ii) x2 – 2x = (–2) (3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
(iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
(vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x (x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3
Solutions:
(i) Given,
(x + 1)2 = 2(x – 3)
By using the formula for (a+b)= a2+2ab+b2
⇒ x2 + 2x + 1 = 2x – 6
⇒ x2 + 7 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(ii) Given, x2 – 2x = (–2) (3 – x)
⇒ x 2x = -6 + 2x
⇒ x– 4x + 6 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3)
By multiplication
⇒ x– x – 2 = x+ 2x – 3
⇒ 3x – 1 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
(iv) Given, (x – 3)(2x +1) = x(x + 5)
By multiplication
⇒ 2x– 5x – 3 = x+ 5x
⇒  x– 10x – 3 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)
By multiplication
⇒ 2x– 7x + 3 = x+ 4x – 5
⇒ x– 11x + 8 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is quadratic equation.
(vi) Given, x2 + 3x + 1 = (x – 2)2
By using the formula for (a-b)2=a2-2ab+b2
⇒ x2 + 3x + 1 = x2 + 4 – 4x
⇒ 7x – 3 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
(vii) Given, (x + 2)3 = 2x(x2 – 1)
By using the formula for (a+b)= a3+b3+3ab(a+b)
⇒ x3 + 8 + x2 + 12x = 2x3 – 2x
⇒ x3 + 14x – 6x2 – 8 = 0
Since the above equation is not in the form of ax2 + bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(viii) Given, x3 – 4x2 – x + 1 = (x – 2)3
By using the formula for (a-b)= a3-b3-3ab(a-b)
⇒  x3 – 4x2 – x + 1 = x3 – 8 – 6x + 12x
⇒ 2x2 – 13x + 9 = 0
Since the above equation is in the form of ax2 + bx + c = 0.
Therefore, the given equation is a quadratic equation.

1. Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m2. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken
Solutions:
(i) Let us consider,
The breadth of the rectangular plot = x m
Thus, the length of the plot = (2x + 1) m.
As we know,
Area of rectangle = length × breadth = 528 m2
Putting the value of length and breadth of the plot in the formula, we get,
(2x + 1) × x = 528
⇒ 2x2 + x =528
⇒ 2x2 + x – 528 = 0
Therefore, the length and breadth of the plot, satisfies the quadratic equation, 2x2 + x – 528 = 0, which is the required representation of the problem mathematically.
(ii) Let us consider,
The first integer number = x
Thus, the next consecutive positive integer will be = x + 1
Product of two consecutive integers = x × (x +1) = 306
⇒ x+ x = 306
⇒ x+ x – 306 = 0
Therefore, the two integers x and x+1 satisfy the quadratic equation, x+ x – 306 = 0, which is the required representation of the problem mathematically.
(iii) Let us consider,
Age of Rohan’s = x  years
Therefore, as per the given question,
Rohan’s mother’s age = x + 26
After 3 years,
Age of Rohan’s = x + 3
Age of Rohan’s mother will be = x + 26 + 3 = x + 29
The product of their ages after 3 years will be equal to 360, such that
(x + 3)(x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 32x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x2 + 32x – 273 = 0, which is the required representation of the problem mathematically.
(iv) Let us consider,
The speed of train = x  km/h
And
Time taken to travel 480 km = 480/x km/hr
As per second condition, the speed of train = (x – 8) km/h
Also given, the train will take 3 hours to cover the same distance.
Therefore, time taken to travel 480 km = (480/x)+3 km/h
As we know,
Speed × Time = Distance
Therefore,
(x – 8)(480/x )+ 3 = 480
⇒ 480 + 3x – 3840/x – 24 = 480
⇒ 3x – 3840/x = 24
⇒ x– 8x – 1280 = 0
Therefore, the speed of the train, satisfies the quadratic equation, x– 8x – 1280 = 0, which is the required representation of the problem mathematically.