NCERT Solutions for Class 10 Maths Chapter 4
NCERT Solutions for class 10 Maths Chapter 4 Quadratic Equations all exercises in Hindi Medium as well as English medium PDF format free to download. Uttar Pradesh Board, Prayagraj also implemented NCERT Books for High School Students for the academic session 2021-2022. UP Board students also take the benefits of these solutions and study material. Download here UP Board Solutions for Class 10 Maths Chapter 4 all exercises. 10th Maths Chapter 4 solutions are online as well as download in PDF format for sesion 2021-22. Offline Apps and NCERT Solutions 2021-22 are as per Latest CBSE syllabus for class 10 2021-2022 MP Board, UP board, Gujrat board and CBSE Board exams.Download Class 10 Maths NCERT Solutions Offline Apps in Hindi or English for offline use. For any educational help, you may contact us without any hesitation. We will try to help you as soon as possible.
NCERT Solutions for class 10 Maths Chapter 4
10th Maths Chapter 4 Exercise 4.1
10th Maths Chapter 4 Exercise 4.2
10th Maths Chapter 4 Exercise 4.3
10th Maths Chapter 4 Exercise 4.4
|Class: 10||Maths (English and Hindi Medium)|
|Chaper 4:||Quadratic Equations|
10th Maths Chapter 4 Solutions
NCERT Solutions for class 10 Maths Chapter 4 are given below in PDF form or view online. Solutions are in Hindi and English Medium. Uttar Pradesh students also can download UP Board Solutions for Class 10 Maths Chapter 4 here in Hindi Medium. It is very essential to learn quadratic equations, because it have wide applications in other branches of mathematics, physics, in other subjects and also in real life situations. Download NCERT books 2021-22, revision books and solutions from the links given below.
Class 10 Maths Chapter 4 Exercise 4.1 Solutions in Videos
Class 10 Maths Chapter 4 Exercise 4.2 Solutions in Videos
Class 10 Maths Chapter 4 Exercise 4.3 Solutions in Videos
Class 10 Maths Chapter 4 Exercise 4.4 Solutions in Videos
Previous Year’s CBSE Questions
- Two marks questions
Find the roots of the quadratic equation √2 x² + 7x + 5√2 = 0. [CBSE 2017] 2. Find the value of k for which the equation x²+k(2x + k – 1) + 2 = 0 has real and equal roots. [CBSE 2017]
- Three marks questions
If the equation (1 + m² ) x² +2mcx + c² – a² = 0 has equal roots then show that c² = a² (1 + m² ).
- . Four marks questions
Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream. [CBSE 2017]
The word quadratic is derived from the Latin word “Quadratum” which means “A square figure”.
Brahmagupta (an ancient Indian Mathematician)(A.D. 598-665) gave an explicit formula to solve a quadratic equation. Later Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula, for solving a quadratic equation by the method of completing the square. An Arab mathematician Al-khwarizni(about A.D. 800) also studied quadratic equations of different types. It is believed that Babylonians were the first to solve quadratic equations. Greek mathematician Euclid developed a geometrical approach for finding lengths, which are nothing but solutions of quadratic equations.
Important Questions on Class 10 Maths Chapter 4
Check whether the following is quadratic equation: (x + 1)² = 2(x – 3)
(x + 1)² = 2(x – 3)
Simplifying the given equation, we get (x + 1)² = 2(x – 3)
⇒ x² + 2x + 1 = 2x – 6
⇒ x² + 7 = 0 or x² + 0x + 7 = 0
This is an equation of type ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.
Represent the following situation in the form of quadratic equation: The product of two consecutive positive integers is 306. We need to find the integers.
Let the first integer = x
Therefore, the second integer = x + 1
Hence, the product = x(x + 1)
According to questions,
x(x + 1) = 306
⇒ x² + x = 306
⇒ x² + x – 306 = 0
Hence, the two consecutive integers satisfies the quadratic equation x² + x – 306 = 0.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Let, Shefali’s marks in Mathematics = x
Therefore, Shefali’s marks in English = 30 – x
If she got 2 marks more in Mathematics and 3 marks less in English,
Marks in Mathematics = x + 2
Marks in English = 30 – x – 3
According to questions,
Product = (x + 2)(27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒(-x)² + 25x – 156 = 0
⇒ x² – 25x + 156 = 0
⇒ x² – 12x – 13x + 156 = 0
⇒ x(x – 12) – 13(x – 12) = 0
⇒ (x – 12)(x – 13) = 0
⇒ (x – 12) = 0 or (x – 13) = 0
Either x = 12 or x = 13
If x = 12 then, marks in Maths = 12 and marks in English = 30 – 12 = 18
If x = 13 then, marks in Maths = 13 and marks in English = 30 – 13 = 1
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Let the larger number = x
Let the smaller number = y
Therefore, y² = 8x
According to question,
x² – y² = 180
⇒ x² – 8x = 180 [As y² = 8x]
⇒ x² – 8x – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x – 18)(x + 10) = 0
⇒ (x – 18) = 0 or (x + 10) = 0
Either x = 18 or x = -10
But x ≠ -10 , as x is the larger of two numbers.
So, x = 18 Therefore, the larger number = 18
Hence, the smaller number = y = √144 = 12
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Let the side of larger square = x m
Let the side of smaller square = y m
According to question,
x² + y² = 468 …(i)
Difference between perimeters,
4x – 4y = 24
⇒ x – y = 6
⇒ x = 6 + y … (ii)
Putting the value of x in equation (i), we get
(y + 6)² + y² = 468
⇒ y² + 12y + 36 + y² = 468
⇒ (2y)² +12y – 432 = 0
⇒ y² + 6y – 216 = 0
⇒ y² + 18y – 12y – 216 = 0
⇒ y(y + 18) – 12(y + 18) = 0
⇒ (y + 18)(y – 12) = 0
⇒ (y + 18) = 0 or (y – 12) = 0
Either y = -18 or y = 12
But, y ≠ -18 , as x is the side of square, which can’t be negative.
So, y = 12
Hence, the side of smaller square = 12 m Putting the value of y in equation (ii), we get Side of larger square = x = y + 6 = 12 + 6 = 18 m
How can I get good marks in Class 10 Maths Chapter 4 Quadratic Equations?
Student should know the methods of factorization to a quadratic equation. It will help a lot during the solution of questions in 10th Maths chapter 4. Quadratic formula is the ultimate trick to find the roots of difficult or easy format of any quadratic equation. So if someone has practiced well the factorization method and quadratic formula method, he will score better then ever in chapter 4 of class 10 mathematics.
How a quadratic polynomial is different from a quadratic equation in 10th Maths Chapter 4?
A polynomial of degree two is called a quadratic polynomial. When a quadratic polynomial is equated to zero, it is called a quadratic equation. A quadratic equation of the form ax² + bx + c = 0, a > 0, where a, b, c are constants and x is a variable is called a quadratic equation in the standard format.
In Class 10 Maths Chapter 4, which exercise is considered as the most difficult to solve?
Class 10 Maths, exercise 4.1, 4.2 and 4.4 are easy to solve and having less number of questions. Exercise 4.3 is not only lengthy but the tricky to find the solutions and answers also. In this exercise most of the questions are based on application of quadratic equations.
What is meant by zeros of a quadratic equation in Chapter 4 of 10th Maths?
A zero of a polynomial is that real number, which when substituted for the variable makes the value of the polynomial zero. In case of a quadratic equation, the value of the variable for which LHS and RHS of the equation become equal is called a root or solution of the quadratic equation. There are three algebraic methods for finding the solution of a quadratic equation. These are (i) Factor Method (ii) Completing the square method and (iii) Using the Quadratic Formula.
What are the main topics to study in Class 10 Maths chapter 4?
In chapter 4 (Quadratic equations) of class 10th mathematics, Students will study
- 1) Meaning of Quadratic equations
- 2) Solution of a quadratic equation by factorization.
- 3) Solution of a quadratic equation by completing the square.
- 4) Solution of a quadratic equation using quadratic formula.
- 5) Nature of roots.
How many exercises are there in chapter 4 of Class 10th Maths?
There are in all 4 exercises in class 10 mathematics chapter 4 (Quadratic equations).
In first exercise (Ex 4.1), there are only 2 questions (Q1 having 8 parts and Q2 having 4 parts).
In second exercise (Ex 4.2), there are in all 6 questions.
In third exercise (Ex 4.3), there are in all 11 questions.
In fourth exercise (Ex 4.4), there are in all 5 questions.
So, there are total 24 questions in class 10 mathematics chapter 4 (Quadratic equations).
In this chapter there are in all 18 examples. Examples 1, 2 are based on Ex 4.1, Examples 3, 4, 5, 6 are based on Ex 4.2, Examples 7, 8, 9, 10, 11, 12, 13, 14, 15 are based on Ex 4.3, Examples 16, 17, 18 are based on Ex 4.4.
Does chapter 4 of class 10th mathematics contain optional exercise?
No, chapter 4 (Quadratic equations) of class 10th mathematics doesn’t contain any optional exercise. All the four exercises are compulsory for the exams.
How much time required to complete chapter 4 of 10th Maths?
Students need maximum 3-4 days to complete chapter 4 (Quadratic equations) of class 10th mathematics. But even after this time, revision is compulsory to retain the way to solving questions.