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Class 10 Mathematics Quadratic Equation Assignments

We have provided below free printable Class 10 Mathematics Quadratic Equation Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 10 Mathematics Quadratic Equation . These Assignments for Grade 10 Mathematics Quadratic Equation cover all important topics which can come in your standard 10 tests and examinations. Free printable Assignments for CBSE Class 10 Mathematics Quadratic Equation , school and class assignments, and practice test papers have been designed by our highly experienced class 10 faculty. You can free download CBSE NCERT printable Assignments for Mathematics Quadratic Equation Class 10 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Quadratic Equation Class 10. Students can click on the links below and download all Pdf Assignments for Mathematics Quadratic Equation class 10 for free. All latest Kendriya Vidyalaya Class 10 Mathematics Quadratic Equation Assignments with Answers and test papers are given below.

Mathematics Quadratic Equation Class 10 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 10 Mathematics Quadratic Equation . Students and teachers can download and save all free Mathematics Quadratic Equation assignments in Pdf for grade 10th. Our expert faculty have covered Class 10 important questions and answers for Mathematics Quadratic Equation as per the latest syllabus for the current academic year. All test papers and question banks for Class 10 Mathematics Quadratic Equation and CBSE Assignments for Mathematics Quadratic Equation Class 10 will be really helpful for standard 10th students to prepare for the class tests and school examinations. Class 10th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 10 Mathematics Quadratic Equation Download in Pdf

Advantages of Class 10 Mathematics Quadratic Equation Assignments

• As we have the best and largest collection of Mathematics Quadratic Equation assignments for Grade 10, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Mathematics Quadratic Equation textbooks for Class 10 .
• All Mathematics Quadratic Equation assignments for Class 10 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 10 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Mathematics Quadratic Equation chapter wise worksheets and assignments for free in Pdf
• Class 10 Mathematics Quadratic Equation question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 10 Mathematics Quadratic Equation students

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Revised NCERT Solutions for class 10 Maths Chapter 4 Quadratic Equations in Hindi and English Medium updated for board exams 2024. The question answers and explanation of chapter 4 of 10th Maths are based on NCERT textbooks published for 2024-25. Class 10 Maths Chapter 4 Solutions for CBSE Board Class 10 Maths Chapter 4 Exercise 4.1 Class 10 Maths Chapter 4 Exercise 4.2 Class 10 Maths Chapter 4 Exercise 4.3 Class 10 Maths Chapter 4 Important Questions

Class 10 Maths Chapter 4 Solutions for State Boards Class 10 Maths Exercise 4.1 Class 10 Maths Exercise 4.2 Class 10 Maths Exercise 4.3 Class 10 Maths Exercise 4.4

Get the free Hindi Medium solutions for the academic session 2024-25. Download here UP Board Solutions for Class 10 Maths Chapter 4 all exercises. 10th Maths Chapter 4 solutions are online or download in PDF format. Download Assignments for practice with Solutions 10th Maths Chapter 4 Assignment 1 10th Maths Chapter 4 Assignment 2 10th Maths Chapter 4 Assignment 3 10th Maths Chapter 4 Assignment 4

10th Maths Chapter 4 NCERT Solutions follows the latest CBSE syllabus. Students of MP, UP, Gujarat board and CBSE can use it for Board exams. Class 10 Maths NCERT Solutions Offline Apps in Hindi or English for offline use. For any scholarly help, you may contact us. We will try to help you in the best possible ways.

NCERT Solutions for class 10 Maths Chapter 4 are given below in PDF format 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 2024-25, revision books and solutions from the links given below.

Previous Year’s CBSE Questions

1. 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] 2. Three marks questions If the equation (1 + m² ) x² +2mcx + c² – a² = 0 has equal roots then show that c² = a² (1 + m²). 3. 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 to Revise 10th Maths Chapter 4 Quadratic Equations for Exams

Schools and institutions across the world promptly acted according to a pandemic, by moving online. Tech advancement helped institutions transition physical classrooms to virtual ones in record time. For as long as I can remember, I have liked everything about mathematics – especially teaching young schoolers, I have seen some of the brilliant minds grapple to comprehend the concepts. I have seen hard-working bright-eyed students losing to perform better than average in the classroom. In this article, you will read some of the effective practices that helped many students score 100% in math of 10th standard chapter Quadratic equations.

Step 1: NCERT Solutions for Class 10 Maths Chapter 4 helps to practice real life based Problems in Exercise 4.3.

Step 2: class 10 maths chapter 4 solutions provides the fundamental facts of quadratic equations., step 3: ncert solutions class 10 maths chapter 4 by applying the perfect formula for answers., step 4: class 10 maths chapter 4 needs regular practice session after short intervals., step 5: practice class 10 maths chapter 4 from ncert textbook for exams..

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, and 4.2 are easy to solve and having less number of questions. Exercise 4.3 is 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 fourth exercise (Ex 4.3), there are in all 5 questions. So, there are total 13 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 16, 17, 18 are based on Ex 4.3.

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.

« Chapter 3: Pair of Linear Equations in Two Variables

Chapter 5: arithmetic progression ».

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NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations

NCERT solutions for class 10 maths chapter 4 Quadratic Equations deal with the concept of quadratic equations and the different ways of finding their roots. A quadratic equation is represented as ax 2 + bx + c = 0 , where a, b, c are the values of real numbers, and the value of ‘a’ is not equal to zero. This is also known as the standard form of the quadratic equation. An interesting fact to note is that many people believe that Babylonians were the first to solve quadratic equations. For instance, they knew how to find two positive numbers with a given positive sum and a given positive product, and this problem is equivalent to solving a quadratic equation. Moreover, the Greek mathematician Euclid developed a geometrical approach for finding out lengths which, in our present-day terminology, are solutions of quadratic equations The NCERT solutions class 10 maths chapter 4 Quadratic Equations teaches kids how to solve these equations by the factorization method and completing the square method. Students will come across important formulas like the quadratic formula for finding the roots of the equation.

The major takeaways from this chapter would be that for a quadratic equation two real roots will be distinct, if b 2 – 4ac > 0 ; two coincident roots will be obtained, if b 2 – 4ac = 0 ; and no roots will be there, if b 2 – 4ac < 0. The students will also explore how quadratic equations can be applied to real-life situations. The pdf file of the class 10 maths NCERT Solutions Chapter 4 Quadratic Equations in detail can be found below and also you can find some of these in the exercises given below.

• NCERT Solutions Class 10 Maths Chapter 4 Ex 4.1
• NCERT Solutions Class 10 Maths Chapter 4 Ex 4.2
• NCERT Solutions Class 10 Maths Chapter 4 Ex 4.3
• NCERT Solutions Class 10 Maths Chapter 4 Ex 4.4

NCERT Solutions for Class 10 Maths Chapter 4 PDF

The quadratic formula was discovered in 1025 AD by Sridharacharya. Since then, the quadratic equations have been widely used not just in mathematics but for real-life situations as well. This chapter explains the concept of solving quadratic equations with the help of real world practical examples so as to keep the learning process engaging and interesting. Students can have a quick glance of each section in the NCERT solutions for class 10 maths chapter 4 Quadratic Equations from the below-mentioned links :

☛ Download Class 10 Maths NCERT Solutions Chapter 4 Quadratic Equations

NCERT Class 10 Maths Chapter 4   Download PDF

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Owing to the applicability of quadratic equations in various real-life problems, it becomes an important topic that needs to be understood well. The chapter showcases enough solved examples to ensure the coverage of the topic properly. Students will learn the basics of the various methods involved in finding the roots of quadratic equations along with the graphical representation of the equation. A brief exercise-wise analysis of NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations can be seen below :

• Class 10 Maths Chapter 4 Ex 4.1 - 2 Questions
• Class 10 Maths Chapter 4 Ex 4.2 - 6 Questions
• Class 10 Maths Chapter 4 Ex 4.3 - 11 Questions
• Class 10 Maths Chapter 4 Ex 4.4 - 5 Questions

Topics Covered: The topics covered in the class 10 maths NCERT Solutions Chapter 4 Quadratic Equations are the definition of quadratic equations , standard form of a quadratic equation, nature of roots , the concept of discriminant, quadratic formula, solution of a quadratic equation by the factorization method , and completing the square method.

Total Questions: Class 10 maths chapter 4 Quadratic Equations consists of a total of 24 questions, out of which 15 are straightforward, 5 are intermediate level questions, and 4 are difficult problems. These questions are explained in a step-wise manner. The important points are written in lucid language to encourage better comprehension.

List of Formulas in NCERT Solutions Class 10 Maths Chapter 4

NCERT solutions for class 10 maths chapter 4 will help students deal with different methods of solving a quadratic equation, and these methods involve the use of various formulas, which have been explained step by step in the chapter. Along with the formulas, kids also need to understand their implications. For example, as mentioned above the value of the discriminant can help us gain important insights into the practical applications of that particular quadratic equation. It is a good idea to make a formula sheet that will give kids a quick overview of these points as and when required. A few important formulas are as given below :

• Standard form of a quadratic equation: ax² + bx + c = 0, a ≠ 0
• Quadratic formula : [-b±√(b²-4ac)]/(2a) to find the solution of a quadratic equation.
• Discriminant : b 2 – 4ac

Important Questions for Class 10 Maths NCERT Solutions Chapter 4

Video Solutions for Class 10 Maths NCERT Chapter 4

FAQs on NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations

Why are ncert solutions class 10 maths chapter 4 vital for scoring well.

The NCERT solutions class 10 maths chapter 4 has been designed by scholars in their respective fields. After a lot of research, they have compiled the content into this textbook, making it a precious resource for study. The quadratic equations have widespread applications in our daily lives; hence, to understand them properly, it is advisable for the students to go through this chapter thoroughly. Also, the CBSE board recommends following these solutions, thereby making them of utmost importance.

Do I Need to Practice all Questions Given in Class 10 Maths NCERT Solutions Quadratic Equations?

The NCERT Solutions Class 10 Maths Quadratic Equations covers a wide range of topics that are important from the board exam's perspective. Problems based on topics such as the roots of a quadratic equation and the factorization method are often asked in exams. Hence, it is worthwhile to go through the entire theory and the solved examples from this chapter to get a better grasp of the concepts.

What are the Important Topics Covered in NCERT Solutions Class 10 Maths Chapter 4?

The important topics covered in the NCERT Solutions Class 10 Maths Chapter 4 are how to represent the given problem statements mathematically, what is the standard form of a quadratic equation, how to solve quadratic equations by factoring and completing the squares which is an essential topic requiring regular practice. Solving questions related to these topics will help the students score well in their board exams.

How Many Questions are there in NCERT Solutions Class 10 Maths Chapter 4?

The NCERT Solutions Class 10 Maths Chapter 4 contains a total of 24 well-researched questions by the subject experts. These 24 problems include both theoretical as well as graph-based questions. All the solutions are elaborated well in the NCERT solutions and are self-explanatory. The questions are solved in more than one method to strengthen the understanding of basic quadratic equation concepts.

How CBSE Students can utilize NCERT Solutions Class 10 Maths Chapter 4 effectively?

Students can utilize the NCERT Solutions Class 10 Maths Chapter 4 effectively by reviewing the principles and theorems explained in each section of the lesson on a frequent basis. They must then solve the exercise questions after practicing all of the examples and reviewing fundamental formulas related to factorization as well as completion of squares method of quadratic equations. This will help them build up their problem-solving approach thereby building confidence in maths.

Why Should I Practice NCERT Solutions Class 10 Maths Chapter 4?

Quadratic equations is one of the topics that is not just important for mathematics but also plays a fundamental role in a lot of real-life scenarios. If you have to calculate the length and breadth of a garden, you can use the quadratic equation. Based on this information, you can plan the quantity of a grass carpet needed for the garden. Quadratic equations are also used in fields such as astronomy, science, architecture. Owing to its vast range of applications, practicing the NCERT Solutions Class 10 Maths Chapter 4 in detail becomes very important for the students. Thus, to perfect this lesson practice is key.

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations

Ncert solutions for class 10 maths chapter 4 – quadratic equations pdf.

Free PDF of NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations includes all the questions provided in NCERT Books prepared by Mathematics expert teachers as per CBSE NCERT guidelines from Mathongo.com. To download our free pdf of Chapter 4 Quadratic Equations Maths NCERT Solutions for Class 10 to help you to score more marks in your board exams and as well as competitive exams.

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Chapter 4 Class 10 Quadratic Equations

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Updated for Latest NCERT for 2023-2024 Boards.

Get NCERT Solutions for all exercise questions and examples of Chapter 4 Class 10 Quadratic Equations free at Teachoo. Answers to each and every question is provided video solutions. 

In this chapter, we will learn

• What is a Quadratic Equation
• What is the Standard Form of a Quadratic Equation
• Solution of a Quadratic Equation by Factorisation ( Splitting the Middle Term method)
• Solving a Quadratic Equation by Completing the Square
• Solving a Quadratic Equation using D Formula (x = -b ± √b 2 - 4ac / 2a)
• Checking if roots are real, equal or no real roots (By Checking the value of D = b 2 - 4ac)

This chapter is divided into two parts - Serial Order Wise, Concept Wise

In Serial Order Wise, the chapter is divided into exercise questions and examples.

In Concept Wise, the chapter is divided into concepts. First the concepts are explained, and then the questions of the topic are solved - from easy to difficult.

We suggest you do the Chapter from Concept Wise - it is the Teachoo (टीचू) way of learning.

Note: When you click on a link, the first question of the exercise will open. To open other question of the exercise, go to bottom of the page. There is a list with arrows. It has all the questions with Important Questions also marked.

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• NCERT Solutions
• NCERT Solutions for Class 10
• NCERT Solutions for Class 10 Maths
• Chapter 4: Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Ncert solutions class 10 maths chapter 4 – cbse free pdf download.

NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations  contain all the solutions to the problems provided in the Class 10 Maths NCERT textbook for CBSE exam preparations. The questions from every section are framed and solved accurately by the subject experts. NCERT Solutions for Class 10 are detailed and step-by-step guides to all the queries of the students. The exercises present in the chapter should be dealt 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 ax 2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0. That is, ax 2 + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 4 Quadratic Equations

Download most important questions for class 10 maths chapter – 4 quadratic equations.

Quadratic equations arise in several situations around us. Hence, students should give special attention to learning the concepts related to this chapter of the latest CBSE Syllabus for 2023-24 thoroughly to excel in Class 10 Maths examinations. NCERT Solutions help the students in learning these concepts as well as in evaluating themselves. Practising 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.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability
• Exercise 4.1
• Exercise 4.2
• Exercise 4.3
• Exercise 4.4

NCERT Solutions Class 10 Maths Chapter 4 – Quadratic Equations

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Access answers to NCERT Class 10 Maths Chapter 4 – Quadratic Equations

Exercise 4.1 page: 73.

1. Check whether the following are quadratic equations:

(i) (x + 1) 2 = 2(x – 3)

(ii) x 2 – 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) x 2 + 3x + 1 = (x – 2) 2

(vii) (x + 2) 3 = 2x (x 2 – 1)

(viii) x 3 – 4x 2 – x + 1 = (x – 2) 3

(x + 1) 2 = 2(x – 3)

By using the formula for (a+b) 2 = a 2 +2ab+b 2

⇒ x 2  + 2x + 1 = 2x – 6

⇒ x 2  + 7 = 0

The above equation is in the form of ax 2 + bx + c = 0

Therefore, the given equation is a quadratic equation.

(ii) Given, x 2 – 2x = (–2) (3 – x)

⇒ x 2  –   2x = -6 + 2x

⇒ x 2  – 4x + 6 = 0

(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3)

By multiplication,

⇒ x 2  – x – 2 = x 2  + 2x – 3

⇒ 3x – 1 = 0

The above equation is not in the form of ax 2 + bx + c = 0

Therefore, the given equation is not a quadratic equation.

(iv) Given, (x – 3)(2x +1) = x(x + 5)

⇒ 2x 2  – 5x – 3 = x 2  + 5x

⇒  x 2  – 10x – 3 = 0

(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1)

⇒ 2x 2  – 7x + 3 = x 2  + 4x – 5

⇒ x 2  – 11x + 8 = 0

The above equation is in the form of ax 2  + bx + c = 0.

(vi) Given, x 2  + 3x + 1 = (x – 2) 2

By using the formula for (a-b) 2 =a 2 -2ab+b 2

⇒ x 2  + 3x + 1 = x 2  + 4 – 4x

⇒ 7x – 3 = 0

(vii) Given, (x + 2) 3  = 2x(x 2  – 1)

By using the formula for (a+b) 3  = a 3 +b 3 +3ab(a+b)

⇒ x 3  + 8 + x 2  + 12x = 2x 3  – 2x

⇒ x 3  + 14x – 6x 2  – 8 = 0

(viii) Given, x 3  – 4x 2  – x + 1 = (x – 2) 3

By using the formula for (a-b) 3  = a 3 -b 3 -3ab(a-b)

⇒  x 3  – 4x 2  – x + 1 = x 3  – 8 – 6x 2   + 12x

⇒ 2x 2  – 13x + 9 = 0

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

(i) The area of a rectangular plot is 528 m 2 . The length of the plot (in metres) 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

(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 m 2

Putting the value of the length and breadth of the plot in the formula, we get,

(2x + 1) × x = 528

⇒ 2x 2  + x =528

⇒ 2x 2  + x – 528 = 0

Therefore, the length and breadth of the plot satisfy the quadratic equation, 2x 2  + 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 2  + x = 306

⇒ x 2  + x – 306 = 0

Therefore, the two integers x and x+1 satisfy the quadratic equation, x 2  + 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

⇒ x 2  + 29x + 3x + 87 = 360

⇒ x 2  + 32x + 87 – 360 = 0

⇒ x 2  + 32x – 273 = 0

Therefore, the age of Rohan and his mother satisfies the quadratic equation, x 2  + 32x – 273 = 0, which is the required representation of the problem mathematically.

(iv) Let us consider,

The speed of the train = x   km/h

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

Speed × Time = Distance

( x  – 8)(480/ x  )+ 3 = 480

⇒ 480 + 3 x  – 3840/ x  – 24 = 480

⇒ 3 x  – 3840/ x  = 24

⇒ x 2  – 8 x  – 1280 = 0

Therefore, the speed of the train satisfies the quadratic equation, x 2  – 8 x  – 1280 = 0, which is the required representation of the problem mathematically.

Exercise 4.2 Page: 76

1. Find the roots of the following quadratic equations by factorisation:

(i) x 2  – 3x – 10 = 0 (ii) 2x 2  + x – 6 = 0 (iii) √2 x 2  + 7x + 5√2 = 0 (iv) 2x 2  – x +1/8 = 0 (v) 100x 2  – 20x + 1 = 0

(i) Given,  x 2  – 3 x  – 10 =0

Taking L.H.S.,

=> x 2  – 5 x  + 2 x  – 10

=> x ( x  – 5) + 2( x  – 5)

=>( x  – 5)( x  + 2)

The roots of this equation, x 2  – 3 x  – 10 = 0 are the values of x for which ( x  – 5)( x  + 2) = 0

Therefore,  x  – 5 = 0 or  x  + 2 = 0

=>  x  = 5 or  x  = -2

(ii) Given, 2 x 2  +  x  – 6 = 0

=> 2 x 2  + 4 x  – 3 x  – 6

=> 2 x ( x  + 2) – 3( x  + 2)

=> ( x  + 2)(2 x  – 3)

The roots of this equation, 2 x 2  +  x  – 6=0 are the values of x for which ( x x  + 2)(2 x  – 3) = 0

Therefore,  x  + 2 = 0 or 2 x  – 3 = 0

=>  x  = -2 or  x  = 3/2

(iii) √2  x 2  + 7 x  + 5√2=0

=> √2  x 2  + 5 x  + 2 x  + 5√2

=>  x  (√2 x  + 5) + √2(√2 x  + 5)= (√2 x  + 5)( x  + √2)

The roots of this equation, √2  x 2  + 7 x  + 5√2=0 are the values of x for which (√2 x  + 5)( x  + √2) = 0

Therefore, √2 x  + 5 = 0 or  x  + √2 = 0

=>  x  = -5/√2 or  x  = -√2

(iv) 2 x 2  –  x  +1/8 = 0

=1/8 (16 x 2   – 8 x  + 1)

= 1/8 (16 x 2   – 4 x  -4 x  + 1)

= 1/8 (4 x (4 x    – 1) -1(4 x  – 1))

= 1/8 (4 x  – 1) 2

The roots of this equation, 2 x 2  –  x  + 1/8 = 0, are the values of x for which (4 x  – 1) 2 = 0

Therefore, (4 x  – 1) = 0 or (4 x  – 1) = 0

⇒  x  = 1/4 or  x  = 1/4

(v) Given, 100x 2  – 20x + 1=0

= 100x 2  – 10x – 10x + 1

= 10x(10x – 1) -1(10x – 1)

= (10x – 1) 2

The roots of this equation, 100x 2  – 20x + 1=0, are the values of x for which (10x – 1) 2 = 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

2. Solve the problems given in Example 1.

Represent the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.

(i) Let us say the number of marbles John has = x

Therefore, the number of marble Jivanti has = 45 – x

After losing 5 marbles each,

Number of marbles John has = x  – 5

Number of marble Jivanti has = 45 – x  – 5 = 40 –  x

Given that the product of their marbles is 124.

∴ ( x  – 5)(40 –  x ) = 124

⇒  x 2  – 45 x  + 324 = 0

⇒  x 2  – 36 x  – 9 x  + 324 = 0

⇒  x ( x  – 36) -9( x  – 36) = 0

⇒ ( x  – 36)( x  – 9) = 0

Thus, we can say,

x  – 36 = 0 or  x  – 9 = 0

⇒  x  = 36 or  x  = 9

If John’s marbles = 36

Then, Jivanti’s marbles = 45 – 36 = 9

And if John’s marbles = 9

Then, Jivanti’s marbles = 45 – 9 = 36

(ii) Let us say the number of toys produced in a day is x .

Therefore, cost of production of each toy = Rs(55 –  x )

Given the total cost of production of the toys = Rs 750

∴  x (55 –  x ) = 750

⇒  x 2  – 55 x  + 750 = 0

⇒  x 2  – 25 x  – 30 x  + 750 = 0

⇒  x ( x  – 25) -30( x  – 25) = 0

⇒ ( x  – 25)( x  – 30) = 0

Thus, either x  -25 = 0 or  x  – 30 = 0

⇒  x  = 25 or  x  = 30

Hence, the number of toys produced in a day will be either 25 or 30.

3. Find two numbers whose sum is 27 and product is 182.

Let us say the first number is x, and the second number is 27 – x.

Therefore, the product of two numbers

x(27 – x) = 182

⇒ x 2  – 27x – 182 = 0

⇒ x 2  – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

Thus, either, x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

Therefore, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

4. Find two consecutive positive integers, the sum of whose squares is 365.

Let us say the two consecutive positive integers are x  and  x  + 1.

Therefore, as per the given questions,

x 2  + ( x  + 1) 2  = 365

⇒  x 2  +  x 2  + 1 + 2 x  = 365

⇒ 2 x 2  + 2x – 364 = 0

⇒  x 2  +  x  – 182 = 0

⇒  x 2  + 14 x  – 13 x  – 182 = 0

⇒  x ( x  + 14) -13( x  + 14) = 0

⇒ ( x  + 14)( x  – 13) = 0

Thus, either, x  + 14 = 0 or  x  – 13 = 0,

⇒  x  = – 14 or  x  = 13

Since the integers are positive, x can be 13 only.

∴  x  + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Let us say the base of the right triangle is x cm.

Given, the altitude of right triangle = (x – 7) cm

From Pythagoras’ theorem, we know,

Base 2  + Altitude 2  = Hypotenuse 2

∴ x 2  + (x – 7) 2  = 13 2

⇒ x 2  + x 2  + 49 – 14x = 169

⇒ 2x 2  – 14x – 120 = 0

⇒ x 2  – 7x – 60 = 0

⇒ x 2  – 12x + 5x – 60 = 0

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12)(x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

⇒ x = 12 or x = – 5

Since sides cannot be negative, x can only be 12.

Therefore, the base of the given triangle is 12 cm, and the altitude of this triangle will be (12 – 7) cm = 5 cm.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and the cost of each article.

Let us say the number of articles produced is x .

Therefore, cost of production of each article = Rs (2 x  + 3)

Given the total cost of production is Rs.90

∴  x (2 x  + 3) = 90

⇒ 2 x 2  + 3 x  – 90 = 0

⇒ 2 x 2  + 15 x  -12 x  – 90 = 0

⇒  x (2 x  + 15) -6(2 x  + 15) = 0

⇒ (2 x  + 15)( x  – 6) = 0

Thus, either 2 x  + 15 = 0 or  x  – 6 = 0

⇒  x  = -15/2 or  x  = 6

As the number of articles produced can only be a positive integer, x can only be 6.

Hence, the number of articles produced = 6

Cost of each article = 2 × 6 + 3 = Rs 15

Exercise 4.3 Page: 87

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2 x 2  – 7 x  +3 = 0

(ii) 2 x 2  +  x  – 4 = 0 (iii) 4 x 2  + 4√3 x  + 3 = 0

(iv) 2 x 2  +  x  + 4 = 0

(i) 2 x 2  – 7 x  + 3 = 0

⇒ 2 x 2  – 7 x  = – 3

Dividing by 2 on both sides, we get

⇒ x 2 -7x/2 = -3/2

⇒ x 2 -2 × x ×7/4 = -3/2

On adding (7/4) 2 to both sides of the equation, we get

⇒ (x) 2 -2×x×7/4 +(7/4) 2 = (7/4) 2 -3/2

⇒ (x-7/4) 2 = (49/16) – (3/2)

⇒(x-7/4) 2 = 25/16

⇒(x-7/4) 2 = ±5/4

⇒  x  = 7/4 ± 5/4

⇒  x  = 7/4 + 5/4 or x = 7/4 – 5/4

⇒ x = 12/4 or x = 2/4

⇒ x = 3 or x = 1/2

(ii) 2x 2  + x – 4 = 0

⇒ 2x 2  + x = 4

Dividing both sides of the equation by 2, we get

⇒ x 2  +x/2 = 2

Now on adding (1/4) 2  to both sides of the equation, we get,

⇒ (x) 2  + 2 × x × 1/4 + (1/4) 2  = 2 + (1/4) 2

⇒ (x + 1/4) 2  = 33/16

⇒ x + 1/4 = ± √33/4

⇒ x = ± √33/4 – 1/4

⇒ x = (± √33-1)/4

Therefore, either x = (√33-1)/4 or x = (-√33-1)/4

(iii) 4x 2  + 4√3x + 3 = 0

Converting the equation into a 2 +2ab+b 2 form, we get,

⇒ (2x) 2  + 2 × 2x × √3 + (√3) 2  = 0

⇒ (2x + √3) 2  = 0

⇒ (2x + √3) = 0 and (2x + √3) = 0

Therefore, either x = -√3/2 or x = -√3/2

(iv) 2x 2  + x + 4 = 0

⇒ 2x 2  + x = -4

⇒ x 2  + 1/2x = 2

⇒ x 2  + 2 × x × 1/4 = -2

By adding (1/4) 2  to both sides of the equation, we get

⇒ (x) 2  + 2 × x × 1/4 + (1/4) 2  = (1/4) 2  – 2

⇒ (x + 1/4) 2  = 1/16 – 2

⇒ (x + 1/4) 2  = -31/16

As we know, the square of numbers cannot be negative.

Therefore, there is no real root for the given equation, 2x 2 + x + 4 = 0

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

(i) 2x 2  – 7x + 3 = 0

On comparing the given equation with ax 2  + bx + c = 0, we get,

a = 2, b = -7 and c = 3

By using the quadratic formula, we get,

⇒ x = (7±√(49 – 24))/4

⇒ x = (7±√25)/4

⇒ x = (7±5)/4

⇒ x = (7+5)/4 or x = (7-5)/4

⇒ x = 12/4 or 2/4

∴  x = 3 or 1/2

(ii) 2x 2  + x – 4 = 0

a = 2, b = 1 and c = -4

⇒x = (-1±√1+32)/4

⇒x = (-1±√33)/4

∴ x = (-1+√33)/4 or x = (-1-√33)/4

(iii) 4x 2  + 4√3x + 3 = 0

On comparing the given equation with ax 2  + bx + c = 0, we get

a = 4, b = 4√3 and c = 3

⇒ x = (-4√3±√48-48)/8

⇒ x = (-4√3±0)/8

∴ x = -√3/2 or x = -√3/2

(iv) 2x 2  + x + 4 = 0

a = 2, b = 1 and c = 4

By using the quadratic formula, we get

⇒ x = (-1±√1-32)/4

⇒ x = (-1±√-31)/4

As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.

3. Find the roots of the following equations:

(i) x-1/x = 3, x ≠ 0 (ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7

(i) x-1/x = 3

⇒ x 2  – 3x -1 = 0

a = 1, b = -3 and c = -1

⇒ x = (3±√9+4)/2

⇒ x = (3±√13)/2

∴ x = (3+√13)/2 or x = (3-√13)/2

(ii) 1/x+4 – 1/x-7 = 11/30 ⇒ x-7-x-4/(x+4)(x-7) = 11/30

⇒ -11/(x+4)(x-7) = 11/30

⇒ (x+4)(x-7) = -30

⇒ x 2  – 3x – 28 = 30

⇒ x 2  – 3x + 2 = 0

We can solve this equation by factorisation method now.

⇒ x 2  – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 2)(x – 1) = 0

⇒ x = 1 or 2

4. The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Let us say the present age of Rahman is x  years.

Three years ago, Rehman’s age was ( x  – 3) years.

Five years after, his age will be ( x  + 5) years.

Given the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3.

∴ 1/ x -3 + 1/ x -5 = 1/3

(x +5+ x -3)/( x -3)( x +5) = 1/3

(2 x +2)/( x -3)( x +5) = 1/3

⇒ 3(2 x  + 2) = ( x -3)( x +5)

⇒ 6 x  + 6 =  x 2  + 2 x  – 15

⇒  x 2  – 4 x  – 21 = 0

⇒  x 2  – 7 x  + 3 x  – 21 = 0

⇒  x ( x  – 7) + 3( x  – 7) = 0

⇒ ( x  – 7)( x  + 3) = 0

⇒  x  = 7, -3

As we know, age cannot be negative.

Therefore, Rahman’s present age is 7 years.

5. 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 us say the marks of Shefali in Maths be x.

Then, the marks in English will be 30 – x.

As per the given question,

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

⇒ -x 2  + 25x + 54 = 210

⇒ x 2  – 25x + 156 = 0

⇒ x 2  – 12x – 13x + 156 = 0

⇒ x(x – 12) -13(x – 12) = 0

⇒ (x – 12)(x – 13) = 0

⇒ x = 12, 13

Therefore, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18, and if the marks in Maths are 13, then marks in English will be 30 – 13 = 17 .

6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Let us say the shorter side of the rectangle is x  m.

Then, larger side of the rectangle = ( x  + 30) m

As given, the length of the diagonal is = x + 30 m

⇒  x 2  + ( x  + 30) 2  = ( x  + 60) 2

⇒  x 2  +  x 2  + 900 + 60 x  =  x 2  + 3600 + 120 x

⇒  x 2  – 60 x  – 2700 = 0

⇒  x 2  – 90 x  + 30 x  – 2700 = 0

⇒  x ( x  – 90) + 30( x  -90) = 0

⇒ ( x  – 90)( x  + 30) = 0

⇒  x  = 90, -30

However, the side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m.

And the length of the larger side will be (90 + 30) m = 120 m.

7. The difference of the squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Let us say the larger and smaller number be x  and  y,  respectively.

As per the question given,

x 2  –  y 2  = 180 and  y 2  = 8 x

⇒  x 2  – 8 x  = 180

⇒  x 2  – 8 x  – 180 = 0

⇒  x 2  – 18 x  + 10 x  – 180 = 0

⇒  x ( x  – 18) +10( x  – 18) = 0

⇒ ( x  – 18)( x  + 10) = 0

⇒  x  = 18, -10

However, the larger number cannot be considered a negative number, as 8 times the larger number will be negative, and hence, the square of the smaller number will be negative, which is not possible.

Therefore, the larger number will be 18 only.

∴  y 2  = 8x = 8 × 18 = 144

⇒  y  = ±√144 = ±12

∴ Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and -12.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

It is given that

Distance = 360 km

Consider x as the speed, then the time taken

If the speed is increased by 5 km/h, the speed will be (x + 5) km/h.

Distance will be the same.

t = 360/(x + 5)

We know that

Time with original speed – Time with increased speed = 1

360/x – 360/(x + 5) = 1

LCM = x (x + 5)

360 x + 1800 – 360x = x (x + 5)

x 2  + 5x = 1800

x 2  + 5x – 1800 = 0

x 2  + 45x – 40x – 1800 = 0

x (x + 45) – 40 (x + 45) = 0

(x – 40) (x + 45) = 0

x = 40 km/hr

As we know, the value of speed cannot be negative.

Therefore, the speed of the train is 40 km/h.

Let the time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = ( x  – 10) hr

Part of the tank filled by smaller pipe in 1 hour = 1/ x

Part of the tank filled by larger pipe in 1 hour = 1/( x  – 10)

1/ x  + 1/ x -10 = 8/75

x -10+ x / x ( x -10) = 8/75

⇒ 2 x -10/ x ( x -10) = 8/75

⇒ 75(2 x  – 10) = 8 x 2  – 80 x

⇒ 150 x  – 750 = 8 x 2  – 80 x

⇒ 8 x 2  – 230 x  +750 = 0

⇒ 8 x 2  – 200 x  – 30 x  + 750 = 0

⇒ 8 x ( x  – 25) -30( x  – 25) = 0

⇒ ( x  – 25)(8 x  -30) = 0

⇒  x  = 25, 30/8

Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.

Therefore, the time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours, respectively.

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Let us say the average speed of the passenger train =   x  km/h.

Average speed of express train = ( x  + 11) km/h

Given the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,

(132/x) – (132/(x+11)) = 1

132(x+11-x)/(x(x+11)) = 1

132 × 11 /(x(x+11)) = 1

⇒ 132 × 11 =  x ( x  + 11)

⇒  x 2  + 11 x  – 1452 = 0

⇒  x 2  +  44 x  -33 x  -1452 = 0

⇒  x ( x  + 44) -33( x  + 44) = 0

⇒ ( x  + 44)( x  – 33) = 0

⇒  x  = – 44, 33

As we know, speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.

11. Sum of the areas of two squares is 468 m 2 . If the difference between their perimeters is 24 m, find the sides of the two squares.

Let the sides of the two squares be  x  m and  y  m.

Therefore, their perimeter will be 4 x  and 4 y,  respectively

And the area of the squares will be x 2  and  y 2,  respectively.

4 x  – 4 y  = 24

x  –  y  = 6

x  =  y  + 6

Also,  x 2  +   y 2  = 468

⇒ (6 +  y 2 ) +  y 2  = 468

⇒ 36 +  y 2  + 12 y  +  y 2  = 468

⇒ 2 y 2  + 12 y  + 432 = 0

⇒  y 2  + 6y – 216 = 0

⇒  y 2  + 18 y  – 12 y  – 216 = 0

⇒  y ( y  +18) -12( y  + 18) = 0

⇒ ( y  + 18)( y  – 12) = 0

⇒  y  = -18, 12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

Exercise 4.4 Page: 91

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them. (i) 2 x 2  – 3 x  + 5 = 0 (ii) 3 x 2  – 4√3 x  + 4 = 0 (iii) 2 x 2  – 6 x  + 3 = 0

2x 2  – 3 x  + 5 = 0

Comparing the equation with  ax 2  +  bx  +  c  = 0, we get

a  = 2,  b  = -3 and  c  = 5

We know, Discriminant =  b 2  – 4 ac

=  ( – 3) 2  – 4 (2) (5) = 9 – 40

= – 31

As you can see, b 2  – 4ac < 0

Therefore, no real root is possible for the given equation, 2x 2  – 3 x + 5 = 0

(ii) 3 x 2  – 4√3 x  + 4 = 0

a  = 3,  b  = -4√3 and  c  = 4

= (-4√3) 2  – 4(3)(4)

= 48 – 48 = 0

As  b 2  – 4 ac  = 0,

Real roots exist for the given equation, and they are equal to each other.

Hence, the roots will be – b /2 a  and – b /2 a .

– b /2 a  = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3

Therefore, the roots are 2/√3 and 2/√3.

(iii) 2 x 2  – 6 x  + 3 = 0

a  = 2,  b  = -6,  c  = 3

As we know, Discriminant =  b 2  – 4 ac

= (-6) 2  – 4 (2) (3)

= 36 – 24 = 12

As  b 2  – 4 ac  > 0,

Therefore, there are distinct real roots that exist for this equation, 2 x 2  – 6 x + 3 = 0

= (-(-6) ± √(-6 2 -4(2)(3)) )/ 2(2)

= (6±2√3 )/4

Therefore, the roots for the given equation are (3+√3)/2 and (3-√3)/2

2. Find the values of  k for each of the following quadratic equations so that they have two equal roots. (i) 2 x 2  +  kx  + 3 = 0 (ii)  kx  ( x  – 2) + 6 = 0

(i) 2 x 2  +  kx  + 3 = 0

Comparing the given equation with  ax 2  +  bx  +  c  = 0, we get,

a  = 2,  b  = k and  c  = 3

= ( k ) 2  – 4(2) (3)

=  k 2  – 24

For equal roots, we know,

Discriminant = 0

k 2  – 24 = 0

k = ±√24 = ±2√6

(ii)  kx ( x  – 2) + 6 = 0

or  kx 2  – 2 kx  + 6 = 0

Comparing the given equation with  ax 2  +  bx  +  c  = 0, we get

a  =  k ,  b  = – 2 k  and  c  = 6

= ( – 2 k ) 2 – 4 ( k ) (6)

= 4 k 2 – 24 k

b 2 – 4 ac  = 0

4 k 2 – 24 k  = 0

4 k  ( k  – 6) = 0

Either 4 k  = 0 or  k  = 6 = 0

k  = 0 or  k  = 6

However, if  k  = 0, then the equation will not have the terms ‘ x 2 ‘ and ‘ x ‘.

Therefore, if this equation has two equal roots,  k  should be 6 only.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m 2 ? If so, find its length and breadth.

Let the breadth of the mango grove be l .

The length of the mango grove will be 2 l .

Area of the mango grove = (2 l ) ( l )= 2 l 2

2 l 2  = 800

l 2  = 800/2 = 400

l 2  – 400 =0

a  = 1,  b  = 0,  c  = 400

=> (0) 2  – 4 × (1) × ( – 400) = 1600

Here,  b 2  – 4 ac  > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

As we know, the value of length cannot be negative.

Therefore, the breadth of the mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.

Let’s say the age of one friend is x years.

Then, the age of the other friend will be (20 – x) years.

Four years ago,

Age of First friend = ( x  – 4) years

Age of Second friend = (20 –  x  – 4) = (16 –  x ) years

As per the given question, we can write,

( x  – 4) (16 –  x ) = 48

16 x – x 2  – 64 + 4 x  = 48

 – x 2  + 20 x –  112 = 0

x 2  – 20 x +  112 = 0

a  =  1 ,  b  = -2 0  and  c  = 112

Discriminant =  b 2  – 4 ac

= (- 20 ) 2  – 4 × 112

= 400 – 448 = -48

b 2  – 4 ac  < 0

Therefore, there will be no real solution possible for the equations. Hence, the condition doesn’t exist.

5. Is it possible to design a rectangular park of perimeter 80 and an area of 400 m2? If so, find its length and breadth.

Let the length and breadth of the park be l and b.

Perimeter of the rectangular park = 2 ( l + b ) = 80

So, l + b  = 40

Or,  b  = 40 –  l

Area of the rectangular park =  l×b = l(40 – l) =  40 l  –  l 2 = 400

l 2   –    40 l  + 400   = 0, which is a quadratic equation.

a  = 1,  b  = -40,  c  = 400

Since, Discriminant =  b 2  – 4 ac

=(- 40 ) 2  – 4 × 400

= 1600 – 1600 = 0

Thus, b 2  – 4 ac  = 0

Therefore, this equation has equal real roots. Hence, the situation is possible.

The root of the equation,

l  = – b /2 a

l  = -(-40)/2(1) = 40/2 = 20

Therefore, the length of the rectangular park, l  = 20 m

And the breadth of the park, b  = 40 –  l  = 40 – 20 = 20 m.

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations

A 1-mark question was asked from Chapter 4 Quadratic Equations in the year 2018. However, in the year 2017, a total of 13 marks were asked from the topic Quadratic Equations. Therefore, students need to have a thorough understanding of the topic. The topics and sub-topics provided in this chapter include:

4.1 Introduction

If we equate the polynomial  ax 2 + bx + c, a ≠ 0 to zero, we get a quadratic equation. Quadratic equations come up when we deal with many real-life situations. In this chapter, students will study quadratic equations and various ways of finding their roots. They will also see some applications of quadratic equations in daily life situations.

4.2 Quadratic Equations

A quadratic equation in the variable x is an equation of the form ax 2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0. In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax 2 + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.

4.3 Solution of Quadratic Equations by Factorisation

A real number α is called a root of the quadratic equation ax 2  + bx + c = 0, a ≠ 0 if a α 2 + bα + c = 0. We also say that x = α is a solution of the quadratic equation or that α satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax 2 + bx + c and the roots of the quadratic equation ax 2  + bx + c = 0 are the same.

4.4 Solution of a Quadratic Equation by Completing the Square

Finding the value that makes a quadratic equation a square trinomial is called  completing the square. The square trinomial can then be solved easily by factorising.

4.5 Nature of Roots

If b 2  – 4ac < 0, then there is no real number whose square is b 2  – 4ac. Therefore, there are no real roots for the given quadratic equation in this case. Since b 2  – 4ac determines whether the quadratic equation ax 2  + bx + c = 0 has real roots or not, b 2  – 4ac is called the discriminant of this quadratic equation. So, a quadratic equation ax 2  + bx + c = 0 has (i) two distinct real roots, if b 2  – 4ac > 0, (ii) two equal real roots, if b 2 – 4ac = 0, (iii) no real roots, if b 2 – 4ac < 0. 4.6

List of exercises we covered in NCERT Solutions for Class 10 Maths Chapter 4

Exercise 4.1 Solutions – 2 Questions Exercise 4.2 Solutions – 6 Questions Exercise 4.3 Solutions – 11 Questions Exercise 4.4 Solutions – 5 Questions

In a quadratic equation, x represents an unknown form, and a, b, and c are the known values. An equation to be quadratic “a” should not be equal to 0. The equation is of the form ax 2 + bx + c = 0. The values of a, b, and c are always real numbers. A quadratic equation can be calculated by completing the square. A quadratic equation has

• Two different real roots
• No real roots
• Two equal roots

Key Features of NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations

• The highly experienced faculty at BYJU’S designs NCERT Solutions with utmost care.
• The solutions are 100% accurate and can be used by the students while preparing for their CBSE board exams.
• All the minute concepts are also covered to help students face other competitive exams more confidently.
• The exercise questions present in the NCERT Textbook have been answered in a step-wise manner so that students attain good scores not only on the final answer but also on each step.

For more questions to practise, students can refer to the other study materials which are given at BYJU’S.

• RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations

Disclaimer – 

Dropped Topics –  4.4 Solution of a quadratic equation by completing the squares

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VERY SHORT ANSWER TYPE QUESTIONS

1. State which of the following equations are quadratic equations :

(i) 3x + 1/x – 8 = 0       (ii) 18x2 – 6x = 0     (iii) x 2 – 5x = 7 – 6x 3     (iv) x 2 = 25     (v) 6x 5 +3x 2 –7=0     (vi) x + 1/x 2 = 3 (vii) 5x 2 + 6x =7     (viii) 5x 3 – 2x – 3 = 0

2. Represent each of the following situations in the form of a quadratic equation : (i) The sum of the squares of two consecutive positive integers is 545. We need to find the integers. (ii) The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. We need to find the lengths of these sides. (iii) One year ago, the father was 8 times as old as his son. Now his age is square of the son's age. We need to find their present ages. (iv) Ravi and Raj together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. (v) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.

3. In each of the following determine whether the given values are the solutions of the given equation or not :

4. In each of the following, find the value of k for which the given value is a solution of the given equation :

(i) (x + 3) (2x – 3k) = 0 : x = 6

(ii) 3√7x 2 – 4x + k = 0 ; x =√7/3

5. Find the values of p and q for which x = 2/3 and  x = – 3 are the roots of the equation px 2 + 7x + q = 0.

SHORT ANSWER TYPE QUESTIONS

1. 27x 2 – 12 = 0

3. 16(x – 4) 2 = 9 (x + 3) 2

4.  x 2 – 300 = 0

5. x 2 + (a–b) x = ab

6. (3x+a) (3x+b)=ab

7. x 2 –1+ √2x+√2 =0

8. 3√7x 2 + 4x – √7= 0

9. √3y 2  + 11y + 6√3 = 0

10. abx 2 – (a 2 + b 2 )x + ab = 0

Find the roots of each of the following quadratic equations by the method of completing the squares

25. x 2 – 6x + 4 = 0

26. 2x 2 – 5x + 3 = 0

27. √5x 2 + 9x + 4 √5 = 0 28. (5z + 2a) (3z + 4b) = 8ab

29. 2 √2x 2 + 15x + 2 = 0

30. Find the solutions of 3x 2 – 2√6x + 2 = 0 by the method of completing the squares when

(i) x is a rational number

(ii) x is a real number

31. Find the solutions of 15x 2 + 3 = 17x, when (i) x is a rational number(ii) x is a real number.

32. Find the solutions of 5x 2 – 6x – 2 = 0, when (i) x is a rational number (ii) x is a real number

Find the roots of each of the following quadratic equations by using the quadratic formula

1. Equations in question No. (i), (ii), (iv), (vii), (ix), (xi),(xiii), (xiv), (xvi) and (xvii) are quadratic equations. 2. (i) x2 + x – 272 = 0, where x is the smaller integer. (ii) x 2 + 5x – 300 = 0, where x is the length of one side. (iii) x 2 – 8x + 7 = 0, where x (in years) is the present age of son. (iv) x 2 – 45x + 324 = 0, where x is the number of marbles with Ravi. (v) x 2 – 55x + 750 = 0, where x (in km/h) is the speed of the train. 3. (a) (i) Both are solution (ii) x = -√2 is a solution but x = -2√2 is not a solution. (iii) x = 1/2 is a solution but x = –1/2 is not a solution. (iv) Both are solution (b) (i) Both are solution (ii) Both are solution 4. (i) k = 4, (ii) k =  -√7

5. p = 3, q = – 6

1. 2/3 , – 2/3

2. 4, –8

4. 10√3, -10√3

5. b, – a

7. 1, √2

14. –a, – b

15. – 1

16. 12, – 2

18. -5/2, 3/2

19. 5, – 1

20. 6, 40/13

21. –10, – 1/5

22. – 4/3 , 1/8

23. 11/5, 5/8

24. 3, – 7/11

25. 3 ± √5

26. 1, 3/2 

29. No solution

30. (i) No solution (ii) √2/3

33. No solution

36. 1/√2

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NCERT Books for Class 10 Maths Chapter 4 Quadratic Equations

Quadratic Equations Class 10 NCERT Book: If you are looking for the best books of Class 10 Maths then NCERT Books can be a great choice to begin your preparation. NCERT Books for Class 10 Maths Chapter 4 Quadratic Equations can be of extreme use for students to understand the concepts in a simple way. NCERT Textbooks for Class 10 Maths are highly recommended as they help cover the entire syllabus. Class 10 Maths NCERT Books PDF Provided will help you during your preparation for both school exams as well as competitive exams.

NCERT Class 10 Maths Chapter 4 Books will give you authentic information and you can rely on them during your preparation. Prepare from the NCERT Class 10 Chapter 4 Books PDF download as they contain all sets of questions so that you can have a strong foundation of basics. Try practicing the previous papers and sample questions attached in the NCERT Books of Class 10 Maths Chapter 4 Quadratic Equations to solve the questions in your exam easily.

NCERT Books for Class 10 Maths Chapter 4 Quadratic Equations PDF Download

To ace in your exam preparation, you can refer to the 10th Class NCERT Solutions prevailing in NCERT e-Book. NCERT Books for Class 10 Maths Quadratic Equations will have illustrative problems and solutions. Brush up the concepts prevailing in Maths from Class 10 NCERT Books and learn the fundamentals.  Students can understand the concepts written in NCERT Class 10 Maths Textbooks for Ch 4 Quadratic Equations well as all of them are written in a comprehensive matter.

NCERT Textbooks of 10th Class Maths Chapter 4 Quadratic Equations are free of cost and you can access them through the quick links available.

• Quadratic Equations Class 10 NCERT Books
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Quadratic Equations Class 10 Notes CBSE Maths Chapter 4 (Free PDF Download)

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• Chapter 4 Quadratic Equations

Class 10 Maths Revision Notes for Quadratic Equations of Chapter 4 - Free PDF Download

CBSE Class 10 Maths Notes Chapter 4 curated by Vedantu are most preferred by students during their exam preparation. The Class 10 Maths Chapter 4 Notes come with shortcut techniques along with step by step explanations of all topics. Quadratic Equations Class 10 Notes curated by subject experts are available as PDF downloads. The Quadratic Equations Notes are as per the syllabus of upcoming CBSE board exams. The study guides for all subjects of 10 th standard can be easily availed using the download option. Vedantu is a platform that also provides free NCERT Solutions and other study materials for students. You can download NCERT Solutions Class 10 Maths to help you to revise the complete Syllabus and score more marks in your examinations. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution Class 10 Science , Maths solutions, and solutions of other subjects that are available on Vedantu only.

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Access Class 10 Maths Chapter 4-Quadratic Equation

Definition of the quadratic equation: .

A quadratic equation in the variable $x$ is an equation of the form $a{{x}^{2}}+bx+c=0$, where $a,b,c$ are real numbers, $a\ne 0$ . 

For example, $2{{x}^{2}}+x-300=0$ is a quadratic equation

The Standard Form of the Quadratic Equation:

Any equation of the form $p\left( x \right)=0$, where $p\left( x \right)$ is a polynomial of degree $2$ , is a quadratic equation. 

But when we write the terms of $p\left( x \right)$ in descending order of their degrees, then we get the standard form of the equation.

 That is, $a{{x}^{2}}+bx+c=0,a\ne 0$ is called the standard form of a quadratic equation.

Roots of the Quadratic Equation:

A solution of the equation $p\left( x \right)=a{{x}^{2}}+bx+c=0$, with $a\ne 0$ is called a root of the quadratic equation.

A real number $\alpha $  is called a root of the quadratic equation $a{{x}^{2}}+bx+c=0,a\ne 0$ if $a{{\alpha }^{2}}+b\alpha +c=0$.

It means $x=\alpha $ satisfies the quadratic equation or $x=\alpha $ is the root of the quadratic equation.

The zeroes of the quadratic polynomial $a{{x}^{2}}+bx+c$ and the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ are the same.

Method of Solving a Quadratic Equation:

Factorization Method

Factorize the quadratic equation by splitting the middle term.

After splitting the middle term, convert the equation into linear factors by taking common terms out.

Then on equating each factor to zero the roots are determined. 

For example: 

$\Rightarrow 2{{x}^{2}}-5x+3$            (Split the middle term)

$\Rightarrow 2{{x}^{2}}-2x-3x+3$    (Take out common terms to determine linear factors)

$\Rightarrow 2x\left( x-1 \right)-3\left( x-1 \right)$ 

$\Rightarrow \left( x-1 \right)\left( 2x-3 \right)$        (Equate to zero)

$\Rightarrow \left( x-1 \right)\left( 2x-3 \right)=0$

When $\left( x-1 \right)=0$ , $x=1$

When $\left( 2x-3 \right)=0$ , $x=\dfrac{3}{2}$ 

So, the roots of $2{{x}^{2}}-5x+3$ are $1$ and $\dfrac{3}{2}$ 

Method of completing the square

The solution of a quadratic equation can be found by converting any quadratic equation to perfect square of the form ${{\left( x+a \right)}^{2}}-{{b}^{2}}=0$.

To convert quadratic equation ${{x}^{2}}+ax+b=0$ to perfect square equate $b$ i.e., the constant term to the right side of equal sign then add square of half of $a$  i.e., square of half of coefficient of $x$  both sides.

To convert quadratic equation of form $a{{x}^{2}}+bx+c=0,a\ne 0$ to perfect square first divide the equation by $a$  i.e., the coefficient of ${{x}^{2}}$ then follow the above-mentioned steps.

For example:

$\Rightarrow {{x}^{2}}+4x-5=0$ (Equate constant term $5$  to the right of equal sign)

$\Rightarrow {{x}^{2}}+4x=5$      (Add square of half of $4$ both sides)

$\Rightarrow {{x}^{2}}+4x+{{\left( \dfrac{4}{2} \right)}^{2}}=5+{{\left( \dfrac{4}{2} \right)}^{2}}$

$ \Rightarrow {{x}^{2}}+4x+4=9$

$\Rightarrow {{\left( x+2 \right)}^{2}}=9 $

 $\Rightarrow {{\left( x+2 \right)}^{2}}-{{\left( 3 \right)}^{2}}=0 $

It is of the form ${{\left( x+a \right)}^{2}}-{{b}^{2}}=0$

$\Rightarrow {{\left( x+2 \right)}^{2}}-{{\left( 3 \right)}^{2}}=0$

$ \Rightarrow {{\left( x+2 \right)}^{2}}=9 $

$ \Rightarrow \left( x+2 \right)=\pm 3 $

      \[\Rightarrow x=1\] and $x=-5$  

So, the roots of ${{x}^{2}}+4x-5=0$are $1$ and $-5$ 

By using the quadratic formula

The root of a quadratic equation $a{{x}^{2}}+bx+c=0$ is given by formula 

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ , where $\sqrt{{{b}^{2}}-4ac}$ is known as discriminant. 

If  $\sqrt{{{b}^{2}}-4ac}\ge 0$ then only the root of quadratic equation is given by 

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$\Rightarrow {{x}^{2}}+4x+3$ 

On using quadratic formula, we get

$ \Rightarrow x=\dfrac{-4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\times 1\times 3}}{2\times 1} $

$ \Rightarrow x=\dfrac{-4\pm \sqrt{16-12}}{2}$

$ \Rightarrow x=\dfrac{-4\pm \sqrt{4}}{2}$

$ \Rightarrow x=\dfrac{-4\pm 2}{2} $

$\Rightarrow x=\dfrac{-4+2}{2}$ , $x=-1$ 

$\Rightarrow x=\dfrac{-4-2}{2}$ ,$x=-3$ 

So, the roots of ${{x}^{2}}+4x+3=0$ are $-1$  and $-3$ 

Nature of Roots Based on Discriminant:

If $\sqrt{{{b}^{2}}-4ac}=0$ then the roots are real and equal

If $\sqrt{{{b}^{2}}-4ac}>0$ then the roots are real and distinct 

If $\sqrt{{{b}^{2}}-4ac}<0$then the roots are imaginary

Quadratic Equation Notes – A Short Summary

Sridharacharya derived the quadratic formula in A.D 1025. This chapter is crucial not only from an exam point of view but also to deal with various real-life situations. For instance, you can calculate the length and breadth of a hall using quadratic equations. You can form an equation based on the information that the hall has a carpet area of 300 sq. metres and its length is one metre more than twice its breadth.

In this chapter, you will learn more about quadratic equations and how to find their roots. The Quadratic Equation Class 10 Notes for this chapter have been formulated by expert subject teachers to assist students with their overall preparation for the board exam. For this reason, the revision notes follow the NCERT guidelines to maintain accuracy and a high standard.

The relevant sub-topics under this chapter have been highlighted in our Quadratic Equations Notes to help students go through them quickly while revising. Key concepts and essential definitions come with an in-depth explanation to encourage better comprehension on your part. 

The revision notes are written in lucid language to boost your understanding and help you memorise them quickly.  Read through our revision notes to lend further clarity to your doubts and increase your chances of scoring high grades. Download the Quadratic Equation Class 10 Notes PDF and make them a part of your year-round study schedule.

Quadratic Equation Notes Class 10 – Revision Notes

Quadratic Equations Notes by us are divided into following sub-sections to help you revise efficiently before the examination.

Quadratic Equation Class 10 Notes – Roots of a Quadratic Equation

Under this section, you will learn to calculate the roots of quadratic equations. You should remember that quadratic equations in the variable x are an equation of the form ax 2 + bx + c = 0. Here a, b and c are real numbers. It has been explained in our Quadratic Equations Notes with the help of an example. 

As you go through our Quadratic Equations Notes, you will learn why this form of quadratic equations is called its standard form. Our revision notes will help you to revise how quadratic equations are used to represent real-life situations mathematically based on the information given in the question.

Under this section from our Class 10 Maths Chapter 4 Notes, you will also learn how to find out the roots of quadratic equations using shortcut techniques. While calculating the roots, you must remember that a root is a real number which satisfies the quadratic equation and is not equal to zero.

Quadratic Equation Notes Class 10 – Solution of Quadratic Equations By Factorisation

Quadratic equations can be solved by applying more than one method. In this class, you will learn to factorise quadratic equations to find its roots. However, it would help if you kept in mind that to factorise quadratic polynomials, you must split the middle term. You should also remember that to determine the roots; you must factorise the equation into linear factors and equate each factor to zero.

After solving the quadratic equation and finding its roots, you must verify that these are the roots of the equation given. You can go through our revision notes to review the method correctly. Since this is a very crucial method, you must practise it more than once and refer to our Quadratic Equations Notes to clarify your doubts, if any.

 You can even download our notes for Quadratic Equations Class 10 PDF and make it a part of your exam preparation process. Our notes contain a clear step by step explanation of the factorisation process, which you can easily memorise while revising a day before your board exams.

Class 10 Maths Chapter 4 Notes – Methods of Square

It is yet another method that can be used to solve a quadratic equation and find out its roots. Under this method, the quadratic equation ax 2 + bx + c = 0 is converted to the form (x + a) 2 – b 2 = 0. Now you have to apply your knowledge of square and square roots to solve this equation. Hence, this method is known as the method of completing the square. The terms containing x is completely inside a square, and the roots were found by taking the square roots.

This above method has been explained in our Quadratic Equations Notes with the help of an example. Students are given a quadratic equation x 2 + 4x. Now, employing this method, this will be converted to (x + 2) 2 – 4 = (x + 2) 2 - 2 2 . 

Under this section, you can also recall your knowledge of the quadratic formula. This formula is used to calculate the roots of a quadratic equation. Using this method, you can quickly solve a quadratic equation to find its roots.  So revise this technique again and again to increase your accuracy and speed. Referring to our Quadratic Equation PDF Class 10 will also help you to improve your understanding.

Quadratic Equations Class 10 Notes – Nature of Roots

You will learn about discriminant of a quadratic equation in the form ax 2 + bx + c = 0. Our Quadratic Equations Notes analyses this for your benefit. You can quickly revise the following pointers by downloading our revision notes.

 A quadratic equation ax 2 + bx + c = 0 has

Two distinct real roots, if b 2 – 4ac > 0.

Two equal real roots, if b 2 – 4ac = 0.

No real roots, if b 2 – 4ac < 0.

A clear grasp of this will help you to understand the nature of the roots. Read through our Quadratic Equations Class 10 Notes which has been framed by expert maths teachers to strengthen your understanding of the basic concepts. we have created a table that would let students know everything that they need to know about the nature of the roots of quadratic equations. That table is mentioned below.

The Relationship Between Roots of Quadratic Equations and Coefficients

It is important to understand the relationship between the roots of quadratic equations and coefficients while going through Ch 4 Maths Class 10 Notes. This is why we will go through that exact topic right now.

Let’s begin with the assumption that α and β are roots of a quadratic equation. This quadratic equation is ax 2 + bx + c. This means that:

α + β = -b/a

α – β = ±√[(α + β) 2 – 4αβ]

|α + β| = √D/|a|

Keep all of these equations in mind. From these equations, it can be said that the relationship between the roots and coefficients of a polynomial equation can be derived if one simplifies the given polynomials and substitutes the results. All of this can be depicted by the following equations:

α 2 β + β 2 α = αβ (α + β) = – bc/a 2

α 2 + αβ + β 2 = (α + β) 2 – αβ = (b 2 – ac)/a 2

α 2 + β 2 = (α – β) 2 – 2αβ

α 2 – β 2 = (α + β) (α – β)

α 3 + β 3 = (α + β) 3 + 3αβ(α + β)

α 3 – β 3 = (α – β) 3 + 3αβ(α – β)

(α/β) 2 + (β/α) 2 = α 4 + β 4 /α 2 β 2

The Range of Quadratic Equations

By now, we have looked at almost all the major formulas and theorems related to quadratic equations in these notes of Chapter 4 Maths Class 10. However, the one major topic that is still remaining is finding out the range of quadratic equations. So, let’s dive into this topic now.

Assume that a quadratic expression f (x) = ax 2 + bx + c. In this equation, a is not equal to 0. Also, a, b, and c are real. Hence, we can write the quadratic expression as:

F (x) = x 2 + bax = ca

Now, before moving forward, it is advised that you should look at the images that are given below.The first thing that you need to know about this image is that it shows the intervals in which the roots of a quadratic equation lie. We will look at all of these cases below.

Case 1: When both the roots of the quadratic equation are larger than any given number m

This happens when: b 2 - 4ac = (D) ≥ 0, -b / 2a > m, and f (m) > 0

Case 2: When both the roots of the quadratic expression are less than any given number ‘m’

This is true when b 2 - 4ac = (D) ≥ 0, -b / 2a < m, and f (m) > 0

Case 3: If both the roots of a quadratic equation lie in a particular interval (m 1 , m 2 )

This happens when b 2 - 4ac = (D) ≥ 0, m 1 < -b / 2a > m 2 , f (m 1 ) > 0, and f (m 2 ) > 0

Case 4: This happens when one root of a quadratic equation lies exactly in the given interval (m 1 , m 2 ) and f (m 1 ). f (m 2 ) < 0

Case 5: A given number ‘m’ will lie between the roots of a quadratic equation if f (m) < 0

Case 6: The roots of the quadratic equation will have opposite signs when f (0) < 0

Case 7: Both the roots of the quadratic equations are positive

This is true when b 2 - 4ac = (D) ≥ 0, α + β = -b / a > 0, and α x β = c / a > 0

Case 8: When both the roots of a quadratic equation are negative

This happens when b 2 - 4 ac = (D) ≥ 0, α + β = -b / a < 0, and α x β = c / a < 0

Fun Facts About Quadratic Equations

Did you know that you can use quadratic equations to solve many problems in the real world? This is true as quadratic equations can help you solve problems related to speed and geometry. Many problems related to quadrilateral figures and problems related to distance and time can also be solved by using quadratic equations.

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Other Maths Related Links

The following is a list of links for Maths-related topics that you can check out to improve your current understanding of Class 10 maths chapters and be able to score maximum marks in your exams.  

NCERT Solutions for Class 10

Revision Notes for Class 10

Important Questions for Class 10

CBSE Syllabus for Class 10

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Maths formulas for Class 10

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CBSE Class 10 Revision Notes - Other Chapters

The following are the links to the notes for all the important chapters in CBSE Class 10. We recommend that students visit the mentioned pages to get the most out of what Vedantu has to offer for students who wish to go the extra mile and prepare for their exams with the best materials out on the internet. 

Chapter 1 - Real Numbers Revision Notes

Chapter 2 - Polynomials Revision Notes

Chapter 3 - Pair of Linear Equations in Two Variables Revision Notes

Chapter 5 - Arithmetic Progressions Revision Notes

Chapter 6 - Triangles Revision Notes

Chapter 7 - Coordinate Geometry Revision Notes

Chapter 8 - Introduction to Trigonometry Revision Notes

Chapter 9 - Some Applications of Trigonometry Revision Notes

Chapter 10 - Circles Revision Notes

Chapter 11 - Constructions Revision Notes

Chapter 12 - Areas Related to Circles Revision Notes

Chapter 13 - Surface Areas and Volumes Revision Notes

Chapter 14 - Statistics Revision Notes

Chapter 15 - Probability Revision Notes

Quadratic Equations is a significant chapter in CBSE Class 10 Maths and our experts have worked diligently to come up with study materials and important notes to make sure students appearing for the Class 10 exam are prepared with all the necessary tools and information they need to score desirable marks. We recommend that students must go through these notes as well as the other links provided in this page to make the best out of these chapters.  

FAQs on Quadratic Equations Class 10 Notes CBSE Maths Chapter 4 (Free PDF Download)

1. What is the value of k for a quadratic equation (x - a) (x - 10) + 1 = 0. This equation has integral roots.

We can write the given quadratic equation as x 2 - (10 + k) x + 1 + 10k = 0

This means that, D = b 2 - 4ac = 100 + k 2 + 20k - 40k = k 2 - 20k + 96 = (k - 10) 2 - 4

This quadratic equation will have integral roots if the value of the discriminant > 0. This means that D is a perfect square and a = 1, while b and c are integers

Hence, (k - 10) 2 - D = 4

We know that the discriminant is a perfect square. This further means that the difference between the two perfect squares will be 4 only if D = 0 and (k - 10) 2 = 4

Therefore, k - 10 = ± 2. And the values of k = 8 and 12

2. What is the value of k when the equation p / (x + r) + q / (x - r) = k / 2x has two equal roots?

We can write the given quadratic equation as:

[2p + 2q - k] x 2 - 2r [p - q] x + r 2 k = 0

In case of equal roots, the discriminant or D = 0

This means that b 2 - 4ac = 0

In this equation, a = [2p + 2q - k], b = -2r [p - q], and c = r 2 k

                              = [-2r (p - q)] 2 - 4 (2p + 2q - k) (r 2 k)] = 0

                                  R 2 (p - q) 2 - r 2 k (2p + 2q - k) = 0

We also know that r is not equal to zero.

Hence, (p - q) 2 - k (2p + 2q - k) = 0

K 2 - 2(p + q) k + (p - q) 2

K = 2 (p + q) ± √[4 (p + q) 2 - 4 (p - q)] 2 / 2 = -(p + q) ± √4pq

Therefore, we can conclude that the value of k is (p + q) ± 2 √p x q = (√p ± √q) 2

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Quadratic Equation Worksheet Class 10 PDF with Answers

These Quadratic Equation worksheet PDF can be helpful for both teachers and students. Teachers can track their student’s performance in the chapter Quadratic Equation. Students can easily identify their strong points and weak points by solving questions from the worksheet. Accordingly, students can work on both weak points and strong points. 

All students studying in CBSE class 10th, need to practise a lot of questions for the chapter Quadratic Equation. Students can easily practise questions from the Quadratic Equation problems worksheet PDF. By practising a lot of questions, students can improve their confidence level. With the help of confidence level, students can easily cover all the concepts included in the chapter Quadratic Equation. 

Quadratic Equation Worksheets with Solutions

Solutions is the written reply for all questions included in the worksheet. With the help of Quadratic Equation worksheets with solutions, students can solve all doubts regarding questions. Students can have deep learning in the chapter Quadratic Equation by solving all their doubts. By solving doubts, students can also score well in the chapter Quadratic Equation. 

Quadratic Equation Worksheet PDF

Worksheet is a sheet which includes many questions to solve for class 10th students. The Quadratic Equation worksheet PDF provides an opportunity for students to enhance their learning skills. Through these skills, students can easily score well in the chapter Quadratic Equation. Students can solve the portable document format (PDF) of the worksheet from their own comfort zone. 

How to Download the Quadratic Equation Worksheet PDF? 

To solve questions from the Quadratic Equation worksheet PDF, students can easily go through the given steps. Those steps are- 

• Open Selfstudys website.

• Bring the arrow towards CBSE which can be seen in the navigation bar. 

• Drop down menu will appear, select KVS NCERT CBSE Worksheet. 

• A new page will appear, select class 10th from the given list of classes. 
• Select Mathematics from the given list of subjects. Now click the chapter’s name that is Quadratic Equation. 

Features of the Quadratic Equation Worksheet PDF

Before starting to solve questions from the Quadratic Equation problems worksheet PDF, students need to know everything about the worksheet. Those features are- 

• Variety of questions are included:  The Quadratic Equation Maths Worksheet for Class 10 includes varieties of questions. Those varieties of questions are- one mark questions, two mark questions, three mark questions, etc. 
• Solutions are provided:  Doubts regarding each question can be easily solved through the solutions given. Through solving questions, a student's comprehensive skill can be increased. 
• All concepts are covered:  By solving questions from the Quadratic Equation problems worksheet, students can easily cover all the concepts included in the chapter.  
• Created by Expert:  These worksheets are personally created by the subject experts. These Quadratic Equation worksheet pdf are created with proper research. 
• Provides plenty of questions:  The Quadratic Equation worksheet provides plenty of questions to practise. Through good practice, students can get engaged in the learning process. 

Benefits of the Quadratic Equation Worksheet PDF

With the help of Quadratic Equation problems worksheet PDF, students can easily track their performance. This is the most crucial benefit, other than this there are more benefits. Those benefits are- 

• Builds a strong foundation:  Regular solving questions from the worksheet can help students to build a strong foundation. Through the strong foundation, students can score well in the chapter Quadratic Equation. 
• Improves speed and accuracy:  While solving questions from the chapter Quadratic Equation, students need to maintain the speed and accuracy. Speed and accuracy can be easily maintained and improved by solving questions from the Quadratic Equation worksheet PDF. 
• Acts as a guide:  Quadratic Equation worksheets with solutions acts as guide for both the teachers and students. Through the worksheet, teachers can guide their students according to the answers given by them. Students can also analyse themselves with the help of answers and can improve accordingly. 
• Enhances the learning process:  Regular solving of questions from the worksheet can help students enhance their learning process. According to the learning skills, students can easily understand all topics and concepts included in the chapter Quadratic Equation.  
• Improvisation of grades:  Regular solving of questions from the worksheet can help students to improve their marks and grades. With the help of good marks and good grades, students can select their desired field further. 

Tips to Score Good Marks in Quadratic Equation Worksheet

Students are requested to follow some tips to score good marks in the Quadratic Equation worksheet. Those tips are-

• Complete all the concepts:  First and the most crucial step is to understand all the concepts included in the chapter Quadratic Equation.  
• Practise questions:  Next step is to practise questions from the Quadratic Equation problems worksheet. Through this students can identify all types of questions: easy, moderate, difficult, etc.  
• Note down the mistakes:  After practising questions, students need to note down the wrong sums that have been done earlier. 
• Rectify the mistakes:  After noting down the mistakes, students need to rectify all the mistakes made. 
• Maintain a positive attitude:  Students are requested to maintain a positive attitude while solving worksheets. By maintaining a positive attitude, students can improve speed and accuracy while solving the worksheets. 
• Remain focused:  Students need to remain focused while solving questions from the Quadratic Equation problems worksheet pdf. As it helps students to solve the questions as fast as possible. 

When should a student start solving the Quadratic Equation Worksheet PDF?

Students studying in class 10 should start solving worksheets after covering each and every concept included in the chapter. Regular solving questions from the Quadratic Equation worksheet PDF, can help students to have a better understanding of the chapter. Better understanding of the chapter Quadratic Equation can help students to score well in the class 10th board exam. 

Regular solving questions from the Quadratic Equation Worksheet PDF can help students to build a strong foundation for the chapter Quadratic Equation. Strong foundation of the chapter Quadratic Equation can help students to understand further chapters. 

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Important Questions for Class 10 Maths Chapter 4 Quadratic Equations

August 3, 2019 by Sastry CBSE

Quadratic Equations Class 10 Important Questions Very Short Answer (1 Mark)

Question 2. If 1 is a root of the equations ay 2 + ay + 3 = 0 and y 2  + y + b = 0, then find the value of ab. (2012D) Solution: ay 2 + ay + 3 = 0 a(1) 2 + a(1) + 3 = 0 2a = -3 a = \(\frac{-3}{2}\)

y 2 + y + b = 0 1 2 + 1 + b = 0 b = -2 ∴ ab =\(\left(\frac{-3}{2}\right)(-2)\) = 3

Question 4. If the quadratic equation px\frac{1}{2} – 2\(\sqrt{5}\) px + 15 = 0 has two equal roots, then find the value of p. (2015OD) Solution: The given quadratic equation can be written as px\frac{1}{2} – 2\(\sqrt{5}\) px + 15 = 0 Here a = p, b = – 2\(\sqrt{5}\) p, c = 15 For equal roots, D = 0 D = b 2 – 4ac – 0 …[∵ Equal roots 0 = (-2\(\sqrt{5}\)p) 2 – 4 × p × 15 0 = 4 × 5p 2 – 60p 0 = 20p 2 – 60p => 20p 2 = 60p p = \(\frac{60 p}{20 p}\) = 3 ∴ p = 3

Quadratic Equations Class 10 Important Questions Short Answer-I (2 Marks)

Question 5. Find the value of p so that the quadratic equation px(x – 3) + 9 = 0 has two equal roots. (2011D, 2014OD) Solution: We have, px (x – 3) + 9 = 0 px 2 – 3px + 9 = 0 Here a = p, b = -3p, D = 0 b 2 – 4ac = 0 ⇒ (-3p) 2 – 4(p)(9) = 0 ⇒ 9p 2 – 36p = 0 ⇒ 9p (p – 4) = 0 ⇒ 9p = 0 or p – 4= 0 p = 0 (rejected) or p = 4 ∴ p = 4 ……..(∵ Coeff. of x 2 cannot be zero

Question 7. Find the value of m so that the quadratic equation mx (x – 7) + 49 = 0 has two equal roots. (2011OD) Solution: We have, mx (x – 7) + 49 = 0 mx 2 – 7mx + 49 = 0 Here, a = m, b = – 7m, c = 49 D = b 2 – 4ac = 0 …[For equal roots ⇒ (-7m) 2 – 4(m) (49) = 0 ⇒ 49m 2 – 4m (49) = 0 ⇒ 49m (m – 4) = 0 ⇒ 49m = 0 or m – 4 = 0 m = 0 (rejected) or m = 4 ∴ m = 4

Question 8. Solve for x: 36x 2 – 12ax + (a 2 – b 2 ) = 0 (2011OD) Solution: We have, 36x 2 – 12ax + (a 2 – b 2 ) = 0 ⇒ (36x 2 – 12ax + a 2 ) – b 2 = 0 ⇒ [(6x) 2 – 2(6x)(a) + (a) 2 ] – b 2 = 0 ⇒ (6x – a) 2 – (b) 2 = 0 …[∵ x 2 – 2xy + y 2 = (x – y) 2 ⇒ (6x – a + b) (6x – a – b) = 0 „[∵ x 2 – y 2 = (x + y)(x – y) ⇒ 6x – a + b = 0 or 6x – a – b = 0 ⇒ 6x = a – b or 6x = a + b ⇒ x = \(\frac{a-b}{6}\) or \(\frac{a+b}{6}\)

Question 9. Find the value(s) of k so that the quadratic equation x 2 – 4kx + k = 0 has equal roots. (2012D) Solution: We have, x 2 – 4kx + k = 0 Here a = 1, b = -4k:, c = k D = 0 …[Since, Equal roots As b 2 – 4ac = 0 ⇒ (-4k) 2 – 4(1) (k) = 0 ⇒ 16k 2 – 4k = 0 ⇒ 4k(4k – 1) = 0 ⇒ 4k = 0 or 4k – 1 = 0 k = 0 (rejected) or 4k = 1 ∴ k = \(1 / 4\)

Question 10. Find the value of k for which the equation x 2 + k(2x + k – 1) + 2 = 0 has real and equal roots. (2017D) Solution: We have, x 2 + k(2x + k – 1) + 2 = 0 x 2 + 2kx + k 2 – k + 2 = 0 Here a = 1, b = 2k, c = k 2 – k + 2 D = 0 …[real and equal roots ∴ b 2 – 4ac = 0 ⇒ (2k) 2 – 4 × 1(k 2 – k + 2) = 0 ⇒ 4k 2 – 4 (k 2 – k + 2) = 0 ⇒ 4(k 2 – k 2 + k – 2) = 0 ⇒ 4(k – 2) = 0 ⇒ k – 2 = 0 ⇒ k = 2

Question 11. Find the value of p for which the roots of the equation px(x – 2) + 6 = 0, are equal. (2012OD) Solution: We have , px(x – 2) + 6 = 0 px 2 – 2px + 6 = 0, p ≠ 0 Two equal roots …[Given b 2 – 4ac = 0 ….[a = p, b = -2p, c = 6 ∴ (-2p) 2 – 4(p)(6) = 0 4p 2 – 24p = 0 ⇒ 4p(p – 6) = 0 4p = 0 or p – 6 = 0 p = 0 (rejected) or p = 6 Since p cannot be equal to 0. …[Standard form of a quad. eq. ax 2 + bx + c = 0, a ≠ 0 ∴ P = 6

Question 14. Solve the quadratic equation 2x 2 + ax – a 2 = 0 for x. (2014D) Solution: We have, 2x 2 + ax – a 2 = 0 2x 2 + 2ax – ax – a 2 = 0 2x(x + a) – a(x + a) = 0 (x + a) (2x – a) = 0 x + a = 0 or 2x – a = 0 ∴ x = -a or x = \(\frac{a}{2}\) Alternatively: First calculate D = b 2 – 4ac Then apply x = \(\frac{-b \pm \sqrt{D}}{2 a}\) We get x = -a, x = \(\frac{a}{2}\)

Question 15. Find the values of p for which the quadratic equation 4x 2 + px + 3 = 0 has equal roots. (2014OD) Solution: Given: 4x 2 + px + 3 = 0 Here a = 4, b = p. (= 3 … [Equal roots D = 0 (Equal roots) As b 2 – 4ac = 0 ∴ (p) 2 – 4(4)(3) = 0 = p 2 – 48 = 0 ⇒ p 2 = 48 ∴ p = \(\pm \sqrt{16 \times 3}=\pm 4 \sqrt{3}\)

Question 16. Solve the following quadratic equation for x: 4x 2 – 4a 2 x + (a 4 – b 4 ) = 0. (2015D) Solution: The given quadratic equation can be written as, 4x 2 – 4a 2 x + (a4 4 – b 4 ) = 0 (4x 2 – 4a 2 x + a 4 ) – b 4 = 0 or (2x – a 2 ) 2 – (b 2 ) 2 = 0 ⇒ (2x – a 2 + b 2 ) (2x – a 2 – b 2 ) = 0 ⇒ (2x – a 2 + b 2 ) = 0 or (2x – a 2 – b 2 ) = 0 ∴ x = \(\frac{a^{2}-b^{2}}{2}\) or x = \(\frac{a^{2}+b^{2}}{2}\)

Question 19. Solve the following quadratic equation for x: x 2 – 2ax – (4b 2 – a 2 ) = 0) (2015OD) Solution: Given quadratic equation can be written as x 2 – 2ax – 4b 2 + a 2 = 0. (x 2 – 2ax + a 2 ) – 4b 2 = 0 or (x – a) 2 – (2b) 2 = 0 As we know, [a 2 – b 2 = (a + b)(a – b)] ∴ (x – a + 2b) (x – a – 2b) = 0 ⇒ x – a + 2b = 0 or x – a – 2b = 0 ⇒ x = a – 2b or x = a + 2b ⇒ x = a – 2b and x = a + 2b

Question 22. If -5 is a root of the quadratic equation 2x 2 + px – 15 = 0 and the quadratic equation p(x 2 + x) + k = 0 has equal roots, find the value of k. (2016OD) Solution: We have, 2x 2 + px – 15 =0 Since (-5) is a root of the given equation ∴ 2(-5) 2 + p(-5) – 15 = 0 ⇒ 2(25) – 5p – 15 = 0 ⇒ 50 – 15 = 5p ⇒ 35 = 5p ⇒ p = 7 …(i) Now, p(x 2 + x) + k ⇒ px 2 + px + k = 0 7x 2 + 7x + k = 0 …[From (i) Here, a = 7, b = 7, c = k D = 0 …[Roots are equal b 2 – 4ac = 0 ⇒ (7) 2 – 4(7)k = 0 ⇒ 49 – 28k = 0 ⇒ 49 = 28k ∴ k = \(\frac{49}{28}=\frac{7}{4}\)

Question 23. Solve for x: \(\sqrt{2 x+9}\) + x = 13. (20160D) Solution: \(\sqrt{2 x+9}\) + x = 13 …(i) ⇒ \(\sqrt{2 x+9}\) = 13 – x ⇒ (\(\sqrt{2 x+9}\)) 2 = (13 – x) 2 …[Squaring both sides ⇒ 2x + 9 = 169 + x 2 2 – 26x ⇒ 0 = 169 + x 2 2 – 26x – 2x – 9 ⇒ x 2 – 28x + 160 = 0 ⇒ x 2 – 20x – 8x + 160 = 0 ⇒ x(x – 20) – 8(x – 20) = 0 ⇒ (x – 20) (x – 8) = 0 ⇒ x – 8 = 0 or x – 20 = 0 ⇒ x = 8 or x = 20 Checking, When x = 8 in (i) \(\sqrt{2(8)+9}\) + 8 = 13 \(\sqrt{25}\) + 8 = 13 5 + 8 = 13 ⇒ 13 = 13 …[True ∴ x = 8 is the solution. Checking, When x = 20 in (i), \(\sqrt{2(20)+9}\) + 20 = 13 … [From (1) \(\sqrt{49}\) + 20 = 13 7 + 20 ≠ 13 …[False ∴ x = 20 is not a solution. Therefore, x = 8 is the only solution.

Question 24. Solve for x: \(\sqrt{6 x+7}\) – (2x – 7) = 0 (20160D) Solution: \(\sqrt{6 x+7}\) – (2x – 7) = 0…(i) ⇒ \(\sqrt{6 x+7}\) = 2x – 7 ⇒ (\(\sqrt{6 x+7}\)) 2 = (2x – 7) 2 …[Squaring both sides ⇒ 6x + 7 = 4x 2 – 28x + 49 ⇒ 0 = 4x 2 – 28x – 6x – 7 + 49 ⇒ 4x 2 – 34x + 42 = 0 ⇒ 2x 2 – 17x + 21 = 0 …[Dividing by 2 ⇒ 2x 2 – 14x – 3x + 21 = 1 ⇒ 2x(x – 7) – 3(x – 7) = 0 ⇒ (x – 7) (2x – 3) = 0 ⇒ x – 7 = 0 or 2x – 3 = 0 = x= 7 or x = \(\frac{3}{2}\) Checking: When x = 7 in (i), \(\sqrt{6(7)+7}\) – [2(7) – 7] = 0 \(\sqrt{49}\) – (14 – 7) = 0 7 – 7 = 0 … [True Checking: When x = \(\frac{3}{2}\) in (i), \(\sqrt{6\left(\frac{3}{2}\right)+7}-\left(2\left(\frac{3}{2}\right)-7\right)\) = 0 \(\sqrt{9+7}\) – (3 – 7) = 0 \(\sqrt{16}\) – (-4) = 0 4 + 4 ≠ 0 ∴ x = 7 is the only solution.

Quadratic Equations Class 10 Important Questions Short Answer-II (3 Marks)

Question 26. Solve for x: 4x 2 – 4ax + (a 2 – b 2 ) = 0 (2011OD) Solution: 4x 2 – 4ax + (a 2 – b 2 ) = 0 ⇒ [4x 2 – 4ax + a 2 ] – b 2 = 0 ⇒ [(2x) 2 – 2(2x)(a) + (a) 2 ] – b 2 = 0 ⇒ (2x – a) 2 – (b) 2 = 0 ⇒ (2x – a + b) (2x – a – b) = 0 ⇒ 2x – a + b = 1 or 2x – a – b = 0 2x = a – b or 2x = a + b ∴ x = \(\frac{a-b}{2}\) or x = \(\frac{a+b}{2}\)

Question 27. Solve for x: 3x 2 } – 2\(\sqrt{6}\) x + 2 = 0 (2012D) Solution: 3x 2 } – 2\(\sqrt{6}\) x + 2 = 0 ⇒ 3x 2 – \(\sqrt{6}\)x – \(\sqrt{6}\)x + 2 = 0 ⇒ \(\sqrt{3}\)x (\(\sqrt{3}\)x – \(\sqrt{2}\) – \(\sqrt{2}\)(\(\sqrt{3}\)x – \(\sqrt{2}\)) = 0 ⇒ (\(\sqrt{3}\)x – \(\sqrt{2}\))(\(\sqrt{3}\)x – \(\sqrt{2}\)) = 0 ⇒ \(\sqrt{3}\)x – \(\sqrt{2}\) = 0 ⇒ x = \(\frac{\sqrt{2}}{\sqrt{3}}\) ∴ x = \(\frac{\sqrt{6}}{3}\) ….[\(\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}\)

Question 28. Find the value(s) of k so that the quadratic equation 2x 2 + kx + 3 = 0 has equal roots. (2012D) Solution: Given: 2x 2 + kx + 3 = 0 Here a = 2, b = k, c= 3 D = 0 … [Since roots are equal As b 2 – 4ac = 0 ∴ K 2 – 4(2)(3) = 0 K 2 – 24 = 0 or k 2 = 24 ∴ k = \(\sqrt{2 \times 2 \times 6}=\pm 2 \sqrt{6}\)

Question 29. Find the value(s) of k so that the quadratic equation 3x 2 – 2kx + 12 = 0 has equal roots. (2012D) Solution: Given: 3x 2 – 2kx + 12 = 0 Here a = 3, b = -2k, c = 12 D = 0 … [Since roots are equal As b 2 – 4ac = 0 ∴ (-2k) 2 – 4(3) (12) = 0 ⇒ 4k 2 – 144 = 0 ⇒ k 2 = \(\frac{144}{4}\) = 36 ∴ k = \(\pm \sqrt{36}=\pm 6\)

Question 30. Solve the following quadratic equation for x: x 2 – 4ax – b 2 + 4a 2 = 0 (2012OD) Solution: We have, x 2 – 4ax – b 2 + 4a 2 = 0 ⇒ x 2 – 4ax + 4a 2 – b\frac{144}{4} = 0 ⇒ [(x)\frac{144}{4} – 2(x)(2a) + (2a)\frac{144}{4}] – (b) 2 = 0 (x – 2a) 2 – (b) 2 = 0 (x – 2a + b) (x – 2a – b) = 0 x – 2a + b = 0 or x – 2a – b = 0 ∴ x = 2a – b or x = 2a + b

Question 31. Find the value of k for which the roots of the equation kx(3x – 4) + 4 = 0, are equal. (20120D) Solution: We have, kx(3x – 4) + 4 = 0 3kx 2 – 4kx + 4 = 0 Here a = 3k, b = -4k, c = 4 D = 0 … [Since roots are equal b 2 – 4ac = 0 ∴ (-4k) 2 – 4(3k) (4) = 0 16k 2 – 48k = 0 16k (k – 3) = 0 16k = 0 or k – 3 = 0 k = 0 or k = 3 …[Rejecting k = 0, as coeff. of x 2 cannot be zero ∴ k = 3

Question 32. Find the value of m for which the roots of the equation. mx (6x + 10) + 25 = 0, are equal. (2012OD) Solution: We have, mx(6x + 10) + 25 = 0 6mx 2 + 10mx + 25 = 0 Here a = 6m, b = 10m, c = 25 D = 0 … Since roots are equal b 2 – 4ac = 0 ∴ (10m) 2 – 4(6m) (25) = 0 100m 2 – 600m = 0 ⇒ 100m (m – 6) = 0 100m = 0 or m – 6 = 0 m = 0 or m = 6 …[Rejecting m = 0, as coeff. of x 2 cannot be zero ∴ m = 6

Question 33. For what value of k, the roots of the quadratic equation kx(x – 2\(\sqrt{5}\)) + 10 = 0, are equal? (2013D) Solution: We have, kx(x – 2\(\sqrt{5}\)) + 10 = 0 kx 2 – 2\(\sqrt{5}\)kx + 10 = 0 Here a = k, b= -2\(\sqrt{5}\)k, c= 10 D = 0 …[∵ Roots are equal As b 2 – 4ac = 0 ∴ (-2 \(\sqrt{5}\)k) 2 – 4(k)(10) = 0 20k 2 – 40k = 0 ⇒ 20k(k – 2) = 0 ∴ 20k = 0 or k – 2 = 0 k = 0 (rejected) or k = 2 …[∵ Coeff. of x 2 cannot be zero ∴ k= 2

Question 34. For what values of k, the roots of the quadratic equation (k + 4)x 2 + (k + 1)x + 1 = 0 are equal? (2013D) Solution: We have, (k + 4) x 2 + (k + 1) x + 1 = 0 Here, a = k + 4, b = k + 1, c = 1 D =0 …[∵ Roots are equal b 2 – 4ac = 0 ∴ (k + 1) 2 – 4(k + 4)(1) = 0 k 2 + 2k + 1 – 4k – 16 = 0 k 2 – 2k – 15 = 0 k 2 – 5k + 3k – 15 = 0 k(k – 5) + 3(k – 5) = 0 (k – 5)(k + 3) = 0 k – 5 = 0 or k + 3= 0 k = 5 or k = -3 ∴ k = 5 and -3

Question 35. For what value of k, are the roots of the quadratic equation: (k – 12)x 2 + 2(k – 12)x + 2 = 0 equal? (2013OD) Solution: We have, (k – 12)x 2 + 2(k – 12)x + 2 = 0 The given quadratic equation will have equal roots if D = 0 ⇒ b 2 – 4ac = 0 Here a = (k – 12), b = 2(k – 12), c = 2 b 2 – 4ac = 0 0 = 4(k – 12) 2 – 4 × (k – 12) × 2 0 = (k – 12)[4(k – 12) – 4 × 2] 0 = (k – 12) 4[k – 12 – 2] 0 = 4(k – 12) (k – 14) ∴ 4(k – 12)(k – 14) = 0 ∴ k = 12 (rejected) or k = 14 But k cannot be equal to 12 because in that case the given equation will imply 2 = 0 which is not true. ∴ k = 14

Question 36. For what value of k, are the roots of the quadratic equation y 2 + k 2 = 2 (k + 1)y equal? (2013OD) Solution: y 2 + k 2 = 2(k + 1)y y 2 – 2(k + 1)y + k 2 = 0 Here a = 1, b = -2(k + 1), c = k 2 D = 0 … [Roots are equal b 2 – 4ac = 0 ∴ [-2(k + 1)] 2 – 4 × (1) × (k 2 ) = 0 ⇒ 4(k 2 + 2k + 1) – 4k 2 = 0 ⇒ 4k 2 + 8k + 4 – 4k 2 = 0 ⇒ 8k + 4 = 0 ⇒ 8k = -4 ∴ k = \(\frac{-4}{8}=\frac{-1}{2}\)

Question 37. Find that non-zero value of k, for which the quadratic equation kx 2 + 1 – 2(k – 1)x + x 2  = 0 has equal roots. Hence find the roots of the equation. (2015D) Solution: The given quadratic equation can be written as kx 2 + x 2 – 2(k – 1)x + 1 = 0 (k + 1) x 2 – 2 (k – 1) x + 1 = 0 …(i) Here, a = (k + 1), b = -2(k – 1), c = 1 For equal roots, D = 0 D = b 2 – 4ac ⇒ (-2(k – 1)] 2 – 4 × (k + 1) × 1 = 0 ⇒ 4(k − 1) 2 – 4(k + 1) = 0 ⇒ 4k 2 + 4 – 8k – 4k – 4 = 0 ⇒ 4k 2 – 12k = 0 ⇒ 4k(k – 3) = 0 k = 3 or k = 0 (rejected) ∴ k = 3 Putting k = 3 put in equation (i), we get ⇒ 4x 2 – 4x + 1 = 0 ⇒ 4x 2 – 2x – 2x + 1 = 0 ⇒ 2x(2x – 1) – 1(2x – 1) = 0 ⇒ (2x – 1) (2x – 1) = 0 ⇒ 2x – 1 = 0 or 2x – 1 = 0 ⇒ x = \(\frac{1}{2}\) or x = \(\frac{1}{2}\) Roots are \(\frac{1}{2} \text { and } \frac{1}{2}\)

Question 38. Find that value of p for which the quadratic equation (p + 1)x 2 – 6(p + 1)x + 3 (p + 9) = 0, p ≠ -1 had equal roots. (2015D( Solution: For the given quadratic equation to have equal roots, D = 0 Here a = (p + 1), b = -6(p + 1), c = 3(p + 9) D = b 2 – 4ac ⇒ [-6(p + 1)] 2 – 4(p + 1).3 (p + 9) = 0 ⇒ 36(p + 1) 2 – 12(p + 1) (p + 9) = 0 ⇒ 12(p + 1) (3p + 3 – p – 9) = 0 ⇒ 12(p + 1)(2p – 6) = 0 ⇒ 24(p + 1)(p – 3) = 0 ⇒ p + 1 = 0 or p – 3 = 0 ⇒ p = -1 (rejected) or p = 3 ∴ p = 3

Question 41. If the roots of the quadratic equation (a – b)x 2 + (b – c)x + (c – a) = 0 are equal, prove that 2a = b + c. (2016OD) Solution: Here’a’ = a – b, ‘b’ = b – c, ‘c’ = c – a D = 0 ….[Roots are equal b 2 – 4ac = 0 ⇒ (b – c) 2 – 4(a – b)(c – a) = 0 ⇒ b 2 + c 2 – 2bc – 4(ac – a 2 – bc + ab) = 0 ⇒ b 2 + c 2 – 2bc – 4ac + 4a 2 + 4bc – 4ab = 0 ⇒ 4a 2 + b2 2 + c 2 – 4ab + 2bc – 4ac = 0 ⇒ (-2a) 2 + (b) 2 + (c)2 2 + 2(-2a)(b) + 2(b)(c) + 2(c)(-2a) = 0 ⇒ [(-2a) + (b) + (c)] 2 = 0 ….[∵ x 2 + y 2 + z 2 + 2xy + 2yz + 2zx = (x + y + z) 2 Taking square-root on both sides -2a + b + c = 0 ⇒ b + c = 2a ∴ 2a = b + c

Question 49. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers. (2016OD) Solution: Let three consecutive natural numbers are x, x + 1, x + 2. According to the question, (x + 1) 2 – [(x + 2) 2 – x 2 ] = 60 ⇒ x 2 + 2x + 1 – (x 2 + 4x + 4 – x 2 ) = 60 ⇒ x 2 + 2x + 1 – 4x – 4 – 60 = 0 ⇒ x 2 – 2x – 63 = 0 ⇒ x 2 – 9x + 7x – 63 = 0 ⇒ x(x – 9) + 7(x – 9) = 0 ⇒ (x – 9) (x + 7) = 0 ⇒ x – 9 = 0 or x + 7 = 0 ⇒ x = 9 or x = -7 Natural nos. can not be -ve, ∴ x = 9 ∴ Numbers are 9, 10, 11.

Question 50. If the sum of two natural numbers is 8 and their product is 15, find the numbers. (2012OD) Solution: Let the numbers be x and (8 – x). According to the Question, x(8 – x) = 15 ⇒ 8x – x 2 = 15 ⇒ 0 = x 2 – 8x + 15 ⇒ x 2 – 5x – 3x + 15 = 0 ⇒ x(x – 5) – 3(x – 5) = 0 ⇒ (x – 3)(x – 5) = 0 x – 3 = 0 or x – 5 = 0 x = 3 or x = 5 When x = 3, numbers are 3 and 5. When x = 5, numbers are 5 and 3.

Quadratic Equations Class 10 Important Questions Long Answer (4 Marks)

Question 51. Find the values of k for which the quadratic equation (3k + 1)x 2 + 2(k + 1)x + 1 = 0 has equal roots. Also find the roots. (2014D) Solution: (3k + 1)x 2 + 2(k + 1) + 1 = 0 Here, a = 3k + 1, b = 2(k + 1), c = 1 D = 0 …[∵ Roots are equal As b 2 – 4ac = 0 ∴ [2(k + 1)] 2 – 4(3k + 1)(1) = 0 4(k + 1) 2 – 4(3k + 1) = 0 4(k 2 + 2k + 1 – 3k – 1) = 0 (k 2 – k) = \(\frac{0}{4}\) ⇒ k(k – 1) = 0 k = 0 or k – 1 = 0 ∴ k = 0 or k = 1 Roots are x = \(\frac{-b}{2 a}\) ..[As equal roots (Given) x = \(\frac{-2(k+1)}{2(3 k+1)}\) ⇒ x = \(\frac{-(k+1)}{(3 k+1)}\) When k = 0, x = \(\frac{-(0+1)}{0+1}\) = -1 ∴ Equal roots are -1 and -1 When k = 1, x = \(\frac{-(1+1)}{3+1}\) x= \(\frac{-2}{4}=\frac{-1}{2}\) ∴ Equal roots are \(\frac{-1}{2} \text { and } \frac{-1}{2}\)

Question 63. Sum of the areas of two squares is 400 cm 2 . If the difference of their perimeters is 16 cm, find the sides of the two squares. (2013D) Solution: Let the side of Large square = x cm Let the side of small square = y cm According to the Question, x 2 + y 2 = 400… (i) …[∵ area of square = (side) 2 4x – 4y = 16 …[∵ Perimeter of square = 4 sides ⇒ x – y = 4 … [Dividing both sides by 4 ⇒ x = 4 + y …(ii) Putting the value of x in equation (i), (4 + y) 2 + y2 2 = 400 ⇒ y 2 + 8y + 16 + y 2 – 400 = 0 ⇒ 2y 2 + 8y – 384 = 0 ⇒ y 2 + 4y – 192 = 0 … [Dividing both sides by 2 ⇒ y 2 + 16y – 12y – 192 = 0 ⇒ y(y + 16) – 12(y + 16) = 0 ⇒ (y – 12)(y + 16) = 0 ⇒ y – 12 = 0 or y + 16 = 0 ⇒ y = 12 or y = -16 … [Neglecting negative value ∴ Side of small square = y = 12 cm and Side of large square = x = 4 + 12 = 16 cm

Question 64. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field. (2015OD) Solution: Let the length of shorter side be x m. ∴ length of diagonal = (x + 16) m and length of longer side = (x + 14) m Using pythagoras theorem, (l) 2 + (b) 2 = (h) 2 ∴ x 2 + (x + 14)2 2 = (x + 16) 2 ⇒ x 2 + x 2 + 196 + 28x = x 2 + 256 + 32x ⇒ x 2 – 4x – 60 = 0 ⇒ x 2 – 10x + 6x – 60 = 0 ⇒ x(x – 10) + 6(x – 10) = 0 ⇒ (x – 10) (x + 6) = 0 ⇒ x – 10 = 0 or x + 6 = 0 ⇒ x = 10 or x = -6 (Reject) ⇒ x = 10 m …[As length cannot be negative Length of shorter side = x = 10 m Length of diagonal = (x + 16) m = 26 m Length of longer side = (x + 14)m = 24m ∴ Length of sides are 10 m and 24 m.

Question 65. The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers. (2016D) Solution: Let three numbers in A.P. are a – d, a, a + d. a – d + a + a + d = 12 ⇒ 3a = 12 ⇒a = 4 (a – d) 3 + (a) 3 + (a + d) 3 = 288 ⇒ a 3 – 3a 2 d + 3ad 2 – d 3 + a 3 + a 3 + 3a 2 d + 3ad 2 + d 3 = 288 ⇒ 3a 3 + 6ad 2 = 288 ⇒ 3a(a 2 + 2d 2 ) = 288 ⇒ 3 × 4(4 2 + 2d 2 ) = 288 ⇒ (16 + 2d 2 ) = \(\frac{288}{12}\) ⇒ 2d 2 = 24 – 16 = 8 ⇒ d 2 = 4 ⇒ d = ± 2 When, a = 4, d = 2, numbers are – a – d, a, a + d, i.e., 2, 4, 6 When, a = 4, d = -2, numbers are – a – d, a, a + d, i.e., 6, 4, 2

Question 67. The sum of two numbers is 9 and the sum of their reciprocals is \(\frac{1}{2}\). Find the numbers. (2012D) Solution: Let the numbers be x and 9 – x. According to the Question, \(\frac{1}{x}+\frac{1}{9-x}=\frac{1}{2}\) \(\frac{9-x+x}{x(9-x)}=\frac{1}{2}\) ⇒ 18 = 9x – x 2 ⇒ x 2 – 9x + 18 = 0 ⇒ x 2 – 3x – 6x + 18 = 0 ⇒ x(x – 3) – 6(x – 3) = 0 ⇒ (x – 3) (x – 6) = 0 ⇒ x – 3 = 0 or x – 6 = 0 ⇒ x = 3 or x = 6 When x = 3, nos. are 3 and 6. When x = 6, nos. are 6 and 3.

Question 72. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the rectangular park. (2016OD) Solution: Let length of the rectangular park = x m, breadth of the rectangular park = (x -3)m ∴ Area of the rectangular park = x(x – 3)m 2 … (i) Base of an isosceles triangle = (x – 3)m Altitude of an isosceles triangle = 12 m ∴ Area of isosceles triangle = 1/2 × base × altitude = 1/2 × (x – 3) × 12 = 6(x – 3) …(ii) According to the question, Ar.(rectangle) – Ar.(isosceles ∆) = 4 m 2 ⇒ x(x – 3) – 6(x – 3) = 4 … [From (i) & (ii) ⇒ x 2 – 3x – 6x + 18 – 4 = 0 ⇒ x 2 – 9x + 14 = 0 ⇒ x 2 – 7x – 2x + 14 = 0 ⇒ x(x – 7) – 2(x – 7) = 0 ⇒ (x – 2) (x – 7) = 0 ⇒ x – 2 = 0 or x – 7 = 0 ⇒ x = 2 or x = 7 When x = 2, breadth of rectangle becomes -ve, so this is not possible. ∴ Length of the rectangular park, x = 7 m and Breadth = (x – 3) = 4 m.

Important Questions for Class 10 Maths

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100+ Maths Questions for Class 9 with Answers for Competitive Exams: Download Free PDF

• Updated on  
• Sep 20, 2024

This blog post presents a collection of over 100 carefully curated math questions for Class 9 for competitive exams. These questions cover a wide range of topics, from basic arithmetic to more advanced concepts like algebra, geometry, and trigonometry. By practising these questions, you can enhance your problem-solving skills, improve your understanding of mathematical concepts, and gain confidence in your ability to tackle challenging problems. Also, you can download the Maths questions for Class 9 with answers PDF here. Keep reading!!

Table of Contents

• 1.1 Number Systems
• 1.2 Polynomials
• 1.3 Coordinate Geometry
• 1.4 Lines and Angles
• 1.5 Triangles
• 1.6 Quadratic Equations
• 1.7 Area of Parallelogram and Triangle
• 1.8 Mensuration
• 1.9 Statistics
• 1.10 Probability
• 2 Maths Questions for Class 9: PDF Link Available
• 3 Importance of Maths Questions for Class 9 Students
• 4 Tips for Preparing Maths Questions for Class 9

Sample Maths Questions for Class 9 with Answers

Solve Maths Questions for Class 9 of number systems, polynomial equations, coordinate geometry, lines and angles, triangles, quadratic equations, menstruation, statistics, and probability. 

Number Systems

1. Simplify: 3√27​×3√8​

Solution: 3×2=6

2. Rationalize the denominator: 1/(√5​−2)​

Solution: √5+2 / (√5​−2)(√5​+2) = √5−4/5​+2​ = √5​+2

Polynomials

3. Find the product: (x+3)(x−2)

Solution: x 2 −2x+3x−6 = x 2 +x−6

4. Factorize: x 2 −9

Solution: (x+3)(x−3)

5. Divide: (x 3 −3x 2 +2x+5)÷(x−2)

Solution: Using synthetic division, we get a quotient of x 2 −x+1 and a remainder of 7.

Coordinate Geometry

6. Find the distance between points A(2, 3) and B(5, 7).

Solution: d = √(5−2) 2 +(7−3) 2 ​ = √9+16 ​= 5

7. Find the midpoint of the line segment joining points A(3, 5) and B(7, 9).

Solution: M=((3+7)/2​,(5+9)/2​) = (5,7)

Lines and Angles

8. If two lines intersect, prove that the vertically opposite angles are equal.

Solution: This is a standard geometry proof that involves using the properties of linear pairs and adjacent angles.

9. If the angles in a triangle are in the ratio 2:3:7, find the angles.

Solution: Let the angles be 2x, 3x, and 7x. Since the sum of angles in a triangle is 180°, we get 12x = 180, so x = 15. Therefore, the angles are 30°, 45°, and 105°.

10. In a right-angled triangle, the hypotenuse is 13 cm and one of the legs is 5 cm. Find the length of the other leg.

Solution: Using the Pythagorean theorem, we get a 2 +5 2 =132, so a 2 =144 and a = 12 cm.

11. Prove that the angles opposite to equal sides of an isosceles triangle are equal.

Solution: This is another standard geometry proof that involves drawing a perpendicular bisector of the base.

Quadratic Equations

13. Solve the quadratic equation: x 2 −5x+6=0

Solution: Factoring, we get (x−2)(x−3)=0, so x = 2 or x = 3.

14. Find the discriminant of the quadratic equation: 2x 2 −3x+1=0

Solution: The discriminant is b 2 −4ac=(−3) 2 −4(2)(1)=1.

Area of Parallelogram and Triangle

15. Find the area of a parallelogram with base 12 cm and height 8 cm.

Solution: Area = base × height = 12 cm × 8 cm = 96 cm²

16. The area of a triangle is 24 cm². If the base is 8 cm, find the height.

Solution: Height = (2 × Area) / base = (2 × 24 cm²) / 8 cm = 6 cm

Mensuration

17. Find the volume of a cube with a side 5 cm.

Solution: Volume = side³ = 5³ cm³ = 125 cm³

18. Find the total surface area of a sphere with a radius 7 cm.

Solution: Total surface area = 4πr² = 4 × π × 7² cm² = 616π cm²

19. Find the mean, median, and mode of the data: 2, 4, 5, 3, 2, 5, 4, 6.

Solution: Mean = (2+4+5+3+2+5+4+6)/8 = 3.75, Median = 4, Mode = 4 and 5.

20. Draw a histogram for the following frequency distribution:

| Class Interval | Frequency | |—|—| | 0-10 | 5 | | 10-20 | 8 | | 20-30 | 12 | | 30-40 | 6 |

Solution: A histogram can be drawn using the given data, with the x-axis representing the class intervals and the y-axis representing the frequency.

Probability

21. A coin is tossed twice. Find the probability of getting two heads.

Solution: The total possible outcomes are HH, HT, TH, and TT. The probability of getting two heads is 1/4.

Maths Questions for Class 9: PDF Link Available

Looking for a fun way to master Maths questions for Class 9? This Maths Questions PDF is perfect for 9th graders! It includes a wide range of interesting and challenging questions on topics like algebra, geometry, trigonometry, and more. Download it today to sharpen your math skills and make learning both enjoyable and rewarding. Click the link to download and start your journey toward becoming a math expert!

Importance of Maths Questions for Class 9 Students

The importance of Maths questions for Class 9 students cannot be overstated! Here’s why they are essential:

• Builds Strong Foundation : Class 9 is an important year as it sets the base for higher-level math in Class 10 and beyond. Practicing maths questions helps strengthen this foundation.
• Improves Problem-Solving Skills : Maths questions encourage students to think critically and find solutions to complex problems. This improves their problem-solving skills, which is useful in both academics and daily life.
• Boosts Confidence for Exams : Regular practice with a variety of questions helps students feel more prepared and confident when facing their exams.
• Helps with Logical Thinking : Maths is all about logic and reasoning. By solving different types of questions, students sharpen their ability to think logically and make better decisions.
• Prepares for Competitive Exams : For students aiming for competitive exams like Olympiads or entrance tests, solving maths questions in Class 9 is an excellent practice.

Tips for Preparing Maths Questions for Class 9

Here are some quick tips for preparing Maths questions for Class 9 students:

• Know the Syllabus : Make sure the questions cover important topics like algebra, geometry, and statistics.
• Mix Topics : Include questions from different chapters to give students complete practice.
• Start Easy : Begin with simple questions, then gradually increase difficulty.
• Use Word Problems : Add real-life problems to help students apply what they’ve learned.
• Vary Question Types : Use multiple-choice, short answer, and long problems for variety.
• Practice Formulas : Include questions that help students remember and use important formulas.
• Provide Solutions : Give clear, step-by-step answers to help students understand.
• Focus on Mistakes : Include questions that correct common errors.
• Use Old Papers : Look at past exam papers to create similar questions.
• Add Tough Questions : Challenge students with a few harder problems for extra practice.

Class 9 maths can be challenging, but with consistent practice and a good understanding of the fundamentals, it can be manageable. Some topics like geometry and algebra can be particularly tricky, but with focused effort, you can overcome these hurdles.

The most important thing in Class 9 math is to build a strong foundation in fundamental concepts. This includes understanding number systems, polynomials, coordinate geometry, linear equations, and basic geometry. A solid grasp of these topics will be crucial for your future studies in mathematics.

Class 9 maths covers topics like number systems, algebra, geometry, coordinate geometry, mensuration, and statistics. These topics build on the foundation of basic arithmetic and introduce more complex concepts.

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Mohit Rajak

Mohit Rajak, a soul entwined with the rhythm of words, finds solace in crafting verses that dance between the lines of poetry. With a pen as his wand, he weaves intricate tales and musings, breathing life into the blank canvas of pages. Through the art of blogging, Mohit embraces the world, sharing his thoughts, emotions, and unique perspective with those who venture into the realm of his written expressions. Each word, a brushstroke painting the canvas of his literary journey.

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