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Case Study Questions Class 10 Maths Quadratic Equations

Case study questions class 10 maths chapter 4 quadratic equations.

CBSE Class 10 Case Study Questions Maths Quadratic Equations. Term 2 Important Case Study Questions for Class 10 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Quadratic Equations.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 10 Maths Quadratic Equations

CASE STUDY 1:

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of x km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

[ CBSE Question Bank ]

4.) How much time took Ajay to travel 400 km?

Answer – d) 16 hour

1.) What will be the distance covered by Ajay’s car in two hours?

a) 2(x +5)km

b) (x – 5)km

c) 2(x + 10)km

d) (2x + 5)km

Answer – a) 2(x +5) km

3.) What is the speed of Raj’s car?

a) 20 km/hour

b) 15 km/hour

c) 25 km/hour

d) 10 km/hour

Answer – a) 20 km/hour

CASE STUDY 2 –

Q.2) Nidhi and Riya are very close friends. Nidhi’s parents have a Maruti Alto. Riya ‘s parents have a Toyota. Both the families decided to go for a picnic to Somnath Temple in Gujarat by their own car. Nidhi’s car travels x km/h, while Riya’s car travels 5km/h more than Nidhi’s car. Nidhi’s car took 4 hours more than Riya’s car in covering 400 km.

[ KVS Raipur 2021 – 22 ]

(i) What will be the distance covered by Riya’s car in two hours? How much time took Riya to travel 400 km?

Answer- 2(x+5)km

(ii) Write the quadratic equation describe the speed of Nidhi’s car. What is the speed of Nidhi’s car?

Answer –  x 2 +5x -500= 0

We hope that above case study questions will help you for your upcoming exams. To see more click below – 

• CBSE Class 10 Maths (standard)
• CBSE Class 10 Maths (Basic)

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
• Assertion and Reason Questions of Class 10th Science
• Assertion and Reason Questions of Class 10th Social Science

Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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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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Case Study on Quadratic Equations Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Quadratic Equations Class 10 Maths can use this page to download the PDF file. 

The case study questions on Quadratic Equations are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Quadratic Equations case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Quadratic Equations Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Quadratic Equations, therefore, they prepared a set of solutions along with the case study questions on Quadratic Equations.

The case study on Quadratic Equations Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Quadratic Equations case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Quadratic Equations Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Quadratic Equations case study questions on Class 10 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Quadratic Equations Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Quadratic Equations case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Quadratic Equations Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Quadratic Equations as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Quadratic Equations?

Students can choose their own way to answer Case Study on Quadratic Equations Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Quadratic Equations Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Quadratic Equations questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Quadratic Equations Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Quadratic Equations?

 A few essential things to know to solve Case Study Questions on Class 10 Quadratic Equations are -

• Basic Formulas of Quadratic Equations: One of the most important things to know to solve Case Study Questions on Class 10 Quadratic Equations is to learn about the basic formulas or revise them before solving the case-based questions on Quadratic Equations.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Quadratic Equations case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Quadratic Equations Class 10 Maths?

Use Selfstudys.com to find Case Study on Quadratic Equations Class 10 Maths. For ease, here is a step-wise procedure to download the Quadratic Equations Case Study for Class 10 Maths in PDF for free of cost.

Since you are already on this page, you can scroll to the top section of this page to get access to the Case Study on Quadratic Equations. To help others reach this page let them know these steps:

• Open Selfstudys.com on your computer/laptop or Smartphone 
• Once the website gets loaded, click on the navigation button

• Find CBSE from the given menu

• Click on Case Study

• Choose Class 10 
• Search Maths and then navigate to the Quadratic Equations Class 10 Maths Case Study

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Case Study Class 10 Maths Questions

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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• CBSE Class 10 Maths Sample Paper 2020-21
• Class 12 Maths Case Study Questions
• CBSE Reduced Syllabus Class 10 (2020-21)
• Class 10 Maths Basic Sample Paper 2024
• How to Revise CBSE Class 10 Maths in 3 Days
• CBSE Practice Papers 2023
• Class 10 Maths Sample Papers 2024
• Competency Based Learning in CBSE Schools

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NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

• NCERT Solutions
• Chapter 4 Quadratic Equations

Complete Resource of NCERT Class 10 Maths Chapter 4 Quadratic Equations - Free PDF Download

NCERT Solutions for Class 10 Chapter 4 Quadratic Equation , covers a crucial aspect of algebra that every student must grasp. This chapter introduces quadratic equations, detailing methods to solve them, including factoring, using the quadratic formula, and completing the square. Understanding these concepts is key, as they are foundational for higher mathematics and problem solving in science and engineering. Focus on mastering the techniques for finding the roots of quadratic equations and recognizing their practical applications. Clear explanations and step-by-step solutions in Vedantu’s materials help understand complex concepts, making them accessible and understandable.

Glance of NCERT Solutions of Maths Chapter 4 Quadratic Equations for Class 10 | Vedantu

Chapter 4 of Class 10 Maths deals with quadratic equations, which are equations  of the form  ax^2 + bx + c = 0, where a ≠ 0.

Learn about standard forms, where a, b, and c are real numbers.

The chapter focuses on finding the roots/solutions of these equations, which are the values of x that satisfy the equation.

There are different methods for solving quadratics, such as Factorization and by using Quadratic Formula

A key concept is the discriminant (b² - 4ac). It helps determine the nature of the roots:

Distinct Real Roots (D > 0): When the discriminant is positive, there are two distinct real solutions for x.

Equal Real Roots (D = 0): A positive discriminant of zero indicates two equal real roots.

No Real Roots (D < 0): A negative discriminant means there are no real number solutions, but there might be complex solutions.

The chapter also covers forming quadratic equations from word problems and applications of quadratic equations in real-life scenarios.

This article contains chapter notes important questions and Exercises link for Chapter 4 - Quadratic Equations, which you can download as PDFs.

There are four exercises and one miscellaneous exercise (24 fully solved questions) in class 10th maths chapter 4 Quadratic Equations.

Access Exercise Wise NCERT Solutions for Chapter 4 Maths Class 10

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Exercises under NCERT Solutions for Maths Chapter 4 Class 10 Quadratic Equations

Exercise 4.1:.

This exercise covers the introduction to quadratic equations and the standard form of a quadratic equation. It also includes methods for solving quadratic equations by factorisation. In this exercise, students will learn how to identify a quadratic equation, how to convert a quadratic equation into standard form, and how to factorise quadratic equations using different methods. The exercise includes a set of questions that range from easy to difficult, allowing students to gradually build their understanding of the concepts.

Exercise 4.2:

This exercise covers more advanced methods for solving quadratic equations, such as completing the square and using the quadratic formula. It includes the derivation of the quadratic formula and shows how to apply it to solve quadratic equations. In this exercise, students will learn how to complete the square of a quadratic equation to convert it into standard form, and how to use the quadratic formula to solve quadratic equations. The exercise includes questions that require students to use both methods to solve quadratic equations.

Exercise 4.3:

This exercise covers real-life applications of quadratic equations and includes word problems that require students to apply their knowledge of quadratic equations to solve practical problems. The exercise includes problems related to the trajectory of a projectile, finding the distance between two ships, and the dimensions of a garden. Students will learn how to formulate and solve quadratic equations to solve real-life problems. The exercise includes a set of word problems that gradually increase in difficulty, allowing students to develop their problem-solving skills.

Access NCERT Solutions for Class - 10 Maths Chapter 4 – Quadratic Equations

Exercise 4.1.

1. Check whether the following are quadratic equations:  

i. ${{\left( \text{x+1} \right)}^{\text{2}}}\text{=2}\left( \text{x-3} \right)$

Ans : ${{\left( \text{x+1} \right)}^{\text{2}}}\text{=2}\left( \text{x-3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2x+1=2x-6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+7=0}$

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is a quadratic equation.

ii. ${{\text{x}}^{\text{2}}}\text{-2x=}\left( \text{-2} \right)\left( \text{3-x} \right)$

Ans : ${{\text{x}}^{\text{2}}}\text{-2x=}\left( \text{-2} \right)\left( \text{3-x} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2x=-6+2x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-4x+6=0}$

iii. $\left( \text{x-2} \right)\left( \text{x+1} \right)\text{=}\left( \text{x-1} \right)\left( \text{x+3} \right)$

Ans : $\left( \text{x-2} \right)\left( \text{x+1} \right)\text{=}\left( \text{x-1} \right)\left( \text{x+3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-x-2=}{{\text{x}}^{\text{2}}}\text{+2x-3}$

$\Rightarrow \text{3x-1=0}$

Since, it is not in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is not a quadratic equation.

iv. $\left( \text{x-3} \right)\left( \text{2x+1} \right)\text{=x}\left( \text{x+5} \right)$

Ans : $\left( \text{x-3} \right)\left( \text{2x+1} \right)\text{=x}\left( \text{x+5} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-5x-3=}{{\text{x}}^{\text{2}}}\text{+5x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-10x-3=0}$

v. $\left( \text{2x-1} \right)\left( \text{x-3} \right)\text{=}\left( \text{x+5} \right)\left( \text{x-1} \right)$

Ans : $\left( \text{2x-1} \right)\left( \text{x-3} \right)\text{=}\left( \text{x+5} \right)\left( \text{x-1} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=}{{\text{x}}^{\text{2}}}\text{+4x-5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-11x+8=0}$

vi. ${{\text{x}}^{\text{2}}}\text{+3x+1=}{{\left( \text{x-2} \right)}^{\text{2}}}$

Ans : ${{\text{x}}^{\text{2}}}\text{+3x+1=}{{\left( \text{x-2} \right)}^{\text{2}}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+3x+1=}{{\text{x}}^{\text{2}}}\text{+4-4x}$

$\Rightarrow \text{7x-3=0}$

vii. ${{\left( \text{x+2} \right)}^{\text{3}}}\text{=2x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$

Ans : ${{\left( \text{x+2} \right)}^{\text{3}}}\text{=2x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{+8+6}{{\text{x}}^{\text{2}}}\text{+12x=2}{{\text{x}}^{\text{3}}}\text{-2x}$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{-14x-6}{{\text{x}}^{\text{2}}}\text{-8=0}$

viii. ${{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\left( \text{x-2} \right)}^{\text{3}}}$

Ans : ${{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\left( \text{x-2} \right)}^{\text{3}}}$ 

$\Rightarrow {{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\text{x}}^{\text{3}}}\text{-8-6}{{\text{x}}^{\text{2}}}\text{+12x}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-13x+9=0}$

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

i. The area of a rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$. 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.

Ans : Let the breath of the plot be $\text{x m}$.

Thus, length would be-

$\text{Length=}\left( \text{2x+1} \right)\text{m}$

Hence, Area of rectangle $=$$\text{Length }\!\!\times\!\!\text{ breadth}$

So, $\text{528=x}\left( \text{2x+1} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{+x-528=0}$

ii. The product of two consecutive positive integers is $\text{306}$. We need to find the integers.

Ans : Let the consecutive integers be $\text{x}$ and $\text{x+1}$.

Thus, according to question-

$\text{x}\left( \text{x+1} \right)\text{=306}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-306=0}$

iii. Rohan’s mother is $\text{26}$ years older than him. The product of their ages (in years) $\text{3}$ years from now will be $\text{360}$. We would like to find Rohan’s present age.

Ans : Let Rohan’s age be $\text{x}$.

Hence, his mother’s age is $\text{x+26}$ .

Now, after $\text{3 years}$.

Rohan’s age will be $\text{x+3}$.

His mother’s age will be $\text{x+29}$ .

So, according to question-

$\left( \text{x+3} \right)\left( \text{x+29} \right)\text{=360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+3x+29x+87=360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+32x-273=0}$

iv. A train travels a distance of $\text{480 km}$ at a uniform speed. If the speed had been $\text{8km/h}$ less, then it would have taken $\text{3}$ hours more to cover the same distance. We need to find the speed of the train.

Ans : Let the speed of train be $\text{x km/h}$.

Thus, time taken to travel $\text{482 km}$ is $\dfrac{\text{480}}{\text{x}}\text{hrs}$.

Now, let the speed of train $\text{=}\left( \text{x-8} \right)\text{km/h}$.

Therefore, time taken to travel $\text{480 km}$ is $\left( \dfrac{\text{480}}{\text{x}}+3 \right)\text{hrs}$.

Hence, $\text{speed }\!\!\times\!\!\text{ time=distance}$

i.e $\left( \text{x-8} \right)\left( \dfrac{\text{480}}{\text{x}}\text{+3} \right)\text{=480}$

$\Rightarrow \text{480+3x-}\dfrac{\text{3840}}{\text{x}}\text{-24=480}$

$\Rightarrow \text{3x-}\dfrac{\text{3840}}{\text{x}}\text{=24}$

$\Rightarrow \text{3}{{\text{x}}^{\text{2}}}\text{-24x-3840=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x-1280=0}$

Exercise 4.2

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

i. ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

Ans : ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-5x+2x-10}$

$\Rightarrow \text{x}\left( \text{x-5} \right)\text{+2}\left( \text{x-5} \right)$

$\Rightarrow \left( \text{x-5} \right)\left( \text{x+2} \right)$

Therefore, roots of this equation are –

$\text{x-5=0}$ or $\text{x+2=0}$

i.e $\text{x=5}$ or $\text{x=-2}$

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

Ans : $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+4x-3x-6}$

$\Rightarrow 2\text{x}\left( \text{x+2} \right)-3\left( \text{x+2} \right)$

$\Rightarrow \left( \text{x+2} \right)\left( \text{2x-3} \right)$

$\text{x+2=0}$ or $\text{2x-3=0}$

i.e $\text{x=-2}$ or $\text{x=}\dfrac{3}{2}$

iii. $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

Ans : $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

$\Rightarrow \sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+5x+2x+5}\sqrt{\text{2}}$

$\Rightarrow \text{x}\left( \sqrt{\text{2}}\text{x+5} \right)+\sqrt{\text{2}}\left( \sqrt{\text{2}}\text{x+5} \right)$

$\Rightarrow \left( \sqrt{\text{2}}\text{x+5} \right)\left( \text{x+}\sqrt{\text{2}} \right)$

$\sqrt{\text{2}}\text{x+5=0}$ or $\text{x+}\sqrt{\text{2}}\text{=0}$

i.e $\text{x=}\dfrac{-5}{\sqrt{\text{2}}}$ or $\text{x=-}\sqrt{\text{2}}$

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

Ans : $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 16{{\text{x}}^{\text{2}}}-8x+1 \right)\]

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 4x\left( 4x-1 \right)-1\left( 4x-1 \right) \right)\]

$\Rightarrow \dfrac{\text{1}}{\text{8}} {{\left( \text{4x-1} \right)}^{2}}$

$\text{4x-1=0}$ or $\text{4x-1=0}$

i.e $\text{x=}\dfrac{1}{4}$ or $\text{x=}\dfrac{1}{4}$

v. $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

Ans : $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

$\Rightarrow 100{{\text{x}}^{\text{2}}}\text{-10x-10x+1}$

$\Rightarrow 10\text{x}\left( \text{10x-1} \right)-1\left( \text{10x-1} \right)$

\[\Rightarrow \left( \text{10x-1} \right)\left( \text{10x-1} \right)\]

\[\left( \text{10x-1} \right)=0\]or \[\left( \text{10x-1} \right)=0\]

i.e $\text{x=}\dfrac{1}{10}$ or $\text{x=}\dfrac{1}{10}$

2. Solve the problems given in Example 1

i. John and Jivanti together have $\text{45}$ marbles. Both of them lost $\text{5}$ marbles each, and the product of the number of marbles they now have is $\text{124}$. Find out how many marbles they had to start with.

Ans : Let the number of john’s marbles be $\text{x}$.

Thus, number of Jivanti’s marble be $\text{45-x}$.

According to question i.e, 

After losing $\text{5}$ marbles.

Number of john’s marbles be $\text{x-5}$

And number of Jivanti’s marble be $\text{40-x}$.

Therefore, $\left( \text{x-5} \right)\left( \text{40-x} \right)\text{=124}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-45x+324=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-36x-9x+324=0}$

$\Rightarrow \text{x}\left( \text{x-36} \right)\text{-9}\left( \text{x-36} \right)\text{=0}$

$\Rightarrow \left( \text{x-36} \right)\left( \text{x-9} \right)\text{=0}$

Case 1 - If $\text{x-36=0}$ i.e $\text{x=36}$

So, the number of john’s marbles be $\text{36}$.

Thus, number of Jivanti’s marble be $\text{9}$.

Case 2 - If $\text{x-9=0}$ i.e $\text{x=9}$

So, the number of john’s marbles be $9$.

Thus, number of Jivanti’s marble be $36$.

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 $\text{55}$ minus the number of toys produced in a day. On a particular day, the total cost of production was Rs $\text{750}$. Find out the number of toys produced on that day.

Ans: Let the number of toys produced be $\text{x}$.

Therefore, Cost of production of each toy be $\text{Rs}\left( \text{55-x} \right)$.

Thus, $\left( \text{55-x} \right)\text{x=750}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-55x+750=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x-30x+750=0}$

$\Rightarrow \text{x}\left( \text{x-25} \right)-30\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{x-30} \right)\text{=0}$

Case 1 - If $\text{x-25=0}$ i.e $\text{x=25}$

So, the number of toys be $25$.

Case 2 - If $\text{x-30=0}$ i.e $\text{x=30}$

So, the number of toys be $30$.

3. Find two numbers whose sum is $\text{27}$ and product is $\text{182}$ .

Ans: Let the first number be $\text{x}$ ,

Thus, the second number be $\text{27-x}$.

$\text{x}\left( \text{27-x} \right)\text{=182}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-27x+182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-13x-14x+182=0}$

$\Rightarrow \text{x}\left( \text{x-13} \right)-14\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \left( \text{x-13} \right)\left( \text{x-14} \right)\text{=0}$

Case 1 - If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $13$ ,

Thus, the second number be $\text{14}$.

Case 2 - If $\text{x-14=0}$ i.e $\text{x=14}$

So, the first number be $\text{14}$.

Thus, the second number be$13$.

4. Find two consecutive positive integers, sum of whose squares is $\text{365}$.

Ans: Let the consecutive positive integers be $\text{x}$ and $\text{x+1}$.

Thus, ${{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+1} \right)}^{\text{2}}}\text{=365}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+1+2\text{x=365}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+2x-364=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+14x-13x-182=0}$

$\Rightarrow \text{x}\left( \text{x+14} \right)-13\left( \text{x+14} \right)\text{=0}$

$\Rightarrow \left( \text{x+14} \right)\left( \text{x-13} \right)\text{=0}$

Case 1 - If $\text{x+14=0}$ i.e $\text{x=-14}$.

This case is rejected because number is positive.

Case 2 - If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $\text{13}$.

Thus, the second number be $14$.

Hence, the two consecutive positive integers are $\text{13}$ and $14$.

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

Ans: Let the base of the right-angled triangle be $\text{x cm}$.

Its altitude be $\left( \text{x-7} \right)\text{cm}$.

Thus, by pythagores theorem-

$\text{bas}{{\text{e}}^{\text{2}}}\text{+altitud}{{\text{e}}^{\text{2}}}\text{=hypotenus}{{\text{e}}^{\text{2}}}$

\[\therefore {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x-7} \right)}^{\text{2}}}\text{=1}{{\text{3}}^{\text{2}}}\]

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+49-14\text{x=169}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{-14x-120=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x-60=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+12x+5x-60=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)+5\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x+5} \right)\text{=0}$

Case 1 - If $\text{x-12=0}$ i.e $\text{x=12}$.

So, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$

Case 2 - If $\text{x+5=0}$ i.e $\text{x=-5}$

This case is rejected because side is always positive.

Hence, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$.

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 $\text{3}$ more than twice the number of articles produced on that day. If the total cost of production on that day was Rs $\text{90}$, find the number of articles produced and the cost of each article.

Ans:  Let the number of articles produced be $\text{x}$.

Therefore, cost of production of each article be $\text{Rs}\left( \text{2x+3} \right)$.

Thus, $\text{x}\left( \text{2x+3} \right)\text{=90}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+3x-90=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+15x-12x-90=0}$

$\Rightarrow \text{x}\left( \text{2x+15} \right)-6\left( \text{2x+15} \right)\text{=0}$

$\Rightarrow \left( \text{2x+15} \right)\left( \text{x-6} \right)\text{=0}$

Case 1 - If $\text{2x-15=0}$ i.e $\text{x=}\dfrac{-15}{2}$.

This case is rejected because number of articles is always positive.

Case 2 - If $\text{x-6=0}$ i.e $\text{x=6}$

Hence, the number of articles produced be $6$.

Therefore, cost of production of each article be $\text{Rs15}$.

Exercise 4.3

1. Find the nature of the roots of the following quadratic equations.  If the real roots exist, find them-

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$

Ans: For a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Where Discriminant $\text{=}{{\text{b}}^{\text{2}}}\text{-4ac}$

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac>0}$ then there will be two distinct real roots.

Case 2- If ${{\text{b}}^{\text{2}}}\text{-4ac=0}$ then there will be two equal real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac<0}$ then there will be no real roots.

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$ .

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-3}$, $\text{c=5}$.

Discriminant $\text{=}{{\left( \text{-3} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{5} \right)$

$\text{=9-40}$

$\text{=-31}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$.

Therefore, there is no real root for the given equation.

ii. $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac > 0}$ then there will be two distinct real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$ then there will be no real roots.

Thus, for $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$ .

So, $\text{a=3}$, $\text{b=-4}\sqrt{\text{3}}$, $\text{c=4}$.

Discriminant $\text{=}{{\left( \text{-4}\sqrt{\text{3}} \right)}^{\text{2}}}\text{-4}\left( \text{3} \right)\left( \text{4} \right)$

$\text{=48-48}$

$\text{=0}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac=0}$.

Therefore, there is equal real root for the given equation and the roots are-

$\dfrac{\text{-b}}{\text{2a}}$ and $\dfrac{\text{-b}}{\text{2a}}$.

Hence, roots are-

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( \text{-4}\sqrt{\text{3}} \right)}{\text{6}}$

$\text{=}\dfrac{\text{4}\sqrt{\text{3}}}{\text{6}}$

\[\text{=}\dfrac{\text{2}\sqrt{\text{3}}}{3}\]

Therefore, roots are \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\] and \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\].

iii. $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$ .

So, $\text{a=2}$, $\text{b=-6}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=36-24}$

$\text{=12}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac>0}$.

Therefore, distinct real roots exists for the given equation and the roots are-

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

$\text{x=}\dfrac{\text{-}\left( \text{-6} \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{36-24}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{12}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ 2}\sqrt{\text{3}}}{\text{4}}$

$\text{=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{3}}}{\text{2}}$

Therefore, roots are $\dfrac{\text{3+}\sqrt{\text{3}}}{\text{2}}$ and $\dfrac{\text{3-}\sqrt{\text{3}}}{\text{2}}$.

2. Find the values of $\text{k}$ for each of the following quadratic equations, so  that they have two equal roots.

i. $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$

Ans: If a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$ has two equal roots, then its discriminant will be $\text{0}$ i.e., ${{\text{b}}^{\text{2}}}\text{-4ac=0}$

So, for $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$ .

So, $\text{a=2}$, $\text{b=k}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{k} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=}{{\text{k}}^{2}}-24$

For equal roots-

${{\text{b}}^{\text{2}}}\text{-4ac=0}$

$\therefore {{\text{k}}^{\text{2}}}\text{-24=0}$

$\Rightarrow {{\text{k}}^{\text{2}}}\text{=24}$

$\Rightarrow \text{k=}\sqrt{\text{24}}$

$\Rightarrow \text{k=}\pm \text{2}\sqrt{\text{6}}$

ii. $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

So, for $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

$\Rightarrow \text{k}{{\text{x}}^{\text{2}}}\text{-2kx+6=0}$

So, $\text{a=k}$, $\text{b=-2k}$, $\text{c=6}$.

Discriminant $\text{=}{{\left( \text{-2k} \right)}^{\text{2}}}\text{-4}\left( \text{k} \right)\left( \text{6} \right)$

$\text{=4}{{\text{k}}^{\text{2}}}\text{-24k}$

$\therefore \text{4}{{\text{k}}^{\text{2}}}\text{-24k=0}$

$\Rightarrow \text{4k}\left( \text{k-6} \right)\text{=0}$

$\Rightarrow \text{k=0 or k=6}$

But $\text{k}$ cannot be zero. Thus, this equation has two equal roots when $\text{k}$ should be $\text{6}$ .

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

Ans: Let the breadth of mango grove be $\text{x}$.

So, length of mango grove will be $\text{2x}$.

Hence, Area of mango grove is $=\left( \text{2x} \right)\text{x}$

$\text{=2}{{\text{x}}^{\text{2}}}$.

So, $\text{2}{{\text{x}}^{\text{2}}}\text{=800}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-400=0}$

So, $\text{a=1}$, $\text{b=0}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{0} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{-400} \right)$

$\text{=1600}$

Therefore, distinct real roots exist for the given equation and the roots are-

$\text{x=}\dfrac{\text{-}\left( 0 \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( 0 \right)}^{\text{2}}}\text{-4}\left( 1 \right)\left( -400 \right)}}{2}$

$\text{=}\dfrac{\pm \sqrt{\text{1600}}}{2}$

$\text{=}\dfrac{\text{ }\!\!\pm\!\!\text{ 40}}{2}$

$\text{=}\pm \text{20}$

Since, length cannot be negative.

Therefore, breadth of the mango grove is $\text{20m}$.

And length of the mango grove be $\text{2}\left( \text{20} \right)\text{m}$ i.e., $\text{40m}$.

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

Ans: Let the age of one friend be $\text{x years}$.

So, age of the other friend will be $\left( \text{20-x} \right)\text{years}$.

Thus, four years ago, the age of one friend be $\left( \text{x-4} \right)\text{years}$.

And age of the other friend will be $\left( \text{16-x} \right)\text{years}$.

Hence, according to question-

$\left( \text{x-4} \right)\left( \text{16-x} \right)\text{=48}$

$\Rightarrow \text{16x-64-}{{\text{x}}^{\text{2}}}\text{+4x=48}$

$\Rightarrow 20\text{x-112-}{{\text{x}}^{\text{2}}}\text{=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-20x+112-=0}$

So, $\text{a=1}$, $\text{b=-20}$, $\text{c=112}$.

Discriminant $\text{=}{{\left( \text{-20} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{112} \right)$

$\text{=400-448}$

$\text{=-48}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac <0}$.

Therefore, there is no real root for the given equation and hence, this situation is not possible.

5. Is it possible to design a rectangular park of perimeter $\text{80 m}$ and area $\text{400}{{\text{m}}^{\text{2}}}$? If so find its length and breadth.

Ans: Let the length of the park be $\text{x m}$ and breadth of the park be $\text{x m}$.

Thus, $\text{Perimeter=2}\left( \text{x+y} \right)$.

$\text{2}\left( \text{x+y} \right)\text{=80}$

$\Rightarrow \text{x+y=40}$

$\Rightarrow \text{y=40-x}$.

Now, $\text{Area=x }\!\!\times\!\!\text{ y}$.

Substituting value of y.

$\text{Area=x}\left( \text{40-x} \right)$

$\text{x}\left( \text{40-x} \right)\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-40x+400=0}$

So, $\text{a=1}$, $\text{b=-40}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{-40} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( 400 \right)$

$\text{=1600-1600}$

Therefore, there is equal real roots for the given equation and hence, this situation is possible.

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( -40 \right)}{2}$

$\text{=}\dfrac{\text{40}}{2}$

\[\text{=20}\]

Therefore, length of park is $\text{x=20m}$ .

And breadth of park be $\text{y=}\left( \text{40-20} \right)\text{m}$ i.e., $\text{y=20m}$.

Important Points from NCERT Class 10 Quadratic Equations

A quadratic equation can be represented as:

ax 2 + bx + c = 0

Where x is the variable of the equation and a, b and c are the real numbers. Also, a≠0.

The nature of roots of a quadratic equation ax 2 + bx + c = 0 can be find as:

A real number α be root of quadratic equations ax 2 + bx + c = 0 if and only if 

aα 2 + bα + c = 0.

Quadratic equations are very important in real-life situations. Learn all the concepts deeply and understand each topic conceptually. And, now let us solve questions related to quadratic equations.

Maths Class 10 Quadratic Equations Mind Map

Relation between the zeroes of a quadratic equation and the coefficient of a quadratic equation.

If α and β are zeroes of the quadratic equation $ax^2 + bx + c = 0$, where a, b, and c are real numbers and a ≠ 0, then

$\alpha + \beta = -\dfrac{b}{a}$

$\text{sum of zeros} = -\dfrac{\text{coefficient of x}}{\text{coefficient of }x^2}$

$\alpha \beta = \dfrac{c}{a}$

$\text{product of zeros} = -\dfrac{\text{constant term}}{\text{coefficient of }x^2}$

Methods of Solving a Quadratic Equation

The following are the methods that are used to solve quadratic equations:

(i) Factorization; (ii) Completing the Square; (iii) Quadratic Formula

Methods of Factorization

In this method, we find the roots of a quadratic equation $(ax^2 + bx + c = 0)$ by factorizing LHS into two linear factors and equating each factor to zero, e.g., $6x^2 - x - 2 = 0$ $\Rightarrow 6x^2 + 3x - 4x - 2 = 0$ …(i) $\Rightarrow 3x (2x + 1) - 2(2x + 1) = 0$ $\Rightarrow (3x - 2) (2x + 1) = 0$ $\Rightarrow 3x - 2 = 0$ or $2x + 1 = 0$

Therefore $x = \dfrac{2}{3}$ or $x = -\dfrac{-1}{2}$

Method of Completing the Square

This is the method of converting the LHS of a quadratic equation that is not a perfect square into the sum or difference of a perfect square and a constant by adding and subtracting the terms.

Quadratic Formula

Consider a quadratic equation: ax2 + bx + c = 0. If b2 – 4ac ≥ 0, then the roots of the above equation are given by:

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

Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 4 Quadratic Equation

Class 10 maths chapter 4: exercise breakdown.

NCERT Solutions for Class 10 Maths Chapter 4 - provides a comprehensive guide to understanding quadratic equations. This chapter is important for students because it teaches topics that are fundamental to advanced mathematics. The solutions describe how to solve quadratic equations, apply the quadratic formula, and investigate the nature of the roots using the discriminant. For effective exam preparation, focus on understanding formula derivation and applying the many types of problem-solving approaches described in this chapter. Last year, four to six questions from this area featured in the board exams, demonstrating its importance. These answers are precisely crafted to help students succeed by improving their problem-solving skills and conceptual understanding.

Other Study Materials of CBSE Class 10 Maths Quadratic Equation

Chapter-specific ncert solutions for class 10 maths.

Given below are the chapter-wise NCERT Solutions for Class 10 Maths . Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

The subject’s experts take online classes available at Vedantu. They have lots of experience in their respective subjects. Along with this, they are working in their field with consistency. So, they have been faced with all kinds of problems and learned the tricks on how to come out from it.

In online classes, teachers share all their experiences with the students and teach some essential tricks. These tricks will be useful at the time of solving questions in the examination hall. Apart from the teaching concepts and formulas, they also motivate students and remove the fear of examinations from the student’s mind that is built up somewhere due to some kind of society’s pressure and other things.

2. Instead of Class 10th Maths Chapter 4, What is the Maths Syllabus for Class 10 for 202 4-25 ?

Here is the complete syllabus for Class 10 Mathematics Revised Syllabus 2024-25:-

Unit- I: Number Systems

Real Numbers

Unit II: Algebra

Polynomials

Pair of Linear Equations in Two Variables

Quadratic Equations

 Arithmetic Progressions

Unit III: Coordinate Geometry

Lines (In two-dimensions)

 Unit IV: Geomtry

Constructions

Unit V: Trigonometry

Introduction to Trigonometry

Trigonometric Identities

 Heights and Distances: Angle of elevation, Angle of Depression

Unit VI: Mensuration

Areas Related to Circles

Surface Areas and Volumes

Unit VII: Statistics and Probability

 Statistics

Probability

3. What is the Weightage for Class 10 Mathematics Unit-Wise?

Students who don’t know the weightage then they should go through the table given below. Here, we have mentioned the entire Class 10 Mathematics Unit-Wise Weightage for the knowledge of the students.

4. Mention the important concepts that you learn in NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations.

If you want to score 100 per cent marks in Class 10 Maths, you have to practice daily. Chapter 4 Quadratic Equations is an important chapter. Students can find NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations on Vedantu. There are five exercises in Chapter 4. Students can download the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations to learn the important concepts that will help them understand the topic.

5. How to download Class 10 Maths Quadratic Equations NCERT Textbooks PDF?

Students can easily download Class 10 Maths Quadratic Equations NCERT textbooks PDF online. NCERT Solutions for Class 10 Maths Quadratic Equations are explained in an easy and simple language. Follow the given steps:

Click NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations.

Click on “Download PDF”.

Download and save it.

Students can use the PDF document without having an internet connection and can study Maths Quadratic Equations anytime. The NCERT Solutions give a clear understanding to them. 

6. What is the Quadratic formula Class 10th?

When a quadratic polynomial is equated to a constant, it forms a quadratic equation. An equation such as Ax = D, where Ax is a polynomial of degree two and D is a constant, forms a quadratic equation. The standard quadratic equation is a x 2 +bx+c=0 where a, b, and c are not equal to zero. You can refer to NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations to understand more about the topic. You can also download Vedantu’s app. All the resources are available free of cost. 

7. What are the important topics covered in NCERT Solutions Class 10 Maths Chapter 4?

Chapter 4 Class 10 Maths is based on Quadratic Equations. The chapter includes important concepts about quadratic equations. This chapter includes five exercises that explain the different concepts of quadratic equations. The following topics are covered:

Exercise 4.1- Introduction

Exercise 4.2- Quadratic Equations

Exercise 4.3- Solution of a Quadratic Equation by Factorisation

Exercise 4.4- Solution of a Quadratic Equation by Completing the Square

Exercise 4.5- Nature of Roots

8. How do you solve Quadratic Equations in Class 10?

If you want to learn how to solve quadratic equations in Class 10, you can refer to NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations. All the solutions are prepared by experts in an easy language. Students can understand the equations clearly. Students have to find the roots by using the quadratic formula. They can find the sum and product of both the roots. The method is simple and explained properly for easier understanding.

NCERT Solutions for Class 10 Maths

Ncert solutions for class 10.

NCERT Solutions for Class 10th Maths Chapter 4 Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

• Exercise 4.1
• Exercise 4.2
• Exercise 4.3
• Exercise 4.4

NCERT Solutions for Class 10 Maths Chapters:

How many exercises in chapter 4 quadratic equations, if the roots of the quadratic equation − ax 2 + bx + c = 0 are equal then show that b 2 = 4ac., if 2 is a root of the equation x 2 + kx + 12 = 0 and the equation x 2 + kx + q = 0 has equal roots, find the value of q., what is roots of a quadratic equation., contact form.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations are provided here to help the students of CBSE class 10. Our expert teachers prepared all these solutions as per the latest CBSE syllabus and guidelines. In this chapter, we have discussed how to find the solution of a quadratic equation by – factorisation, completing the square method in details. CBSE Class 10 Maths solutions provide a detailed and step-wise explanation of each answer to the questions given in the exercises of NCERT books.

CBSE Class 10 Maths Chapter 4 Quadratic Equations Solutions

Below we have given the answers to all the questions present in Quadratic Equations in our NCERT Solutions for Class 10 Maths chapter 4. In this lesson, students are introduced to a lot of important concepts that will be useful for those who wish to pursue mathematics as a subject in their future classes. Based on these solutions, students can prepare for their upcoming Board Exams. These solutions are helpful as the syllabus covered here follows NCERT guidelines.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4

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NCERT Solutions for Class 10 Maths Chapter 4 Free PDF Download

Ncert solutions for class 10 maths chapter 4 quadratic equations.

Maths is one of those subjects which students find very hard to understand. Also, according to CBSE Board NCERT books are sufficient but that’s not the case. NCERT Solutions for Class 10 Maths Chapter 4 helps students to know the study materials of the chapter. Apart from that, the solution provides you with the basic study materials that you need. Toppr is the leader of providing the best NCERT Solutions for Class 10 Maths Chapter 4.

The solution is easy to understand. Our team of experts design and solve every question of the chapter to give you the exact idea of how to understand them? NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations solved by our team of an expert helps to capture the solving methods. Download the NCERT Solutions for Class 10 Maths Chapter 4 by clicking on the download button below.

The solution is based on the up-to-date course of study provided ion the NCERT books. Also, the NCERT Solutions for Class 10 Maths Chapter 4 is according to the guidelines of the board. This solution will help you in scoring high marks in the exams and tests. At Toppr the NCERT Solutions for Class 10 Maths Chapter 4 is available for download also, it’s free of cost.

Toppr provides free study materials, last 10 years of question papers, 1000+ hours of video lectures for free. Download Toppr for Android and iOS or signup for free.

Find other subject NCERT solutions for class 10 to download here .

Also, find other Maths chapter of NCERT solutions for download here .

Firstly the chapter introduces you to quadratic equations and how to solve them. Also, it teaches you to find the roots using various methods. In exercise 4.1, you have to solve the question that introduces you to how to solve basic quadratic equations. Also, it will help you to understand if the equations are quadratic or not.

In the other exercises 4.2 to 4.4, you will learn to find roots using quadratic equations. Also, you will come across the nature of roots, How to complete squares and application of factorization and equations.

On the other hand, the chapter is fun to learn and easy to understand if you get the topic. Also, the students will learn to apply the quadratic equation in some daily life activities.

Sub-topics under CBSE Class 10 Maths Chapter 4 Quadratic Equations

• 4.1 Quadratic Equations
• 4.2 Solution of  Quadratic Equation by Factorisation
• 4.3 Solution of Quadratic Equations by Completing the Square
• 4.4 Nature of Roots

You can download NCERT Solutions for Class 10 Maths Chapter 4 by clicking on the button below

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NCERT Solutions for Class 10 Maths

• NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.1
• NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.2
• NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.3
• NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.4
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.1
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.5
• NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6

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CBSE Class 10 Maths Case Study Questions for Chapter 2 - Polynomials (Published by CBSE)

Check the case study questions published by cbse for class 10 maths chapter 2 - polynomials. these questions are important for the preparation of cbse class 10 maths exam 2021-22..

CBSE Class 10 Maths paper in Board Exam 2022 will have some questions based on the case study. These questions are entirely new for the class 10 students. Therefore, the board has released a question bank to help the students get familiarised with the case study questions. We have provided here the case study questions for CBSE Class 10 Maths Chapter 2 - Polynomials. All the questions have sub-questions of MCQ type. You can find the answer (correct option) written against each question. Practice all the case study based questions right after you finish with the chapter - Polynomials. This will help you prepare for your Maths exam easily and effectively.

Case Study Questions for Class 10 Maths Chapter 2 - Polynomials

CASE STUDY 1:

The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.

1. In the standard form of quadratic polynomial, ax 2 + bx + c, a, b and c are

a) All are Polynomials.

b) All are rational numbers.

c) ‘a’ is a non zero real number and b and c are any Polynomials.

d) All are integers.

Answers: c) ‘a’ is a non zero real number and b and c are any Polynomials.

2. If the roots of the quadratic polynomial are equal, where the discriminant D = b 2 – 4ac, then

a) D > 0

b) D < 0

c) D ≥ 0

Answers: d) D = 0

3. If α and 1/α are the zeroes of the quadratic polynomial 2x2 – x + 8k, then k is

c) –1/4

Answers: b) 1/4

4. The graph of x 2 +1 = 0

a) Intersects x‐axis at two distinct points.

b)Touches x‐axis at a point.

c) Neither touches nor intersects x‐axis.

d)Either touches or intersects x‐ axis.

Answers: c) Neither touches nor intersects x‐axis.

5. If the sum of the roots is –p and product of the roots is –1/p, then the quadratic polynomial is

a) k(–px 2 + x/p + 1)

b) k(px 2 – x/p – 1)

c) k(x 2 + px – 1/p)

d) k(x 2 – px + 1/p)

Answers: c) k(x 2 + px – 1/p)

CASE STUDY 2:

An asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.

1. The shape of the poses shown is

d) Parabola

Answer: d) Parabola

2. The graph of parabola opens downwards, if _______

a) a ≥ 0

c) a < 0

d) a > 0

Answer: c) a < 0

3. In the graph, how many zeroes are there for the polynomial?

Answer: c) 2

4. The two zeroes in the above shown graph are

Answer: b) -2, 4

CASE STUDY 3:

Basketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.

1. The shape of the path traced shown is

2. The graph of parabola opens upwards, if _______

b) a < 0

c) a > 0

d) a ≥ 0

Answer: c) a > 0

3. Observe the following graph and answer

In the above graph, how many zeroes are there for the polynomial?

Answer: d) 3

4. The three zeroes in the above shown graph are

b) -2, 3, 1

c) -3, -1, 2

d) -2, -3, -1

Answer: c) -3, -1, 2

5. What will be the expression of the polynomial?

a) x 3 + 2x 2 – 5x – 6

b) x 3 + 2x 2 – 5x + 6

c) x 3 + 2x 2 + 5x – 6

d) x 3 + 2x 2 + 5x + 6

Answer: a) x 3 + 2x 2 – 5x – 6

Also Check:

CBSE Case Study Questions for Class 10 Maths - All Chapters

Tips to Solve Case Study Based Questions Accurately

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• Class 10 Maths MCQs
• Chapter 4 Quadratic Equations

Class 10 Maths Chapter 4 Quadratic Equations MCQs

Class 10 Maths MCQs for Chapter 4 (Quadratic Equations) are available online here with answers. All these objective questions are prepared as per the latest CBSE syllabus (2022 – 2023) and NCERT guidelines. MCQs for Class 10 Maths Chapter 4 are prepared according to the new exam pattern. Solving these multiple-choice questions will help students to score good marks in the board exams, which they can verify with the help of detailed explanations given here. To get chapter-wise MCQs, click here . Also, find the PDF of MCQs to download here for free.

Class 10 Maths MCQs for Quadratic Equations

CBSE board has released the datasheet for the Class 10 Maths exam. It is advised for students to start revising the chapters, for the exam. Here, we have given multiple-choice questions for Chapter 4 quadratic equations, to help students to solve different types of questions, which could appear in the board exam. They can build their problem-solving capacity and boost their confidence level by practising the questions here. Get important questions for class 10 Maths here at BYJU’S.

Click here to download the PDF of additional MCQs for Practice on Quadratic equations, Chapter of Class 10 Maths along with answer key:

Download PDF

Students can also get access to Quadratic equations Class 10 Notes here.

Below are the MCQs for Quadratic Equations

1. Equation of (x+1) 2 -x 2 =0 has number of real roots equal to:

Answer: (a) 1

Explanation: (x+1) 2 -x 2 =0

X 2 +2x+1-x 2 = 0

Hence, there is one real root.

2. The roots of 100x 2 – 20x + 1 = 0 is:

(a) 1/20 and 1/20

(b) 1/10 and 1/20

(c) 1/10 and 1/10

(d) None of the above

Answer: (c) 1/10 and 1/10

Explanation: Given, 100x 2 – 20x + 1=0

100x 2 – 10x – 10x + 1 = 0

10x(10x – 1) -1(10x – 1) = 0

(10x – 1) 2 = 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

3. The sum of two numbers is 27 and product is 182. The numbers are:

(a) 12 and 13

(b) 13 and 14

(c) 12 and 15

(d) 13 and 24

Answer: (b) 13 and 14

Explanation: Let x is one number

Another number = 27 – x

Product of two numbers = 182

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

⇒ x = 13 or x = 14

4. If ½ is a root of the quadratic equation x 2 -mx-5/4=0, then value of m is:

Answer: (b) -2

Explanation: Given x=½ as root of equation x 2 -mx-5/4=0.

(½) 2 – m(½) – 5/4 = 0

¼-m/2-5/4=0

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

(a) Base=10cm and Altitude=5cm

(b) Base=12cm and Altitude=5cm

(c) Base=14cm and Altitude=10cm

(d) Base=12cm and Altitude=10cm

Answer: (b) Base=12cm and Altitude=5cm

Explanation: Let the base be x cm.

Altitude = (x – 7) cm

In a right triangle,

Base 2 + Altitude 2 = Hypotenuse 2 (From Pythagoras theorem)

∴ x 2 + (x – 7) 2 = 13 2

By solving the above equation, we get;

⇒ x = 12 or x = – 5

Since the side of the triangle cannot be negative.

Therefore, base = 12cm and altitude = 12-7 = 5cm

6. The roots of quadratic equation 2x 2 + x + 4 = 0 are:

(a) Positive and negative

(b) Both Positive

(c) Both Negative

(d) No real roots

Answer: (d) No real roots

Explanation: 2x 2 + x + 4 = 0

⇒ 2x 2 + x = -4

Dividing the equation by 2, we get

⇒ 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

The square root of negative number is imaginary, therefore, there is no real root for the given equation.

Answer: (b) 3

Explanation:

Hence, we can write, √(6+x) = x

x 2 -3x+2x-6=0

x(x-3)+2(x-3)=0

(x+2) (x-3) = 0

Since, x cannot be negative, therefore, x=3

8. The sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3. The present age of Rehman is:

Answer: (a) 7

Explanation: Let, x is the present age of Rehman

Three years ago his age = x – 3

Five years later his age = x + 5

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

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

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = x 2 + 2x – 15

⇒ x 2 – 4x – 21 = 0

⇒ x 2 – 7x + 3x – 21 = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

We know age cannot be negative, hence the answer is 7.

9. 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.

(a) 30 km/hr

(b) 40 km/hr

(c) 50 km/hr

(d) 60 km/hr

Answer: (b) 40 km/hr

Explanation: Let x km/hr be the speed of train.

Time required to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)(360-1/x) = 360

⇒ 360 – x + 1800-5/x = 360

⇒ x 2 + 5x + 10x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

Negative value is not considered for speed hence the answer is 40km/hr.

10. If one root of equation 4x 2 -2x+k-4=0 is reciprocal of the other. The value of k is:

Answer: (b) 8

Explanation: If one root is reciprocal of others, then the product of roots will be:

α x 1/α = (k-4)/4

11. Which one of the following is not a quadratic equation?

(a) (x + 2) 2  = 2(x + 3) 

(b) x 2  + 3x = (–1) (1 – 3x) 2

(c) (x + 2) (x – 1) = x 2  – 2x – 3 

(d) x 3  – x 2  + 2x + 1 = (x + 1) 3

Answer: (c) (x + 2) (x – 1) = x 2  – 2x – 3 

We know that the degree of a quadratic equation is 2.

By verifying the options,

x 2  + 4x + 4 = 2x + 6

x 2  + 2x – 2 = 0

This is a quadratic equation.

x 2  + 3x = -1(1 + 9x 2  – 6x)

x 2  + 3x + 1 + 9x 2  – 6x = 0

10x 2  – 3x + 1 = 0

x 2  + x – 2 = x 2  – 2x – 3

x 2  + x – 2 – x 2  + 2x + 3 = 0

This is not a quadratic equation.

12. Which of the following equations has 2 as a root?

(a) x 2  – 4x + 5 = 0 

(b) x 2  + 3x – 12 = 0

(c) 2x 2  – 7x + 6 = 0 

(d) 3x 2  – 6x – 2 = 0

Answer: (c) 2x 2  – 7x + 6 = 0 

If 2 is a root then substituting the value 2 in place of x should satisfy the equation.

Let us verify the given options.

(a) x 2  – 4x + 5 = 0

(2) 2  – 4(2) + 5 = 1 ≠ 0

So, x = 2 is not a root of x 2  – 4x + 5 = 0

(2) 2  + 3(2) – 12 = -2 ≠ 0

So, x = 2 is not a root of x 2  + 3x – 12 = 0

(c) 2x 2  – 7x + 6 = 0

2(2) 2  – 7(2) + 6 = 0

Here, x = 2 is a root of 2x 2  – 7x + 6 = 0

13. A quadratic equation ax 2  + bx + c = 0 has no real roots, if

(a) b 2  – 4ac > 0

(b) b 2  – 4ac = 0

(c) b 2  – 4ac < 0

(d) b 2  – ac < 0

Answer: (c) b 2  – 4ac < 0

A quadratic equation ax 2  + bx + c = 0 has no real roots, if b 2  – 4ac < 0. That means, the quadratic equation contains imaginary roots.

14. The product of two consecutive positive integers is 360. To find the integers, this can be represented in the form of quadratic equation as

(a) x 2  + x + 360 = 0

(b) x 2  + x – 360 = 0

(c) 2x 2  + x – 360

(d) x 2  – 2x – 360 = 0

Answer: (b) x 2  + x – 360 = 0

Let x and (x + 1) be the two consecutive integers.

According to the given,

x(x + 1) = 360

x 2  + x = 360

x 2  + x – 360

15. The equation which has the sum of its roots as 3 is

(a) 2x 2  – 3x + 6 = 0 

(b) –x 2  + 3x – 3 = 0

(c) √2x 2  – 3/√2x + 1 = 0 

(d) 3x 2  – 3x + 3 = 0

Answer: (b) –x 2  + 3x – 3 = 0

The sum of the roots of a quadratic equation ax 2  + bx + c = 0, a ≠ 0 is given by,

Coefficient of x / coefficient of x 2  = –(b/a)

Let us verify the options.

(a) 2x 2  – 3x + 6 = 0

Sum of the roots = – b/a = -(-3/2) = 3/2

(b) -x 2  + 3x – 3 = 0

Sum of the roots = – b/a = -(3/-1) = 3

(c) √2x 2  – 3/√2x + 1=0

2x 2  – 3x + √2 = 0

Sum of the roots = – b/a = -(-3/3) = 1

16. The quadratic equation 2x 2   – √5x + 1 = 0 has

(a) two distinct real roots 

(b) two equal real roots

(c) no real roots 

(d) more than 2 real roots

Answer: (c) no real roots

2x 2  – √5x + 1 = 0

Comparing with the standard form of a quadratic equation,

a = 2, b = -√5, c = 1

b 2  – 4ac = (-√5) 2  – 4(2)(1)

= 5 – 8 

= -3 < 0

Therefore, the given equation has no real roots.

17. The equation (x + 1) 2  – 2(x + 1) = 0 has

(a) two real roots

(b) no real roots 

(c) one real root

(d) two equal roots

Answer: (a) two real roots

(x + 1) 2  – 2(x + 1) = 0

x 2  + 1 + 2x – 2x – 2 = 0

x 2  – 1 = 0

18. The quadratic formula to find the roots of a quadratic equation ax 2  + bx + c = 0 is given by

(a) [-b ± √(b 2 -ac)]/2a

(b) [-b ± √(b 2 -2ac)]/a

(c) [-b ± √(b 2 -4ac)]/4a

(d) [-b ± √(b 2 -4ac)]/2a

Answer: (d) [-b ± √(b 2 -4ac)]/2a

The quadratic formula to find the roots of a quadratic equation ax 2  + bx + c = 0 is given by [-b ± √(b 2 -4ac)]/2a.

19. The quadratic equation x 2  + 7x – 60 has

(a) two equal roots

(b) two real and unequal roots

(b) no real roots

(c) two equal complex roots

Answer: (b) two real and unequal roots

x 2  + 7x – 60 = 0

Comparing with the standard form,

a = 1, b = 7, c = -60

b 2  – 4ac = (7) 2  – 4(1)(-60) = 49 + 240 = 289 > 0

Therefore, the given quadratic equation has two real and unequal roots.

20. The maximum number of roots for a quadratic equation is equal to

Answer: (b) 2

The maximum number of roots for a quadratic equation is equal to 2 since the degree of a quadratic equation is 2.

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