NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations (Ex 4.3) Exercise 4.3

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Mathematics is one of the most important subjects in the Class 10 curriculum. Students are required to focus more on practising Mathematics questions on a regular basis. All the chapters in Class 10 are very significant for building a foundation for advanced topics that students will be learning in future courses. It is advised that students access study materials that are helpful in scoring well in Mathematics. Students can download all the essential study materials from Extramarks.

Mathematics is a core subject in the Class 10 curriculum. To learn the topics in Class 10, students are advised to hone their analytical abilities. All students’ ability to develop analytical thinking can benefit from an understanding of Mathematics concepts. Practice is the most important aspect of preparing for the Mathematics examination. Class 10 students are encouraged to practise chapter-based questions on a regular basis in order to improve their knowledge. In order to improve their learning process, students must constantly assess their Mathematics preparation level. The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 are thus accessible on the Extramarks website and mobile application. With the NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3, you can easily solve all of the questions in Exercise 4.3.All of the questions in Chapter 4 must be practised numerous times because they may appear on the Mathematics exam paper.Mathematics can help children become more rational and analytical thinkers.The principles of Mathematics come in quite handy in daily life. Calculations involving basic arithmetic are helpful while performing routine chores. It is also essential for fostering mental discipline and improving organisation. The improvement of global comprehension and communication depends on mathematical notions. Basic mathematical operations and calculations are required for a wide variety of official jobs. Understanding mathematical concepts improves students’ employment options. A student who is proficient in Mathematics can find employment as a Business Analyst, Statistician, Mathematician, Information Research Scientist, and other related positions.

The Central Board of Secondary Education was founded by the Ministry of Education (CBSE). The CBSE Board is a national Indian educational board. The CBSE Board annually conducts and manages final examinations for students in Class 10 and Class 12. The CBSE Board syllabus is used in the vast majority of Indian private and public schools. Students of all classes are encouraged to learn the subject according to the CBSE syllabus to save time and effort. Students in Class 10 must improve their ongoing board exam preparations. It is necessary for them to keep referring to the CBSE syllabus for staying on the right track of preparation.

Extramarks offers mock tests for students to test their knowledge of each subject. Mock tests are critical for improving final exam performance. \Students will become familiar with the types of questions that can be framed from the topics in their syllabus after taking mock tests. Extramarks also generates individualised reports and analysis for students to gain a better understanding of their preparation. They are extremely beneficial in improving students’ learning strategies. Extramarks can assist students from kindergarten to class 12 in preparing for their exams.

The past years’ papers and sample papers of Mathematics are accessible on the Extramarks learning website. Students will be able to prepare according to the marking scheme of Mathematics with the help of past years’ papers. It is important to keep revising the formulas and definitions of Chapter 4 in order to solve questions effectively. Students can revise the formulas and definitions specific to questions of Exercise 4.3 by going through the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3. All the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 are very comprehensive. Many of the topics of Chapter 4 are crucial from the examination point of view.

The central theme of Chapter 4 is Quadratic Equations. All the topics are explained well in the NCERT textbook. Reading theoretical portions of Chapter 4 from the NCERT textbook is useful for understanding some of the vital concepts of Algebra. After reading the theories in this chapter, students should focus on solving the questions of Chapter 4. They can download the NCERT solutions for all exercises of Chapter 4 from Extramarks.

Students can take the help of live classes of Extramarks to prepare well for the Mathematics examination. All the chapters of Mathematics can be well prepared with theguidance of teachers present on Extramarks. Live classes will help students in improving their performance in the Mathematics examination. Students can also access practise materials from the Extramarks learning portal. The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 are accessible in PDF format on Extramarks.

In order to answer questions effectively, it is critical to have confidence. With the aid of the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3, students in Class 10 can increase their self-assessment analysis. hey will be encouraged to practise more by using the NCERT Solutions for Class 10 Maths, Chapter 4, Exercise 4.3 to practise questions.The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 can be used to solve questions related to a specific topic. All of the questions in Exercise 4.3 are significant from the board examination perspective, thus students are urged to practise them periodically.

For a high score in the Mathematics examination, revision is crucial. It helps students quickly review the crucial concepts, principles, and theorems from each chapter. Students in Class 10 can obtain the Mathematics revision notes from Extramarks. Since many of the topics are interrelated, it is essential that students continually go over them. They will be able to answer questions quickly when topics are reviewed often. To review all the crucial subjects before the examinations, revision notes are a huge help. It is crucial for students to consistently practise questions in a subject like Mathematics. For some students, practising Mathematics chapters can be challenging. Students need to be helped with their weakest areas in a topic like Mathematics in order to improve. Making a study regimen for chapters in Mathematics may be difficult for students. To familiarise themselves with the concepts and subtopics covered in Class 10 Mathematics, students are urged to access the most recent Mathematics syllabus. Students can get the most updated Class 10 Mathematics syllabus on the Extramarks website and learning application. Every chapter in Mathematics has a specific demand, so it is important to prepare in a way that can meet those requirements.

Chapter 4 is an important chapter in the Class 10 curriculum that will be tested in the board examination. Students must carefully study all of its topics. It is critical for Class 10 students to keep reviewing the topics in order to effectively learn new concepts. Students should devise a strategy for preparing for the Mathematics examination. Accessing the most recent Mathematics syllabus is extremely beneficial in developing a learning strategy for the chapters. Students will be able to effectively prepare for the Mathematics examinations if they go over the topics according to the syllabus. This will assist students in staying on track with their preparation. The topics covered in the Mathematics examinations are used to frame the questions.

All of the topics in Chapter 4 are thoroughly explained in the NCERT Mathematics textbook. When preparing for the Class 10 Mathematics examination, it is best to use the NCERT textbook, which is the CBSE Board recommended textbook. The NCERT textbook thoroughly explains all of the ideas and concepts. Students can easily understand all topics if they refer to textbooks on a regular basis. The NCERT textbook thoroughly explains all of the important topics in Chapter 5. The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 can be crucial in achieving a high score in the Mathematics examination. It is critical for students in Class 10 to keep a timetable for learning subjects.

Students can get assistance from the educators at Extramarks to findeffective solutions to their concerns related to the subjects. The faculty at Extramarks is always accessible to answer questions from students. Through video modules, Extramarks focuses on concept development. These lessons were developed by highly experienced educators from all around India. More than ten thousand schools in India rely on Extramarks. It is a learning platform that links instructors and students. With the aid of Extramarks, all students can develop subject concepts in an engaging way. Students can access reliable study materials on the Extramarks website and mobile application. The Extramarks learning platform allows students to prepare their courses whenever it is most convenient for them. Students can also prepare for competitive examinations with the guidance of Extramarks.

It is important for students to go through the derivations in Chapter 4 to understand the application of the topics. Derivations are significant for learning how formulae can be used to solve problems. All the steps of derivations are important and students should focus well on them. With the help of the NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3, students will be able to revise the derivation related to the solution of a quadratic equation by the method of completing the square.Students need to have access to dependable study materials. Extramarks provides exceptionally reliable and authentic study material for all topics. From the Extramarks website and mobile application, students can easily access them. Every single study tool is updated regularly whenever there is a change in the syllabus. Students can rely on Extramarks to assist them in effectively preparing for their upcoming examinations. Extramarks further provides learning resources for the CAT, CUET, JEE, and other examinations. Study materials are easily accessible to all students, helping them better prepare for examinations. Extramarks is a very credible source for accessing study materials. All the questions of each subject can be solved easily with the help of the NCERT solutions provided by Extramarks. The NCERT solutions are very helpful tools that enable students to solve exercise questions in a comprehensive and logically correct manner. All the important questions related to a topic must be practised well to increase their preparation level for the Mathematics examination. Mathematics is a crucial subject in the Class 10 curriculum and students should practise all the questions in each chapter consistently.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Class 10 Maths Chapter 4 Exercise 4.3 is based on the topic of the Solution of a Quadratic Equation by Completing the Square. To practise questions related to the topic, students can make use of the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3. All the Class 10 Maths Chapter 4 Exercise 4.3 Solutions are given in a format that is easy to understand. Students must keep referring to the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3. They can compare their solutions with the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 and make required changes to them. There are 11 questions in Class 10 Maths Ch 4 Ex 4.3. It is important that students practise all of them a few times, and if they encounter any difficulty, they should refer to the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3. The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 are available in PDF format on the Extramarks learning portal.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3

Students must learn to implement proper formulae in order to solve questions effectively. When they refer to the NCERT Solutions for Class 10 Maths, Chapter 4, Exercise 4.3, they can easily learn to apply the correct formulae.Other than the NCERT Solutions for Class 10 Maths, Chapter 4, Exercise 4.3, students can also access the NCERT solutions for the remaining exercises in Chapter 4.They are also provided in simple language.

Topics to Learn before Solving Exercise 4.3

Students are required to cover the Solution of a Quadratic Equation by Completing the Square topic before trying to solve Exercise 4.3. All the questions of Exercise 4.3 can be readily solved with the help of the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3. They will be able to solve all the 11 questions of Exercise 4.3 in a stepwise manner with the help of the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3.

Exercise 4.3

The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 can help students improve their calculation speed. They can easily score well in Mathematics if they regularly refer to the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3. For them to be able to adequately prepare for the exams, students must be able to retain formulas, definitions, and theorems for a longer period of time. Class 10 students must revise them on a regular basis in order to retain them for a long time. The formulae, definitions, and theorems can all be reviewed using the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3

All the 11 questions of Exercise 4.3 can be solved well with the assistance of the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3. At Extramarks, Central Board of Secondary Education students can access the NCERT Solutions for every subject. The NCERT Solutions Class 12 can be used by students in Class 12 to help them prepare for their board exams. Students in Class 11 can get the NCERT Solutions Class 11 from the Extramarks learning portal to help them study for their final exams. Students will find the NCERT Solutions for Class 10 on Extramarks for Hindi, English, Mathematics, etc. Class 10 students can simply obtain the NCERT Solutions Class 10 for all topics on the Extramarks website and learning app. All students in Class 9 can access the NCERT Solutions Class 9 for Mathematics, English, Hindi, and other subjects via the Extramarks website and mobile application. Class 8 students need access to the NCERT Solutions Class 8 in order to respond to exercise questions for English, Mathematics, Hindi and other subjects. NCERT Solutions Class 7 are available on Extramarks for Class 7 students to use as a study tool while they prepare for the Mathematics board exam. Students in Class 6 can use the NCERT Solutions Class 6 to practise questions from all chapters of each subject in their course.

NCERT Solutions for Class 10 Maths

The NCERT solutions are available chapter-wise on Extramarks. The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 are very comprehensive and detailed. They can easily refer to the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 as they are given in simple format.

NCERT Solutions for Class 10 Maths Chapter 4 Exercises

All the exercises of Chapter 4 can be solved with the help of Extramarks. Students can easily access the NCERT solutions for all exercises of Chapter 4 from Extramarks. The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 are given in a thorough manner for easy understanding. Students can easily gain higher marks in Mathematics with the help of the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3. The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 are given in a well-organised manner. They can access the NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 from the Extramarks website and mobile application in PDF format. The NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.3 are essential study materials for scoring well in the Mathematics examination.

Q.1

$\begin{array}{l} Find the roots of the following quadratic equations, \\ if they exist, by the method of completing the square: \\ (i) 2 x^{2} - 7 x + 3 = 0 (ii) 2 x^{2} + x - 4 = 0 \\ (iii) 4 x^{2} + 4 \sqrt{3} x + 3 = 0 (iv) 2 x^{2} + x + 4 = 0 \end{array}$

Ans.

$\begin{array}{l} (i) \\ 2 x^{2} - 7 x + 3 = 0 \\ \Rightarrow x^{2} - \frac{7}{2} x + \frac{3}{2} = 0 \\ \Rightarrow x^{2} - 2 \times \frac{7}{4} x + {(\frac{7}{4})}^{2} - {(\frac{7}{4})}^{2} + \frac{3}{2} = 0 \\ \Rightarrow {(x - \frac{7}{4})}^{2} - \frac{49}{16} + \frac{3}{2} = 0 \\ \Rightarrow {(x - \frac{7}{4})}^{2} - \frac{25}{16} = 0 \\ \Rightarrow {(x - \frac{7}{4})}^{2} - {(\frac{5}{4})}^{2} = 0 \\ \Rightarrow {(x - \frac{7}{4})}^{2} = {(\frac{5}{4})}^{2} \\ \Rightarrow x - \frac{7}{4} = \pm \frac{5}{4} \\ \Rightarrow x = \frac{7}{4} \pm \frac{5}{4} \\ \Rightarrow x = \frac{12}{4} = 3 or x = \frac{2}{4} = \frac{1}{2} \\ (ii) \\ 2 x^{2} + x - 4 = 0 \\ \Rightarrow x^{2} + \frac{1}{2} x - 2 = 0 \\ \Rightarrow x^{2} + 2 \times \frac{1}{4} x + {(\frac{1}{4})}^{2} = {(\frac{1}{4})}^{2} + 2 \\ \Rightarrow {(x + \frac{1}{4})}^{2} = \frac{1}{16} + 2 = \frac{33}{16} \\ \Rightarrow {(x + \frac{1}{4})}^{2} = {(\frac{\sqrt{33}}{4})}^{2} \\ \Rightarrow x + \frac{1}{4} = \pm \frac{\sqrt{33}}{4} \\ \Rightarrow x = - \frac{1}{4} \pm \frac{\sqrt{33}}{4} \\ \Rightarrow x = \frac{- 1 + \sqrt{33}}{4} or x = \frac{- 1 - \sqrt{33}}{4} \\ (iii) \\ 4 x^{2} + 4 \sqrt{3} x + 3 = 0 \\ \Rightarrow {(2 x)}^{2} + 2 (2 x) (\sqrt{3}) + {(\sqrt{3})}^{2} = 0 \\ \Rightarrow {(2 x + \sqrt{3})}^{2} = 0 \\ \Rightarrow (2 x + \sqrt{3}) (2 x + \sqrt{3}) = 0 \\ \Rightarrow 2 x + \sqrt{3} = 0 or 2 x + \sqrt{3} = 0 \\ \Rightarrow x = - \frac{\sqrt{3}}{2} or x = - \frac{\sqrt{3}}{2} \\ (iv) \\ 2 x^{2} + x + 4 = 0 \\ \Rightarrow x^{2} + \frac{1}{2} x + 2 = 0 \\ \Rightarrow x^{2} + 2 \times \frac{1}{4} x + {(\frac{1}{4})}^{2} - {(\frac{1}{4})}^{2} + 2 = 0 \\ \Rightarrow {(x + \frac{1}{4})}^{2} = \frac{1}{16} - 2 = - \frac{31}{16} \\ Square of a number can not be negative. \\ Therefore, there is no real root for the given equation. \end{array}$

Q.2

$\begin{array}{l} Find the roots of the following quadratic equations, \\ by applying quadratic formula: \\ (i) 2 x^{2} - 7 x + 3 = 0 (ii) 2 x^{2} + x - 4 = 0 \\ (iii) 4 x^{2} + 4 \sqrt{3} x + 3 = 0 (iv) 2 x^{2} + x + 4 = 0 \end{array}$

Ans.

$\begin{array}{l} (i) 2 x^{2} - 7 x + 3 = 0 \\ On comparing the given equation with a x^{2} + bx + c = 0, we get \\ a = 2, b = - 7, c = 3 \\ By using quadratic formula, we get \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ \Rightarrow x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 \times 2 \times 3}}{2 \times 2} \\ \Rightarrow x = \frac{7 \pm \sqrt{49 - 24}}{4} \\ \Rightarrow x = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4} \\ \Rightarrow x = \frac{7 + 5}{4} or \frac{7 - 5}{4} \\ \Rightarrow x = \frac{12}{4} = 3 or x = \frac{2}{4} = \frac{1}{2} \\ (ii) 2 x^{2} + x - 4 = 0 \\ On comparing the given equation with a x^{2} + bx + c = 0, we get \\ a = 2, b = 1, c = - 4 \\ By using quadratic formula, we get \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ \Rightarrow x = \frac{- 1 \pm \sqrt{1^{2} - 4 \times 2 \times (- 4)}}{2 \times 2} \\ \Rightarrow x = \frac{- 1 \pm \sqrt{1^{2} + 32}}{4} \\ \Rightarrow x = \frac{- 1 \pm \sqrt{33}}{4} \\ \Rightarrow x = \frac{- 1 + \sqrt{33}}{4} or x = \frac{- 1 - \sqrt{33}}{4} \\ (iii) 4 x^{2} + 4 \sqrt{3} x + 3 = 0 \\ On comparing the given equation with a x^{2} + bx + c = 0, we get \\ a = 4, b = 4 \sqrt{3}, c = 3 \\ By using quadratic formula, we get \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ \Rightarrow x = \frac{- 4 \sqrt{3} \pm \sqrt{{(4 \sqrt{3})}^{2} - 4 \times 4 \times 3}}{2 \times 4} \\ \Rightarrow x = \frac{- 4 \sqrt{3} \pm \sqrt{48 - 48}}{8} \\ \Rightarrow x = \frac{- 4 \sqrt{3} \pm 0}{8} \\ \Rightarrow x = - \frac{\sqrt{3}}{2} or x = - \frac{\sqrt{3}}{2} \\ (iv) 2 x^{2} + x + 4 = 0 \\ On comparing the given equation with a x^{2} + bx + c = 0, we get \\ a = 2, b = 1, c = 4 \\ By using quadratic formula, we get \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ \Rightarrow x = \frac{- 1 \pm \sqrt{1^{2} - 4 \times 2 \times 4}}{2 \times 2} \\ \Rightarrow x = \frac{- 1 \pm \sqrt{- 31}}{4} \\ Square of a number can not be negative. \\ Therefore, \sqrt{- 31} is not real and so there is no real root for \\ the given equation. \end{array}$

Q.3

$\begin{array}{l} Find the roots of the following equations: \\ (i) x - \frac{1}{x} = 3, x \neq 0 (ii) \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}, x \neq - 4, 7 \end{array}$

Ans.

$\begin{array}{l} (i) \\ x - \frac{1}{x} = 3, x \neq 0 \\ \Rightarrow \frac{x^{2} - 1}{x} = 3 \\ \Rightarrow x^{2} - 1 = 3 x \\ \Rightarrow x^{2} - 3 x - 1 = 0 \\ On comparing this equation with a x^{2} + bx + c = 0, we get \\ a = 1, b = - 3, c = - 1 \\ By using quadratic formula, we get \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ \Rightarrow x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 \times 1 \times (- 1)}}{2 \times 1} \\ \Rightarrow x = \frac{3 \pm \sqrt{9 + 4}}{2} \\ \Rightarrow x = \frac{3 + \sqrt{13}}{2} or x = \frac{3 - \sqrt{13}}{2} \\ (ii) \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}, x \neq - 4, 7 \\ \Rightarrow \frac{x - 7 - x - 4}{(x + 4) (x - 7)} = \frac{11}{30} \\ \Rightarrow \frac{- 11}{x^{2} - 3 x - 28} = \frac{11}{30} \\ \Rightarrow x^{2} - 3 x - 28 = - 30 \\ \Rightarrow x^{2} - 3 x + 2 = 0 \\ On comparing this equation with a x^{2} + bx + c = 0, we get \\ a = 1, b = - 3, c = 2 \\ By using quadratic formula, we get \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ \Rightarrow x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 \times 1 \times 2}}{2 \times 1} \\ \Rightarrow x = \frac{3 \pm \sqrt{9 - 8}}{2} \\ \Rightarrow x = \frac{3 + 1}{2} or x = \frac{3 - 1}{2} \\ \Rightarrow x = 2 or x = 1 \end{array}$

Q.4

$\begin{array}{l} The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago \\ and 5 years from now is \frac{1}{3} . Find his present age. \end{array}$

Ans.

$\begin{array}{l} Let the present age of Rehman is x years. Then 3 years ago, \\ Rehman’s age = (x - 3) years \\ 5 years from now, \\ Rehman’s age = (x + 5) years \\ According to question, \\ \frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \\ \Rightarrow \frac{x + 5 + x - 3}{(x - 3) (x + 5)} = \frac{1}{3} \\ \Rightarrow \frac{2 x + 2}{x^{2} + 2 x - 15} = \frac{1}{3} \\ \Rightarrow 6 x + 6 = x^{2} + 2 x - 15 \\ \Rightarrow x^{2} + 2 x - 15 - 6 x - 6 = 0 \\ \Rightarrow x^{2} - 4 x - 21 = 0 \\ On comparing this equation with a x^{2} + bx + c = 0, we get \\ a = 1, b = - 4, c = - 21 \\ By using quadratic formula, we get \\ x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a} \\ \Rightarrow x = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 \times 1 \times (- 21)}}{2 \times 1} \\ \Rightarrow x = \frac{4 \pm \sqrt{16 + 84}}{2} \\ \Rightarrow x = \frac{4 + 10}{2} or x = \frac{4 - 10}{2} \\ \Rightarrow x = 7 or x = - 3 \\ Age can not be a negative quantity. Therefore, Rehman’s \\ present age is 7 years. \end{array}$

Q.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.

Ans.

$\begin{array}{l} Let Shefali’s marks in Mathematics and English be x and y \\ respectively. \\ According to question, \\ x + y = 30 . .. (1) \\ and \\ (x + 2) (y - 3) = 210 . .. (2) \\ \Rightarrow (x + 2) (30 - x - 3) = 210 [from (1), y = 30 - x] \\ \Rightarrow (x + 2) (27 - x) = 210 \\ \Rightarrow 27 x - x^{2} + 54 - 2 x = 210 \\ \Rightarrow 25 x - x^{2} = 210 - 54 = 156 \\ \Rightarrow x^{2} - 25 x + 156 = 0 \\ \Rightarrow x^{2} - 12 x - 13 x + 156 = 0 \\ \Rightarrow x (x - 12) - 13 (x - 12) = 0 \\ \Rightarrow (x - 12) (x - 13) = 0 \\ \Rightarrow x = 12 or x = 13 \\ On putting these values of x in (1), we get \\ y = 18 when x = 12 \\ and \\ y = 17 when x = 13 \\ Therefore, Shefali’s marks in Mathematics and English are \\ either 12 and 18 or 13 and 17 respectively. \end{array}$

Q.6

$\begin{array}{l} The diagonal of a rectangular field is 60 metres more \\ than the shorter side. If the longer side is 30 metres \\ more than the shorter side, find the sides of the field. \end{array}$

Ans.

$\begin{array}{l} Let length of the shorter side is x . Then length of the \\ longer side is (x + 30) and length of the diagonal is \\ (x + 60) . \\ Using Pythagoras theorem, we have \\ x^{2} + {(x + 30)}^{2} = {(x + 60)}^{2} \\ \Rightarrow x^{2} + x^{2} + 60 x + 900 = x^{2} + 120 x + 3600 \\ \Rightarrow x^{2} - 60 x - 2700 = 0 \\ \Rightarrow x^{2} - 90 x + 30 x - 2700 = 0 \\ \Rightarrow x (x - 90) + 30 (x - 90) = 0 \\ \Rightarrow (x - 90) (x + 30) = 0 \\ \Rightarrow x - 90 = 0 or x + 30 = 0 \\ \Rightarrow x = 90 or x = - 30 \\ Length can not be negative. \\ Thus, x = 90 m \\ and x + 30 = 90 + 30 = 120 m \\ Hence, length of the shorter and longer sides of the rectangular \\ field are 90 m and 120 m respectively. \end{array}$

Q.7 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.

Ans.

$\begin{array}{l} Let smaller and larger number be x and y respectively. \\ Then, we have \\ y^{2} - x^{2} = 180 \\ and \\ x^{2} = 8 y \\ {Replacing x}^{2} with 8y in y^{2} - x^{2} = 180, we get \\ y^{2} - 8 y = 180 \\ \Rightarrow y^{2} - 8 y - 180 = 0 \\ \Rightarrow y^{2} - 18 y + 10 y - 180 = 0 \\ \Rightarrow y (y - 18) + 10 (y - 18) = 0 \\ \Rightarrow (y - 18) (y + 10) = 0 \\ \Rightarrow y - 18 = 0 or y + 10 = 0 \\ \Rightarrow y = 18 or y = - 10 \\ {Negative value of y does not satisfy the equation x}^{2} = 8 y . \\ Therefore, y = 18 \\ and hence \\ x^{2} = 8 y \\ \Rightarrow x^{2} = 8 \times 18 = 144 = 12^{2} \\ \Rightarrow x = \pm 12 \\ Hence larger number is 18 and smaller number is either \\ 12 or - 12 . \end{array}$

Q.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.

Ans.

$\begin{array}{l} Let the speed of the train be x km/h. \\ Time taken to cover 360 km is \frac{360}{x} hours. \\ According to question, \\ (x + 5) (\frac{360}{x} - 1) = 360 \\ \Rightarrow (x + 5) (360 - x) = 360 x \\ \Rightarrow 360 x - x^{2} + 1800 - 5 x = 360 x \\ \Rightarrow - x^{2} + 1800 - 5 x = 0 \\ \Rightarrow x^{2} + 5 x - 1800 = 0 \\ \Rightarrow x^{2} + 45 x - 40 x - 1800 = 0 \\ \Rightarrow x (x + 45) - 40 (x + 45) = 0 \\ \Rightarrow (x + 45) (x - 40) = 0 \\ \Rightarrow x + 45 = 0 or x - 40 = 0 \\ \Rightarrow x = - 45 or x = 40 \\ Here, speed can not be negative. \\ Therefore, speed = x = 40 km/h \end{array}$

Q.9

$\begin{array}{l} \begin{array}{l} Two water taps together can fill a tank in 9 \frac{3}{8} hours. \\ The tap of larger diameter takes 10 hours less than \end{array} \\ \begin{array}{l} the smaller one to fill the tank separately. \\ Find the time in which each tap can separately fill the tank. \end{array} \end{array}$

Ans.

$\begin{array}{l} Let the tap of smaller diameter fills the tank in x hours. \\ So, in 1 hour it fills \frac{1}{x} part of the tank. \\ The tap of larger diameter fills the tank in (x - 10) hours \\ and so in 1 hour it fills \frac{1}{x - 10} part of the tank. \\ The two water taps together fill the tank in 9 \frac{3}{8} hours. \\ So, we have \\ 9 \frac{3}{8} (\frac{1}{x}) + 9 \frac{3}{8} (\frac{1}{x - 10}) = 1 \\ \Rightarrow \frac{75}{8} (\frac{1}{x}) + \frac{75}{8} (\frac{1}{x - 10}) = 1 \\ \Rightarrow \frac{75}{8} (\frac{x - 10 + x}{x (x - 10)}) = 1 \\ \Rightarrow \frac{75}{8} (\frac{2 x - 10}{x^{2} - 10 x}) = 1 \\ \Rightarrow 150 x - 750 = 8 x^{2} - 80 x \\ \Rightarrow 8 x^{2} - 230 x + 750 = 0 \\ \Rightarrow 8 x^{2} - 200 x - 30 x + 750 = 0 \\ \Rightarrow 8 x (x - 25) - 30 (x - 25) = 0 \\ \Rightarrow (8 x - 30) (x - 25) = 0 \\ \Rightarrow 8 x - 30 = 0 or x - 25 = 0 \\ \Rightarrow x = \frac{30}{8} = \frac{15}{4} or x = 25 \\ If x = \frac{15}{4}, then x - 10 = \frac{15}{4} - 10 = \frac{- 25}{4} \\ But time can not be negative. \\ So, here we take x = 25 and so we have x - 10 = 25 - 10 = 15 \\ Therefore, the tap of smaller and larger diameters fill the \\ tank in 25 hours and 15 hours respectively. \end{array}$

Q.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 11km/h more than that of the passenger train, find the average speed of the two trains.

Ans.

$\begin{array}{l} Let average speed of the passenger train be x km/h. \\ Then average speed of the express train is (x + 11) km/h. \\ It is given that the time taken by the express train \\ to cover 132 km is 1 hour less than the passenger \\ train to cover the same distance. \\ Therefore, \\ \frac{132}{x} - \frac{132}{x + 11} = 1 \\ or 132 (x + 11 - x) = x (x + 11) \\ or 1452 = x^{2} + 11 x \\ or x^{2} + 11 x - 1452 = 0 \\ or x^{2} + 44 x - 33 x - 1452 = 0 \\ or x (x + 44) - 33 (x + 44) = 0 \\ or (x - 33) (x + 44) = 0 \\ or x - 33 = 0 or x + 44 = 0 \\ or x = 33 or x = - 44 \\ Speed can not be negative. \\ Therefore, \\ Average speed of the passanger train = x km/h = 33 km/h \\ Average speed of the express train = (x + 11) km/h = 44 km/h \end{array}$

Q.11 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.

Ans.

$\begin{array}{l} Let sides of the two squares are x and y . \\ According to question, \\ 4 x - 4 y = 24 \Rightarrow x - y = 6 \Rightarrow x = y + 6 . .. (1) \\ Also, we have \\ x^{2} + y^{2} = 468 \\ \Rightarrow {(y + 6)}^{2} + y^{2} = 468 \\ \Rightarrow y^{2} + 12 y + 36 + y^{2} = 468 \\ \Rightarrow 2 y^{2} + 12 y + 36 - 468 = 0 \\ \Rightarrow 2 y^{2} + 12 y - 432 = 0 \\ \Rightarrow y^{2} + 6 y - 216 = 0 \\ \Rightarrow y^{2} + 18 y - 12 y - 216 = 0 \\ \Rightarrow y (y + 18) - 12 (y + 18) = 0 \\ \Rightarrow (y - 12) (y + 18) = 0 \\ \Rightarrow y - 12 = 0 or y + 18 = 0 \\ \Rightarrow y = 12 or y = - 18 \\ Side of a square can not be negative. \\ So, y = 12 \\ Substituting this value of y in equation (1), we get \\ x = 12 + 6 = 18 \\ Hence, the sides of the two squares are 12 m and 18 m. \end{array}$

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations (Ex 4.3) Exercise 4.3

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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Topics to Learn before Solving Exercise 4.3

Exercise 4.3

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.3

NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 4 Exercises

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations (Ex 4.3) Exercise 4.3

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NCERT Solutions for Class 10 Maths Chapter 4 Exercises

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