# NCERT Solutions Class 8 Maths Chapter 9 Exercise 9.5

Mathematics is a highly interesting and practical discipline. It is one of the most important subjects that are a part of the Class 8 curriculum. Mathematics is the study of numbers, their relationships, analysis, formulae, quantities, space, and forms. It is an essential component of Engineering, Science, Pharmaceutics, Finance and Accounting, Computer Science, and Medical Science among other fields. Mathematics acts as a cornerstone for numerous occupations, including Engineering, Architecture, Flying, Statistics, Insurance, and many more. Mathematics aids in the development of a variety of abilities that are essential to an individual’s overall growth. It aids in the development of abilities such as Critical Thinking, Problem-Solving, Logical Reasoning, Analysis and Management, Quantitative Aptitude, and so on. Acquiring these abilities can help students clear a variety of competitive examinations like NEET, AFCAT, MAT, JEE Mains, JEE Advance, AIIMS, CAT, CUET, and others. Students must instill these abilities in order to perform well in the relevant examinations. Developing these abilities may help students lay the foundation for a career in a variety of professions; hence, it is critical that students study Mathematics with vigour. Students may make use of the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to develop basic mathematical abilities. The NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 can be accessed by students from the Extramarks website.

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## NCERT Solutions For Class 8 Maths Chapter 9 Algebraic Expressions And Identities (EX 9.5) Exercise 9.5

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### Access NCERT solutions for Class 8 Chapter 9-Algebraic Expressions and Identities

Students must adequately study the NCERT curriculum in order to obtain the greatest possible results in the Senior Secondary Examination. They must understand the significance of performing well in the senior secondary examination. The grades students receive in their senior secondary examinations play a role in determining how far they will advance academically.Students may utilise the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to help them prepare for the senior secondary examination. The NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 have been thoroughly detailed to assist students to comprehend the study material better. One of the most essential exercises from the NCERT Mathematics Chapter 9 Algebraic Expressions and Identities syllabus is Class 8 Maths Exercise 9.5.

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### NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5

Chapter 9 entitled – Algebraic Expressions and Identities is one of the most important chapters in the NCERT Mathematics curriculum for Class 8. This chapter comprises many complex formulas. To master this chapter, students must understand not only the ideas discussed in it but also how to apply these ideas practically. They are, thus, encouraged to practice the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to better understand the practical application required in Chapter 9 Algebraic Expressions and Identities. Students must study and comprehend the mathematical themes taught in the Class 8 Mathematics curriculum. It is critical for students to expand their understanding of mathematical ideas. This has a substantial impact on Class 8 students’ performance in the Senior Secondary Examination. Students can substantially benefit from gaining the knowledge needed for the practical application required in Chapter 9 Algebraic Expressions and Identities, and students are encouraged to practise the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 as they prepare for the Senior Secondary Mathematics Examination.

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The CBSE board guidelines are adhered to in the development of the NCERT curriculum. Students enrolled with the CBSE board are advised to study from the NCERT textbook. It is recommended that students study for their senior secondary examinations using the NCERT exercise problems and solutions. Students in Class 8 can use the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to prepare for the Mathematics examination. The NCERT solutions to all exercise problems from all courses are accessible on the Extramarks website and mobile application. Students may obtain the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 from the Extramarks website or mobile application. Students in Class 8 can participate in a doubt-clearing session on the Extramarks website to get answers to any questions they may have regarding the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5. Extramarks is a website that provides study resources for all boards and classes. Students taking the CBSE board examination can refer to Extramarks’ study resources while preparing for their exam, as they are designed with the CBSE board examination in mind. The NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 available on the Extramarks website, are strongly recommended for Class 8 students preparing for the Senior Secondary Mathematics Examination.

### NCERT Solutions for Class 8

Since the Central Board of Secondary Education question paper pattern is based on the NCERT curriculum, it is recommended that students prepare for the CBSE board examination by studying the NCERT exercise problems and solutions. The NCERT curriculum serves as the foundation for the CBSE board examination framework. Students may get the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 from the Extramarks website to help them prepare for the Senior Secondary Examination of Class 8 Mathematics. To score well in the senior secondary examination, students must master all of the concepts taught in Chapter 9 of the NCERT Mathematics book. Students in Class 8 can use the Extramarks website to get study material based on CBSE board criteria. While studying for the senior secondary examination, students should refer to the study material designed in accordance with the CBSE board standards. The study material and learning resources provided by Extramarks have been prepared in accordance with the CBSE board standards. Students preparing for the Senior Secondary Mathematics Examination can refer to the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 on the Extramarks website and learning application. Students may use the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to help them improve their academic performance.

They should consider the weightage of each topic taught in the concerned curriculum when studying for the senior secondary examination. The topic weightage for the senior secondary examination should be considered during preparation since it can help students determine the value of each topic. The ideas covered in Mathematics Exercise 9.5 of Class 8 are critical in the Senior Secondary Examination of Mathematics. The NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 are useful for studying for the concerned examination. Students must make an effort to comprehend the topics taught in the relevant curriculum in order to be adequately prepared for the final exam. In order to fully prepare for the senior secondary examination, students must stick to a timetable that devotes substantial time to each topic covered in the curriculum. They may use the NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 to help them prepare. Extramarks’ NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 have been formed by experts in the concerned field of study. The NCERT Solutions For Class 8 Maths Chapter 9 Exercise 9.5 have been methodically broken down into steps for the convenience of students.  Students are advised to be sincere in their preparation for the senior secondary examination because the results of the senior secondary examination may influence the professional path that students choose.Students are urged to examine past years’ papers from time to time in order to become acquainted with the question paper format. Knowing the pattern of the question paper might assist students to acquire confidence while answering examination questions.

Q.1

$\begin{array}{l}\mathrm{Use}\text{ }\mathrm{a}\text{ }\mathrm{suitable}\text{ }\mathrm{identity}\text{ }\mathrm{to}\text{ }\mathrm{get}\text{ }\mathrm{each}\text{ }\mathrm{of}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{products}.\\ \left(\mathrm{i}\right)\text{ }\left(\mathrm{x}+3\right)\left(\mathrm{x}+3\right)\text{ }\left(\mathrm{ii}\right)\left(2\mathrm{y}+5\right)\left(2\mathrm{y}+5\right)\\ \left(\mathrm{iii}\right)\text{ }\left(2\mathrm{a}-7\right)\left(2\mathrm{a}-7\right)\text{ }\left(\mathrm{iv}\right)\text{ }\left(3\mathrm{a}-\frac{1}{2}\right)\left(3\mathrm{a}-\frac{1}{2}\right)\\ \left(\mathrm{v}\right)\text{ }\left(1.1\mathrm{m}-0.4\right)\left(1.1\mathrm{m}+0.4\right)\text{ }\left(\mathrm{vi}\right)\text{ }\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\left(-{\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\\ \left(\mathrm{vii}\right)\text{ }\left(6\mathrm{x}-7\right)\left(6\mathrm{x}+7\right)\text{ }\left(\mathrm{viii}\right)\text{ }\left(-\mathrm{a}+\mathrm{c}\right)\left(-\mathrm{a}+\mathrm{c}\right)\\ \left(\mathrm{ix}\right)\text{ }\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)\text{ }\left(\mathrm{x}\right)\text{ }\left(7\mathrm{a}-9\mathrm{b}\right)\left(7\mathrm{a}-9\mathrm{b}\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }\left(\mathrm{x}+3\right)\left(\mathrm{x}+3\right)={\left(\mathrm{x}+3\right)}^{2}\\ ={\left(\mathrm{x}\right)}^{2}+2×3×\mathrm{x}+{\left(3\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a+b)}}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{x}}^{2}+6\mathrm{x}+9\\ \\ \left(\mathrm{ii}\right)\text{ }\left(2\mathrm{y}+5\right)\left(2\mathrm{y}+5\right)={\left(2\mathrm{y}+5\right)}^{2}\\ ={\left(2\mathrm{y}\right)}^{2}+2×5×2\mathrm{y}+{\left(5\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a+b)}}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{y}}^{2}+20\mathrm{y}+25\\ \\ \left(\mathrm{iii}\right)\text{ }\left(2\mathrm{a}-7\right)\left(2\mathrm{a}-7\right)={\left(2\mathrm{a}-7\right)}^{2}\\ ={\left(2\mathrm{a}\right)}^{2}-2×2\mathrm{a}×7+{\left(7\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\left(\text{a}-\text{b}\right)}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{a}}^{2}-28\mathrm{a}+49\\ \\ \left(\mathrm{iv}\right)\text{ }\left(3\mathrm{a}-\frac{1}{2}\right)\left(3\mathrm{a}-\frac{1}{2}\right)={\left(3\mathrm{a}-\frac{1}{2}\right)}^{2}\\ ={\left(3\mathrm{a}\right)}^{2}-2×3\mathrm{a}×\frac{1}{2}+{\left(\frac{1}{2}\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\left(\text{a}-\text{b}\right)}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =9{\mathrm{a}}^{2}-3\mathrm{a}+\frac{1}{4}\\ \left(\mathrm{v}\right)\text{ }\left(1.1\mathrm{m}-0.4\right)\left(1.1\mathrm{m}+0.4\right)\\ ={\left(1.1\mathrm{m}\right)}^{2}-{\left(0.4\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ \text{(a+b)(a}-{\text{b)=a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}\right\}\\ =1.21{\mathrm{m}}^{2}-0.16\\ \\ \left(\mathrm{vi}\right)\text{ }\left({\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\left(-{\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)\\ ={\left({\mathrm{b}}^{2}\right)}^{2}-{\left({\mathrm{a}}^{2}\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ \text{(a+b)(a}-{\text{b)=a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}\right\}\\ ={\mathrm{b}}^{4}-{\mathrm{a}}^{4}\\ \\ \left(\mathrm{vii}\right)\text{ }\left(6\mathrm{x}-7\right)\left(6\mathrm{x}+7\right)\\ ={\left(6\mathrm{x}\right)}^{2}-{\left(7\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ \text{(a+b)(a}-{\text{b)=a}}^{\text{2}}-{\text{b}}^{\text{2}}\end{array}\right\}\\ =36{\mathrm{x}}^{2}-49\\ \\ \left(\mathrm{viii}\right)\text{ }\left(-\mathrm{a}+\mathrm{c}\right)\left(-\mathrm{a}+\mathrm{c}\right)\\ ={\left(-\mathrm{a}+\mathrm{c}\right)}^{2}\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a+b)}}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\text{}\\ ={\mathrm{c}}^{2}+{\mathrm{a}}^{2}-2\mathrm{ac}\end{array}$

$\begin{array}{l}\left(\mathrm{ix}\right)\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)={\left(\frac{\mathrm{x}}{2}+\frac{3\mathrm{y}}{4}\right)}^{2}\\ ={\left(\frac{\mathrm{x}}{2}\right)}^{2}+{\left(\frac{3\mathrm{y}}{4}\right)}^{2}+\text{2}\left(\frac{\mathrm{x}}{2}\right)\left(\frac{3\mathrm{y}}{4}\right)\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\text{(a + b)}}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ \text{=}\frac{{\mathrm{x}}^{2}}{4}+\frac{9{\mathrm{y}}^{2}}{16}+\frac{3\mathrm{xy}}{4}\\ \\ \left(\mathrm{x}\right)\left(7\mathrm{a}-9\mathrm{b}\right)\left(7\mathrm{a}-9\mathrm{b}\right)={\left(7\mathrm{a}-9\mathrm{b}\right)}^{2}\\ ={\left(7\mathrm{a}\right)}^{2}+{\left(9\mathrm{b}\right)}^{2}-\text{2}\left(7\mathrm{a}\right)\left(9\mathrm{b}\right)\text{}\left\{\begin{array}{l}\text{By using the identity}\\ {\left(\text{a}-\text{b}\right)}^{\text{2}}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =49{\mathrm{a}}^{2}+81{\mathrm{b}}^{2}-126\mathrm{ab}\end{array}$

Q.2

$\begin{array}{l}\mathrm{Use}\text{ }\mathrm{the}\text{ }\mathrm{identity}\text{ }\left(\mathrm{x}+\mathrm{a}\right)\left(\mathrm{x}+\mathrm{b}\right)={\mathrm{x}}^{2}+\left(\mathrm{a}+\mathrm{b}\right)\mathrm{x}+\mathrm{ab}\\ \mathrm{to}\text{ }\mathrm{find}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{products}.\\ \left(\mathrm{i}\right)\text{ }\left(\mathrm{x}+3\right)\left(\mathrm{x}+7\right)\\ \left(\mathrm{ii}\right)\text{ }\left(4\mathrm{x}+5\right)\left(4\mathrm{x}+1\right)\\ \left(\mathrm{iii}\right)\text{ }\left(4\mathrm{x}-5\right)\left(4\mathrm{x}-1\right)\\ \left(\mathrm{iv}\right)\text{ }\left(4\mathrm{x}+5\right)\left(4\mathrm{x}-1\right)\\ \left(\mathrm{v}\right)\text{ }\left(2\mathrm{x}+5\mathrm{y}\right)\left(2\mathrm{x}+3\mathrm{y}\right)\\ \left(\mathrm{vi}\right)\text{ }\left(2{\mathrm{a}}^{2}+9\right)\left(2{\mathrm{a}}^{2}+5\right)\\ \left(\mathrm{vii}\right)\text{ }\left(\mathrm{xyz}-4\right)\left(\mathrm{xyz}-2\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }\left(\mathrm{x}+3\right)\left(\mathrm{x}+7\right)={\mathrm{x}}^{2}+\left(3+7\right)\mathrm{x}+3×7\\ ={\mathrm{x}}^{2}+10\mathrm{x}+21\\ \\ \left(\mathrm{ii}\right)\text{ }\left(4\mathrm{x}+5\right)\left(4\mathrm{x}+1\right)={\left(4\mathrm{x}\right)}^{2}+\left(5+1\right)\left(4\mathrm{x}\right)+5×1\\ =16{\mathrm{x}}^{2}+24\mathrm{x}+5\\ \\ \left(\mathrm{iii}\right)\text{ }\left(4\mathrm{x}-5\right)\left(4\mathrm{x}-1\right)={\left(4\mathrm{x}\right)}^{2}+\left[\left(-5\right)+\left(–1\right)\right]\left(4\mathrm{x}\right)+\left(–5\right)×\left(-1\right)\\ =16{\mathrm{x}}^{2}-24\mathrm{x}+5\\ \\ \left(\mathrm{iv}\right)\text{ }\left(4\mathrm{x}+5\right)\left(4\mathrm{x}-1\right)={\left(4\mathrm{x}\right)}^{2}+\left[5+\left(-1\right)\right]\left(4\mathrm{x}\right)+5×\left(-1\right)\\ =16{\mathrm{x}}^{2}+16\mathrm{x}-5\\ \\ \left(\mathrm{v}\right)\text{ }\left(2\mathrm{x}+5\mathrm{y}\right)\left(2\mathrm{x}+3\mathrm{y}\right)={\left(2\mathrm{x}\right)}^{2}+\left(5\mathrm{y}+3\mathrm{y}\right)\left(2\mathrm{x}\right)+5\mathrm{y}×3\mathrm{y}\\ =4{\mathrm{x}}^{2}+16\mathrm{xy}+15{\mathrm{y}}^{2}\\ \\ \left(\mathrm{vi}\right)\text{ }\left(2{\mathrm{a}}^{2}+9\right)\left(2{\mathrm{a}}^{2}+5\right)={\left(2{\mathrm{a}}^{2}\right)}^{2}+\left(9+5\right)\left(2{\mathrm{a}}^{2}\right)+9×5\\ =4{\mathrm{a}}^{4}+28{\mathrm{a}}^{2}+45\\ \\ \left(\mathrm{vii}\right)\text{ }\left(\mathrm{xyz}-4\right)\left(\mathrm{xyz}-2\right)={\left(\mathrm{xyz}\right)}^{2}+\left[\left(-4\right)+\left(–2\right)\right]\left(\mathrm{xyz}\right)+\left(–4\right)×\left(-2\right)\\ ={\mathrm{x}}^{2}{\mathrm{y}}^{2}{\mathrm{z}}^{2}-6\mathrm{xyz}+\end{array}$

Q.3

$\begin{array}{l}\mathrm{Find}\text{ }\mathrm{the}\text{ }\mathrm{followings}\text{ }\mathrm{quares}\text{ }\mathrm{by}\text{ }\mathrm{using}\text{ }\mathrm{the}\text{ }\mathrm{identities}.\\ \left(\mathrm{i}\right)\text{ }{\left(\mathrm{b}-7\right)}^{2}\\ \left(\mathrm{ii}\right)\text{ }{\left(\mathrm{xy}+3\mathrm{z}\right)}^{2}\\ \left(\mathrm{iii}\right)\text{ }{\left(6{\mathrm{x}}^{2}-5\mathrm{y}\right)}^{2}\\ \left(\mathrm{iv}\right)\text{ }{\left(\frac{2}{3}\mathrm{m}+\frac{3}{2}\mathrm{n}\right)}^{2}\\ \left(\mathrm{v}\right)\text{ }{\left(0.4\mathrm{p}-0.5\mathrm{q}\right)}^{2}\\ \left(\mathrm{vi}\right)\text{ }{\left(2\mathrm{xy}+5\mathrm{y}\right)}^{2}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }{\left(\mathrm{b}-7\right)}^{2}\\ ={\left(\mathrm{b}\right)}^{2}-2×\mathrm{b}×7+{\left(7\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{b}}^{2}-14\mathrm{b}+49\\ \\ \left(\mathrm{ii}\right)\text{ }{\left(\mathrm{xy}+3\mathrm{z}\right)}^{2}\\ ={\left(\mathrm{xy}\right)}^{2}+2×\mathrm{xy}×3\mathrm{z}+{\left(3\mathrm{z}\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{x}}^{2}{\mathrm{y}}^{2}+6\mathrm{xyz}+9{\mathrm{z}}^{2}\\ \left(\mathrm{iii}\right)\text{ }{\left(6{\mathrm{x}}^{2}-5\mathrm{y}\right)}^{2}\\ ={\left(6{\mathrm{x}}^{2}\right)}^{2}-2×6{\mathrm{x}}^{2}×5\mathrm{y}+{\left(5\mathrm{y}\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =36{\mathrm{x}}^{4}-60{\mathrm{x}}^{2}\mathrm{y}+25{\mathrm{y}}^{2}\\ \\ \left(\mathrm{iv}\right)\text{ }{\left(\frac{2}{3}\mathrm{m}+\frac{3}{2}\mathrm{n}\right)}^{2}\\ ={\left(\frac{2}{3}\mathrm{m}\right)}^{2}+2×\frac{2}{3}\mathrm{m}×\frac{3}{2}\mathrm{n}+{\left(\frac{3}{2}\mathrm{n}\right)}^{2}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{+}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =\frac{4}{9}{\mathrm{m}}^{2}+2\mathrm{mn}+\frac{9}{4}{\mathrm{n}}^{2}\\ \left(\mathrm{v}\right)\text{ }{\left(0.4\mathrm{p}-0.5\mathrm{q}\right)}^{2}\\ ={\left(0.4\mathrm{p}\right)}^{2}-2×0.4\mathrm{p}×0.5\mathrm{q}+{\left(0.5\mathrm{q}\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =0.16{\mathrm{p}}^{2}-0.4\mathrm{pq}+0.25{\mathrm{q}}^{2}\\ \\ \left(\mathrm{vi}\right)\text{ }{\left(2\mathrm{xy}+5\mathrm{y}\right)}^{2}\\ ={\left(2\mathrm{xy}\right)}^{2}+2×2\mathrm{xy}×5\mathrm{y}+{\left(5\mathrm{y}\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{x}}^{2}{\mathrm{y}}^{2}+20{\mathrm{xy}}^{2}+25{\mathrm{y}}^{2}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Simplify}.\\ \left(\mathrm{i}\right)\text{ }{\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)}^{2}\\ \left(\mathrm{ii}\right)\text{ }{\left(2\mathrm{x}+5\right)}^{2}-{\left(2\mathrm{x}-5\right)}^{2}\\ \left(\mathrm{iii}\right)\text{ }{\left(7\mathrm{m}-8\mathrm{n}\right)}^{2}+{\left(7\mathrm{m}+8\mathrm{n}\right)}^{2}\\ \left(\mathrm{iv}\right)\text{ }{\left(4\mathrm{m}+5\mathrm{n}\right)}^{2}+{\left(5\mathrm{m}+4\mathrm{n}\right)}^{2}\\ \left(\mathrm{v}\right)\text{ }{\left(2.5\mathrm{p}-1.5\mathrm{q}\right)}^{2}-{\left(1.5\mathrm{p}-2.5\mathrm{q}\right)}^{2}\\ \left(\mathrm{vi}\right)\text{ }{\left(\mathrm{ab}+\mathrm{bc}\right)}^{2}-2{\mathrm{ab}}^{2}\mathrm{c}\\ \left(\mathrm{vii}\right)\text{ }{\left({\mathrm{m}}^{2}-{\mathrm{n}}^{2}\mathrm{m}\right)}^{2}+2{\mathrm{m}}^{3}{\mathrm{n}}^{2}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }{\left({\mathrm{a}}^{2}-{\mathrm{b}}^{2}\right)}^{2}\\ ={\left({\mathrm{a}}^{2}\right)}^{2}-2×{\mathrm{a}}^{2}×{\mathrm{b}}^{2}+{\left({\mathrm{b}}^{2}\right)}^{2}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ ={\left(\mathrm{a}\right)}^{4}-2{\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\left(\mathrm{b}\right)}^{4}\\ \\ \left(\mathrm{ii}\right)\text{ }{\left(2\mathrm{x}+5\right)}^{2}-{\left(2\mathrm{x}-5\right)}^{2}\\ =\left[{\left(2\mathrm{x}\right)}^{2}+2×2\mathrm{x}×5+{\left(5\right)}^{2}\right]-\left[{\left(2\mathrm{x}\right)}^{2}-2×2\mathrm{x}×5+{\left(5\right)}^{2}\right]\text{}\\ \text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{+}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{–}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}–2\mathrm{ab}\end{array}\right\}\\ =4{\mathrm{x}}^{2}+20\mathrm{x}+25-4{\mathrm{x}}^{2}+20\mathrm{x}-25\\ =40\mathrm{x}\\ \left(\mathrm{iii}\right)\text{ }{\left(7\mathrm{m}-8\mathrm{n}\right)}^{2}+{\left(7\mathrm{m}+8\mathrm{n}\right)}^{2}\\ =\left[{\left(7\mathrm{m}\right)}^{2}-2×7\mathrm{m}×8\mathrm{n}+{\left(8\mathrm{n}\right)}^{2}\right]+\left[{\left(7\mathrm{m}\right)}^{2}+2×7\mathrm{m}×8\mathrm{n}+{\left(8\mathrm{n}\right)}^{2}\right]\text{}\\ \left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =49{\mathrm{m}}^{2}-112\mathrm{mn}+64{\mathrm{n}}^{2}+49{\mathrm{m}}^{2}+112\mathrm{mn}+64{\mathrm{n}}^{2}\\ =98{\mathrm{m}}^{2}+128{\mathrm{n}}^{2}\\ \\ \left(\mathrm{iv}\right)\text{ }{\left(4\mathrm{m}+5\mathrm{n}\right)}^{2}+{\left(5\mathrm{m}+4\mathrm{n}\right)}^{2}\\ =\left[{\left(4\mathrm{m}\right)}^{2}+2×4\mathrm{m}×5\mathrm{n}+{\left(5\mathrm{n}\right)}^{2}\right]+\left[{\left(5\mathrm{m}\right)}^{2}+2×5\mathrm{m}×4\mathrm{n}+{\left(4\mathrm{n}\right)}^{2}\right]\text{}\\ \text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =\left(16{\mathrm{m}}^{2}+40\mathrm{mn}+25{\mathrm{n}}^{2}\right)+\left(25{\mathrm{m}}^{2}+40\mathrm{mn}+16{\mathrm{n}}^{2}\right)\\ =41{\mathrm{m}}^{2}+80\mathrm{mn}+41{\mathrm{n}}^{2}\\ \\ \left(\mathrm{v}\right)\text{ }{\left(2.5\mathrm{p}-1.5\mathrm{q}\right)}^{2}-{\left(1.5\mathrm{p}-2.5\mathrm{q}\right)}^{2}\\ =\left[{\left(2.5\mathrm{p}\right)}^{2}-2×2.5\mathrm{p}×1.5\mathrm{q}+{\left(1.5\mathrm{q}\right)}^{2}\right]-\left[{\left(1.5\mathrm{p}\right)}^{2}-2×1.5\mathrm{p}×2.5\mathrm{q}+{\left(2.5\mathrm{q}\right)}^{2}\right]\text{}\\ \left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =6.25{\mathrm{p}}^{2}-7.5\mathrm{pq}+2.25{\mathrm{q}}^{2}-\left[2.25{\mathrm{p}}^{2}-7.5\mathrm{pq}+6.25{\mathrm{q}}^{2}\right]\\ =6.25{\mathrm{p}}^{2}-7.5\mathrm{pq}+2.25{\mathrm{q}}^{2}-2.25{\mathrm{p}}^{2}+7.5\mathrm{pq}-6.25{\mathrm{q}}^{2}\\ =4{\mathrm{p}}^{2}-4{\mathrm{q}}^{2}\end{array}$

$\begin{array}{l}\left(\mathrm{vi}\right)\text{ }{\left(\mathrm{ab}+\mathrm{bc}\right)}^{2}-2{\mathrm{ab}}^{2}\mathrm{c}\\ =\left[{\left(\mathrm{ab}\right)}^{2}+2×\mathrm{ab}×\mathrm{bc}+{\left(\mathrm{bc}\right)}^{2}\right]-2{\mathrm{ab}}^{2}\mathrm{c}\text{}\\ \left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{a}}^{2}{\mathrm{b}}^{2}+2{\mathrm{ab}}^{2}\mathrm{c}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}-2{\mathrm{ab}}^{2}\mathrm{c}\\ ={\mathrm{a}}^{2}{\mathrm{b}}^{2}+{\mathrm{b}}^{2}{\mathrm{c}}^{2}\\ \\ \left(\mathrm{vii}\right)\text{ }{\left({\mathrm{m}}^{2}-{\mathrm{n}}^{2}\mathrm{m}\right)}^{2}+2{\mathrm{m}}^{3}{\mathrm{n}}^{2}\\ =\left[{\left({\mathrm{m}}^{2}\right)}^{2}-2×{\mathrm{m}}^{2}×{\mathrm{n}}^{2}\mathrm{m}+{\left({\mathrm{n}}^{2}\mathrm{m}\right)}^{2}\right]+2{\mathrm{m}}^{3}{\mathrm{n}}^{2}\text{}\\ \left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ ={\mathrm{m}}^{4}-2{\mathrm{m}}^{3}{\mathrm{n}}^{2}+{\mathrm{n}}^{4}{\mathrm{m}}^{2}++2{\mathrm{m}}^{3}{\mathrm{n}}^{2}\\ ={\mathrm{m}}^{4}+{\mathrm{n}}^{4}{\mathrm{m}}^{2}\end{array}$

Q.5

$\begin{array}{l}\mathrm{Showthat}.\\ \left(\mathrm{i}\right){\left(3\mathrm{x}+7\right)}^{2}-84\mathrm{x}={\left(3\mathrm{x}-7\right)}^{2}\\ \left(\mathrm{ii}\right){\left(9\mathrm{p}-5\mathrm{q}\right)}^{2}+180\mathrm{pq}={\left(9\mathrm{p}+5\mathrm{q}\right)}^{2}\\ \left(\mathrm{iii}\right){\left(\frac{4}{3}\mathrm{m}-\frac{3}{4}\mathrm{n}\right)}^{2}+2\mathrm{mn}=\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}\\ \left(\mathrm{iv}\right){\left(4\mathrm{pq}+3\mathrm{q}\right)}^{2}-{\left(4\mathrm{pq}-3\mathrm{q}\right)}^{2}=48{\mathrm{pq}}^{2}\\ \left(\mathrm{v}\right)\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)+\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}\right)+\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{c}+\mathrm{a}\right)=0\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(3\mathrm{x}+7\right)}^{2}-84\mathrm{x}\\ ={\left(3\mathrm{x}\right)}^{2}+{7}^{2}+2×\left(3\mathrm{x}\right)×7-84\mathrm{x}\\ =9{\mathrm{x}}^{2}+49+42\mathrm{x}-84\mathrm{x}\\ =9{\mathrm{x}}^{2}+49-42\mathrm{x}\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}={\left(3\mathrm{x}-7\right)}^{2}\\ ={\left(3\mathrm{x}\right)}^{2}+{7}^{2}-2×\left(3\mathrm{x}\right)×7\\ =9{\mathrm{x}}^{2}+49-42\mathrm{x}\\ \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\\ \left(\mathrm{ii}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(9\mathrm{p}-5\mathrm{q}\right)}^{2}+180\mathrm{pq}\\ ={\left(9\mathrm{p}\right)}^{2}+{\left(5\mathrm{q}\right)}^{2}-2×\left(9\mathrm{p}\right)×5\mathrm{q}+180\mathrm{pq}\\ =81{\mathrm{p}}^{2}+25{\mathrm{q}}^{2}-90\mathrm{pq}+180\mathrm{pq}\\ =81{\mathrm{p}}^{2}+25{\mathrm{q}}^{2}+90\mathrm{pq}\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}={\left(9\mathrm{p}+5\mathrm{q}\right)}^{2}\\ ={\left(9\mathrm{p}\right)}^{2}+{\left(5\mathrm{q}\right)}^{2}+2×\left(9\mathrm{p}\right)×5\mathrm{q}\\ =81{\mathrm{p}}^{2}+25{\mathrm{q}}^{2}+90\mathrm{pq}\\ \\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\\ \left(\mathrm{iii}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(\frac{4}{3}\mathrm{m}-\frac{3}{4}\mathrm{n}\right)}^{2}+2\mathrm{mn}\\ ={\left(\frac{4}{3}\mathrm{m}\right)}^{2}+{\left(\frac{3}{4}\mathrm{n}\right)}^{2}-2×\left(\frac{4}{3}\mathrm{m}\right)×\left(\frac{3}{4}\mathrm{n}\right)+2\mathrm{mn}\\ =\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}-2\mathrm{mn}+2\mathrm{mn}\\ =\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{16}{9}{\mathrm{m}}^{2}+\frac{9}{16}{\mathrm{n}}^{2}\\ \therefore \text{\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\end{array}$

$\begin{array}{l}\left(\mathrm{iv}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}={\left(4\mathrm{pq}+3\mathrm{q}\right)}^{2}-{\left(4\mathrm{pq}-3\mathrm{q}\right)}^{2}=48{\mathrm{pq}}^{2}\\ ={\left(4\mathrm{pq}\right)}^{2}+{\left(3\mathrm{q}\right)}^{2}+2×\left(4\mathrm{pq}\right)×\left(3\mathrm{q}\right)-\left[{\left(4\mathrm{pq}\right)}^{2}+{\left(3\mathrm{q}\right)}^{2}-2×m\left(4\mathrm{pq}\right)×\left(3\mathrm{q}\right)\right]\\ =16{\mathrm{p}}^{2}{\mathrm{q}}^{2}+9{\mathrm{q}}^{2}+24{\mathrm{pq}}^{2}-\left[16{\mathrm{p}}^{2}{\mathrm{q}}^{2}+9{\mathrm{q}}^{2}-24{\mathrm{pq}}^{2}\right]\\ =16{\mathrm{p}}^{2}{\mathrm{q}}^{2}+9{\mathrm{q}}^{2}+24{\mathrm{pq}}^{2}-16{\mathrm{p}}^{2}{\mathrm{q}}^{2}-9{\mathrm{q}}^{2}+24{\mathrm{pq}}^{2}\\ =48{\mathrm{pq}}^{2}\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}=48{\mathrm{pq}}^{2}\\ \\ \therefore \text{\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\\ \left(\mathrm{v}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}=\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)+\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}\right)+\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{c}+\mathrm{a}\right)\\ ={\mathrm{a}}^{2}-{\mathrm{b}}^{2}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}+{\mathrm{c}}^{2}-{\mathrm{a}}^{2}\\ =0\\ \text{\hspace{0.17em}}\\ \text{}\mathrm{R}.\mathrm{H}.\mathrm{S}=0\\ \\ \therefore \text{\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\\ \mathrm{Hence}\text{}\mathrm{Proved}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Using}\text{ }\mathrm{identities},\text{ }\mathrm{evaluate}.\\ \left(\mathrm{i}\right)\text{ }{71}^{2}\text{ }\left(\mathrm{ii}\right)\text{ }{99}^{2}\text{ }\left(\mathrm{iii}\right)\text{ }{102}^{2}\text{ }\left(\mathrm{iv}\right)\text{ }{998}^{2}\\ \left(\mathrm{v}\right)\text{ }5.{2}^{2}\text{ }\left(\mathrm{vi}\right)\text{ }297×303\text{ }\left(\mathrm{vii}\right)\text{ }78×82\text{ }\left(\mathrm{viii}\right)\text{ }8.{9}^{2}\\ \left(\mathrm{ix}\right)\text{ }10.5×9.5\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }{71}^{2}={\left(70+1\right)}^{2}\\ ={\left(70\right)}^{2}+2×70×1+{1}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =4900+140+1\\ =5041\\ \left(\mathrm{ii}\right)\text{ }{99}^{2}={\left(100-1\right)}^{2}\\ ={\left(100\right)}^{2}-2×100×1+{1}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =10000-200+1\\ =9801\\ \left(\mathrm{iii}\right)\text{ }{102}^{2}={\left(100+2\right)}^{2}\\ ={\left(100\right)}^{2}+2×100×2+{2}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =10000+400+4\\ =10404\\ \\ \left(\mathrm{iv}\right)\text{ }{998}^{2}={\left(1000-2\right)}^{2}\\ ={\left(1000\right)}^{2}-2×1000×2+{2}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}-\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =1000000-4000+4\\ =996004\end{array}$

$\begin{array}{l}\left(\mathrm{v}\right)\text{ }5{.2}^{2}={\left(5+0.2\right)}^{2}\\ ={\left(5\right)}^{2}+2×5×0.2+{\left(0.2\right)}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}\text{}+\text{}\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+2\mathrm{ab}\end{array}\right\}\\ =25+2+0.04\\ =27.04\\ \\ \left(\mathrm{vi}\right)\text{ }297×303=\left(300-3\right)\left(300+3\right)\\ ={\left(300\right)}^{2}-{3}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ \left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\end{array}\right\}\\ =90000-9\\ =89991\\ \\ \left(\mathrm{vii}\right)\text{ }78×82=\left(80-2\right)\left(80+2\right)\\ ={\left(80\right)}^{2}-{2}^{2}\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ \left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\end{array}\right\}\\ =6400-4\\ =6396\\ \\ \left(\mathrm{viii}\right)\text{ }8{.9}^{2}={\left(9.0-0.1\right)}^{2}\\ =81-1.8+0.01\text{}\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ {\left(\mathrm{a}-\mathrm{b}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}-2\mathrm{ab}\end{array}\right\}\\ =79.21\\ \\ \left(\mathrm{ix}\right)\text{ }10.5×9.5\\ =\left(10+0.5\right)\left(10–0.5\right)\\ =\left[{10}^{2}-{\left(0.5\right)}^{2}\right]\left\{\begin{array}{l}\mathrm{By}\text{}\mathrm{using}\text{}\mathrm{the}\text{}\mathrm{identity}\\ \left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)={\mathrm{a}}^{2}-{\mathrm{b}}^{2}\end{array}\right\}\\ =100–0.25=99.75\end{array}$

Q.7

$\begin{array}{l}{\text{Using a}}^{\text{2}}-{\text{b}}^{\text{2}}\text{= (a + b) (a}-\text{b), find}\\ {\text{(i) 51}}^{\text{2}}\text{}-{\text{49}}^{\text{2}}\text{(ii) (1}{\text{.02)}}^{\text{2}}-\text{(0}{\text{.98)}}^{\text{2}}{\text{(iii) 153}}^{\text{2}}\text{}-{\text{147}}^{\text{2}}\\ \text{(iv) 12}{\text{.1}}^{\text{2}}\text{}-\text{7}{\text{.9}}^{\text{2}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }{51}^{2}-{49}^{2}\\ =\left(51+49\right)\left(51-49\right)\\ =100×2\\ =200\\ \\ {\text{(ii) (1.02)}}^{\text{2}}-{\text{(0.98)}}^{\text{2}}\text{}\\ =\left(1.02+0.98\right)\left(1.02-0.98\right)\\ =2×0.04\\ =0.08\\ \\ \left(\mathrm{iii}\right)\text{ }{153}^{2}-{147}^{2}\\ =\left(153+147\right)\left(153-147\right)\\ =300×6\\ =1800\\ \\ \left(\mathrm{iv}\right)\text{ }12{.1}^{2}-7{.9}^{2}\\ =\left(12.1+7.9\right)\left(12.1-7.9\right)\\ =20.0×4.2\\ =84\\ \end{array}$

Q.8

$\begin{array}{l}\mathrm{Using}\text{ }\left(\mathrm{x}+\mathrm{a}\right)\text{ }\left(\mathrm{x}+\mathrm{b}\right)={\mathrm{x}}^{2}+\left(\mathrm{a}+\mathrm{b}\right)\mathrm{x}+\mathrm{ab},\text{ }\mathrm{find}\\ \left(\mathrm{i}\right)\text{ }103×104\text{ }\left(\mathrm{ii}\right)\text{ }5.1×5.2\text{ }\left(\mathrm{iii}\right)\text{ }103×98\text{ }\left(\mathrm{iv}\right)\text{ }9.7×9.8\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }103×104\\ =\left(100+3\right)\left(100+4\right)\\ ={\left(100\right)}^{2}+\left(3+4\right)×100+3×4\\ =10000+700+12\\ =10712\\ \\ \left(\mathrm{ii}\right)\text{ }5.1×5.2\\ =\left(5+0.1\right)\left(5+0.2\right)\\ ={\left(5\right)}^{2}+\left(0.1+0.2\right)×5+0.1×0.2\\ =25+1.5+0.02\\ =26.52\\ \left(\mathrm{iii}\right)\text{ }103×98\\ =\left(100+3\right)\left(100-2\right)\\ ={\left(100\right)}^{2}+\left[3+\left(-2\right)\right]×100+3×\left(-2\right)\\ =10000+100-6\\ =10094\\ \\ \left(\mathrm{iv}\right)\text{ }9.7×9.8\\ =\left(10-0.3\right)\left(10-0.2\right)\\ ={\left(10\right)}^{2}+\left[\left(-0.3\right)+\left(-0.2\right)\right]×10+\left(-0.3\right)×\left(-0.2\right)\\ =100+\left(-0.5\right)10+0.06\\ =100.06-5=95.06\end{array}$

## FAQs (Frequently Asked Questions)

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### 4. Is the Mathematics Senior Secondary Examination of Class 8 difficult?

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