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

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

Edited By Ramraj Saini | Updated on Mar 01, 2024 06:31 PM IST

Algebraic Expressions and Identities Class 8 Questions And Answers provided here. These NCERT Solutions are prepared by experts team at Careers360 team considering the latest syllabus and pattern of CBSE 2023-24. This chapter will introduce you to the application of algebraic terms and variables to solve various problems. Algebra is the most important branch of mathematics which teaches how to form equations and solving them using different kinds of techniques. Important topics like the product of the equation, finding the coefficient of the variable in the equation, subtraction of the equation and creating quadratic equation by the product of its two roots, and division of the equation are covered in this chapter. Also Practice NCERT solutions for class 8 maths to command the concepts.

Algebraic Expressions and Identities Class 8 Questions And Answers PDF Free Download

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Algebraic Expressions and Identities Class 8 Solutions - Important Formulae

  • (a + b)2 = a2 + 2ab + b2

  • (a - b)2 = a2 - 2ab + b2

  • (a + b)(a - b) = a2 - b2

  • (x + a)(x + b) = x2 + (a + b)x + ab

  • (x + a)(x - b) = x2 + (a - b)x - ab

  • (x - a)(x + b) = x2 + (b - a)x - ab

  • (x - a)(x - b) = x2 - (a + b)x + ab

  • (a + b)3 = a3 + b3 + 3ab(a + b)

  • (a - b)3 = a3 - b3 - 3ab(a - b)

Free download NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities for CBSE Exam.

Algebraic Expressions and Identities Class 8 NCERT Solutions (Intext Questions and Exercise)

NCERT Solutions to Exercises of Chapter 9: Algebraic Expressions and Identities

what are expressions?

Question: 1 Give five examples of expressions containing one variable and five examples of expressions containing two variables.

Answer:

Five examples of expressions containing one variable are:

x^{^{4}}, y, 3z, p^{^{2}}, -2q^{3}

Five examples of expressions containing two variables are:

x + y, 3p-4q,ab,uv^{2},-z^{2}+x^{3}

Question: 2(i) Show on the number line

x

Answer:

x on the number line:

1643105164197

Question: 2(ii) Show on the number line :

x-4

Answer:

x-4 on the number line:

1643105231337

Question: 2(iii) Show on the number line :

2x+1

Answer:

2x+1 on the number line:

c360_4-1


Question: 2(iv) Show on the number line:

3x-2

Answer:

3x - 2 on the number line

1643105272368

Algebraic expressions and identities class 8 solutions - Topic 9.2 Terms, Factors and Coefficients

Question:1 Identify the coefficient of each term in the expression.

x^2y^2-10x^2y+5xy^2-20

Answer:

coefficient of each term are given below

\\The\ coefficient\ of\ x^{2}y^{2}\ is \1\\ \\The\ coefficient\ of\ x^{2}y\ is \ -10\\ \\The\ coefficient\ of\ xy^{2}\ is \5\\

Algebraic expressions and identities class 8 ncert solutions - Topic 9.3 Monomials, Binomials and Polynomials

Question: 1(i) Classify the following polynomials as monomials, binomials, trinomials.

-z+5

Answer:

Binomial since there are two terms with non zero coefficients.

Question: 1(ii) Classify the following polynomials as monomials, binomials, trinomials.

x+y+z

Answer:

Trinomial since there are three terms with non zero coefficients.

Question:1(iii) Classify the following polynomials as monomials, binomials, trinomials.

y+z+100

Answer:

Trinomial since there are three terms with non zero coefficients.

Question: 1(iv) Classify the following polynomials as monomials, binomials, trinomials.

ab-ac

Answer:

Binomial since there are two terms with non zero coefficients.

Question: 1(v) Classify the following polynomials as monomials, binomials, trinomials.

17

Answer:

Monomial since there is only one term.

Question: 2(a) Construct 3 binomials with only x as a variable;

Answer:

Three binomials with the only x as a variable are:

\\ \\x+2,\ x +x^{2},\ 3x^{3}-5x^{4}

Question: 2(b) Construct 3 binomials with x and y as variables;

Answer:

Three binomials with x and y as variables are:

\\ \\x+y,\ x-7y, xy^{2} + 2xy

Question: 2(c) Construct 3 monomials with x and y as variables;

Answer:

Three monomials with x and y as variables are

\\ xy,\ 3xy^{4},\ -2x^{3}y^{2}

Question: 2(d) Construct 2 polynomials with 4 or more terms .

Answer:

Two polynomials with 4 or more terms are:

a+b+c+d, x-3xy+2y+4xy^{2}

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Topic 9.4 Like and Unlike Terms

Question:(i) Write two terms which are like

7xy

Answer:

\\Two\ terms\ like\ 7xy\ are:\\ -3xy\ and\ 5xy

Question:(ii) Write two terms which are like

4mn^2

Answer:

\\Two\ terms\ which\ are\ like\ 4mn^{2}\ are:\\ mn^{2}\ and -3mn^{2.}

we can write more like terms

Question:(iii) Write two terms which are like

2l

Answer:

\\Two\ terms\ which\ are\ like\ 2l\ are:\\ l\ and\ -3l

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Exercise: 9.1

Question:1(i) Identify the terms, their coefficients for each of the following expressions.

5xyz^2-3zy

Answer

following are the terms and coefficient

The terms are 5xyz^{2}\ and\ -3zy and the coefficients are 5 and -3.

Question: 1(ii) Identify the terms, their coefficients for each of the following expressions.

1+x+x^2

Answer:

the following is the solution

\\The\ terms\ are\ 1,\ x,\ and\ x^{2}\ and\ the\ coefficients\ are\ 1,\ 1,\ and\ 1\ respectively.

Question: 1(iv) Identify the terms, their coefficients for each of the following expressions.

3-pq+qr-rp

Answer:

The terms are 3, -pq, qr,and -rp and the coefficients are 3, -1, 1 and -1 respectively.

Question:1(v) Identify the terms, their coefficients for each of the following expressions.

\frac{x}{2}+\frac{y}{2}-xy

Answer:

\\The\ terms\ are\ \frac{x}{2},\ \frac{y}{2}\ and\ -xy\ and\ the\ coefficients\ are\ \frac{1}{2},\ \frac{1}{2}\ and\ -1\ respectively.

Above are the terms and coefficients

Question: 1(vi) Identify the terms, their coefficients for each of the following expressions.

0.3a-0.6ab+0.5b

Answer:

The terms are 0.3a, -0.6ab and 0.5b and the coefficients are 0.3, -0.6 and 0.5.

Question: 3(i) Add the following.

ab-bc , bc -ca, ca-ab

Answer:

ab-bc+bc-ca+ca-ab=0.

Question:3 (ii) Add the following.

a-b+ab, b-c+bc, c-a+ac

Answer:

\\a-b+ab+b-c+bc+c-a+ac\\ =(a-a)+(b-b)+(c-c)+ab+bc+ac\\ =ab+bc+ca

Question:3 (iii) Add the following

2p^2q^2-3pq+4, 5+7pq-3p^2q^2

Answer:

\\2p^{2}q^{2}-3pq+4+5+7pq-3p^{2}q^{2}\\ =(2-3)p^{2}q^{2} +(-3+7)pq +4+5\\ =-p^{2}q^{2}+4pq+9

Question: 3(iv) Add the following.

l^2+m^2+n^2 , n^2+l^2, 2lm+2mn+2nl

Answer:

\\l^{2}+m^{2}+n^{2}+n^{2}+l^{2}+2lm+2mn+2nl\\ =2l^{2}+m^{2}+2n^{2}+2lm+2mn+2nl

Question: 4(a) Subtract 4a-7ab+3b+12 from 12a-9ab+5b-3

Answer:

12a-9ab+5b-3-(4a-7ab+3b+12)
=(12-4)a +(-9+7)ab+(5-3)b +(-3-12)
=8a-2ab+2b-15

Question: 4(b) Subtract 3xy+5yz-7zx from 5xy-2yz-2zx+10xyz

Answer:

\\5xy-2yz-2zx+10xyz-(3xy+5yz-7zx)\\ =(5-3)xy+(-2-5)yz+(-2+7)zx+10xyz\\ =2xy-7yz+5zx+10xyz

Question: 4(c) Subtract 4p^2q - 3pq + 5pq^2 - 8p + 7q - 10 from 18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q

Answer:

\\18-3p-11q+5pq-2pq^{2}+5p^{2}q-(4p^{2}q-3pq+5pq^{2}-8p+7q-10)\\ =18-(-10)-3p-(-8p)-11q-7q+5pq-(-3pq)-2pq^{2}-5pq^{2}+5p^{2}q-4p^{2}q\\ =28+5p-18q+8pq-7pq^{2}+p^{2}q

NCERT class 8 maths chapter 9 question answer - Topic 9.7.2 Multiplying Three or More Monomials

Question:1 Find 4x\times 5y\times 7z . First find 4x\times 5y and multiply it by 7z ; or first find 5y \times 7z and multiply it by 4x .

Answer:

\\4x\times 5y\times 7z\\ =(4x\times 5y)\times 7z\\ =20xy\times 7z\\ =140xyz\\ \\4x\times 5y\times 7z\\ =(5y\times 7z)\times 4x\\ =35yz\times 4x\\ =140xyz

We observe that the result is same in both cases and the result does not depend on the order in which multiplication has been carried out.

Class 8 maths chapter 9 question answer - exercise: 9.2

Question: 1(i) Find the product of the following pairs of monomials.

4,7p

Answer:

4\times 7p=28p

Question:3 Complete the table of products.


First monomial \rightarrow

Second monomial \downarrow

2x

-5y

3x^2

-4xy

7x^2y

-9x^2y^2

2x

4x^2


...

...

...

...

...

-5y

...

...

-15x^2y

...

...

...

3x^2

...

...

...

...

...

...

-4xy

...

...

...

...

...

...

7x^2y

...

...

...

...

...

...

-9x^2y^2

...

...

...

...

...

...

Answer:

First monomial \rightarrow

Second monomial \downarrow

2x

-5y

3x^{2}

-4xy

7x^{2}y

-9x^{2}y^{2}

2x

4x^{2}

-10xy

6x^{3}

-8x^{2}y

14x^{3}y

-18x^{3}y^{2}

-5y

-10xy

25y^{2}

-15x^{2}y

20xy^{2}

-35x^{2}y^{2}

45x^{2}y^{3}

3x^{2}

6x^{3}

-15x^{2}y^{}

9x^{4}

-12x^{3}y

21x^{4}y

-27x^{4}y^{2}

-4xy

-8x^{2}y

20xy^{2}

-12x^{3}y

16x^{2}y^{2}

-28x^{3}y

36x^{3}y^{3}

7x^{2}y

14x^{3}y

-35x^{2}y^{2}

21x^{4}y

-28x^{3}y^{2}

49x^{4}y^{2}

-63x^{4}y^{3}

-9x^{2}y^{2}

-18x^{3}y^{2}

45x^{2}y^{3}

-27x^{4}y^{2}

36x^{3}y^{3}

-63x^{4}y^{3}

81x^{4}y^{4}

Question:4(ii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

2p,4q,8r

Answer:

the volume of rectangular boxes with the following length, breadth and height is

\\Volume=length\times breadth\times height\\ =2p\times 4q\times 8r\\ =8pq\times 8r\\ =64pqr

Question:4(iii) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

xy, 2x^2y, 2xy^2

Answer:

the volume of rectangular boxes with the following length, breadth and height is

\\Volume=length\times breadth\times height\\ =xy\times 2x^{2}y\times 2xy^{2}\\ =2x^{3}y^{2}\times 2xy^{2}\\ =4x^{4}y^{4}

Question:4(iv) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

a, 2b, 3c

Answer:

the volume of rectangular boxes with the following length, breadth and height is

\\Volume=length\times breadth\times height\\ =a\times 2b\times 3c\\ =2ab\times 3c\\ =6abc

Question:5(i) Obtain the product of

xy,yz,zx

Answer:

the product

\\xy\times yz\times zx\\ =xy^{2}z\times zx\\ =x^{2}y^{2}z^{2}

Question:5(ii) Obtain the product of

a,-a^2,a^3

Answer:

the product

\\a\times (-a^{2})\times a^{3}\\ =-a^{^{3}}\times a^{3} =-a^{6}

Question:5(iii) Obtain the product of

2,\ 4y,\ 8y^{2},\ 16y^{3}

Answer:

the product

\\2\times 4y\times 8y^{2}\times 16y^{3}\\ =8y\times 8y^{2}\times 16y^{3}\\ =64y^{3}\times 16y^{3}\\ =1024y^{6}

Question:5(iv) Obtain the product of

a, 2b, 3c, 6abc

Answer:

the product

\\a\times 2b\times 3c\times 6abc\\ =2ab\times 3c\times 6abc\\ =6abc\times 6abc\\ =36a^{2}b^{2}c^{2}

Question:5(v) Obtain the product of

m, -mn, mnp

Answer:

the product

\\m\times (-mn)\times mnp\\ =-m^{2}n\times mnp\\ =-m^{3}n^{2}p

Class 8 maths chapter 9 NCERT solutions - Topic 9.8.1 Multiplying a Monomial by a Binomial

Question:(i) Find the product

2x(3x+5xy)

Answer:

Using distributive law,

2x(3x + 5xy) = 6x^2 + 10x^2y

Question:(ii) Find the product

a^2(2ab-5c)

Answer:

Using distributive law,

We have : a^2(2ab-5c) = 2a^3b - 5a^2c

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Topic 9.8.2 Multiplying A Monomial By A Trinomial

Question:1 Find the product:

(4p^2+5p+7)\times 3p

Answer:

By using distributive law,

(4p^2+5p+7)\times 3p = 12p^3 + 15p^2 + 21p

Class 8 maths chapter 9 NCERT solutions - exercise: 9.3

Question:1(i) Carry out the multiplication of the expressions in each of the following pairs.

4p, q+r

Answer:

Multiplication of the given expression gives :

By distributive law,

(4p)(q+r) = 4pq + 4pr

Question:1(ii) Carry out the multiplication of the expressions in each of the following pairs.

ab, a-b

Answer:

We have ab, (a-b).

Using distributive law we get,

ab(a-b) = a^2b - ab^2

Question:1(iii) Carry out the multiplication of the expressions in each of the following pairs.

a+b, 7a^2b^2

Answer:

Using distributive law we can obtain multiplication of given expression:

(a + b)(7a^2b^2) = 7a^3b^2 + 7a^2b^3

Question:1(iv) Carry out the multiplication of the expressions in each of the following pairs.

a^2-9,4a

Answer:

We will obtain multiplication of given expression by using distributive law :

(a^2 - 9 )(4a) = 4a^3 - 36a

Question:2 Complete the table


First expression

Second expression

Product

(i)

a

b+c+d

...

(ii)

x+y-5

5xy

...

(iii)

p

6p^2-7p+5

...

(iv)

4p^2q^2

p^2-q^2

...

(v)

a+b+c

abc

...


Answer:

We will use distributive law to find product in each case.


First expression

Second expression

Product

(i)

a

b+c+d

ab + ac+ ad

(ii)

x+y-5

5xy

5x^2y + 5xy^2 - 25xy

(iii)

p

6p^2-7p+5

6p^3 - 7p^2 + 5p

(iv)

4p^2q^2

p^2-q^2

4p^4q^2 - 4p^2q^4

(v)

a+b+c

abc

a^2bc + ab^2c + abc^2


Question:3(i) Find the product.

(a^2)\times (2a^{22})\times (4a^{26})


Answer:

Opening brackets :

(a^2)\times (2a^{22})\times (4a^{26}) = (a^2\times2a^{22})\times(4a^{26}) = 2a^{24}\times4a^{26}

or =8a^{50}

Question:3(ii) Find the product.

(\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2)

Answer:

We have,

(\frac{2}{3}xy)\times (\frac{-9}{10}x^2y^2) = \frac{-3}{5}x^3y^3

Question:3(iii) Find the product.

(\frac{-10}{3}pq^3) \times (\frac{6}{5}p^3q)

Answer:

We have

(\frac{-10}{3}pq^3) \times (\frac{6}{5}p^3q) = -4p^4q^4

Question:3(iv) Find the product.

x \times x^2\times x^3\times x^4

Answer:

We have x \times x^2\times x^3\times x^4

x \times x^2\times x^3\times x^4 = (x \times x^2)\times x^3\times x^4

or (x^3)\times x^3\times x^4

= x^{10}

Question:4(a) Simplify 3x(4x-5)+3 and find its values for

(i) \small x=3

Answer:

(a) We have

3x(4x-5)+3 = 12x^2 - 15x + 3

Put x = 3,

We get : 12(3)^2 - 15(3) + 3 = 12(9) - 45 + 3 = 108 - 42 = 66


Question:4(a) Simplify \small 3x(4x-5)+3 and find its values for

(ii) \small x=\frac{1}{2}

Answer:

We have

\small 3x(4x-5)+3 = 12x^2 -15x + 3

Put

x = \frac{1}{2}

. So We get,

12x^2 -15x + 3 = 12(\frac{1}{2})^2 - 15(\frac{1}{2}) + 3 = 6 - \frac{15}{2} = \frac{-3}{2}

Question:4(b) Simplify \small a(a^2+a+1) + 5 and find its value for

(i) \small a =0

Answer:

We have : \small a(a^2+a+1) +5 = a^3 + a^2 + a +5

Put a = 0 : = 0^3 + 0^2 + 0 + 5 = 5

Question:4(b) Simplify \small a(a^2+a+1)+5 and find its value for

(ii) \small a=1

Answer:

We have \small a(a^2+a+1)+5 = a^3 + a^2 + a + 5

Put a = 1 ,

we get : 1^3 + 1^2 + 1 + 5 = 1 + 1 + 1+ 5 = 8

Question:4(b) Simplify \small a(a^2+a+1)+5 and find its value for

(iii) \small a=-1

Answer:

We have \small a(a^2+a+1)+5 .

or \small a(a^2+a+1)+5 = a^3+a^2+a+5

Put a = (-1)

= (-1)^3+(-1)^2+(-1)+5 = -1 + 1 -1 +5 = 4

Question:5(a) Add: p(p-q),q(q-r) and r(r-p)

Answer:

(a)First we will solve each brackets individually.

p(p-q) = p^2 - pq ; q(q-r) = q^2 - qr ; r(r-p) = r^2 - rp

Addind all we get : p^2 - pq + q^2 - qr + r^2 - rp

= p^2 + q^2 + r^2 -pq-qr-rp

Question:5(b) Add: \small 2x(z-x-y) and \small 2y(z-y-x)

Answer:

Firstly, open the brackets:

\small 2x(z-x-y) = 2xz -2x^2-2xy

and \small 2y(z-y-x) = 2yz-2y^2-2xy

Adding both, we get :

\small 2xz -2x^2-2xy +2yz-2y^2-2xy

or \small = -2x^2-2y^2-4xy + 2xz+2yz

Question:5(c) Subtract: \small 3l(l-4m+5n) from \small 4l(10n-3m+2l)

Answer:

At first we will solve each bracket individually,

\small 3l(l-4m+5n) = 3l^2 - 12lm + 15ln

and \small 4l(10n-3m+2l) = 40ln - 12ml + 8l^2

Subtracting:

\small 40ln - 12ml + 8l^2 - (3l^2 - 12lm+15ln)

or \small = 40ln - 12ml + 8l^2 - 3l^2 + 12lm-15ln

or \small = 25ln + 5l^2

Question:5(d) Subtract: \small 3a(a+b+c)-2b(a-b+c) from \small 4c(-a+b+c)

Answer:

Solving brackets :

3a(a+b+c)-2b(a-b+c) = 3a^2+3ab+3ac - 2ab+2b^2-2bc

= 3a^2+ab+3ac+ 2b^2-2bc

and \small 4c(-a+b+c) = -4ac +4bc + 4c^2

Subtracting : \small -4ac +4bc + 4c^2 -(3a^2 + ab + 3ac+2b^2-2bc)

\small = -4ac + 4bc+4c^2-3a^2-ab-3ac-2b^2+2bc

\small =-3a^2 -2b^2+4c^2-ab+ 6bc-7ac

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities-Exercise: 9.4

Question:1(i) Multiply the binomials.

\small (2x+5) and \small (4x-3)

Answer:

We have (2x + 5) and (4x - 3)
(2x + 5) X (4x - 3) = (2x)(4x) + (2x)(-3) + (5)(4x) + (5)(-3)
= 8 x^{2} - 6x + 20x - 15
= 8 x^{2} + 14x -15

Question:1(ii) Multiply the binomials.

\small (y-8) and \small (3y-4)

Answer:

We need to multiply (y - 8) and (3y - 4)
(y - 8) X (3y - 4) = (y)(3y) + (y)(-4) + (-8)(3y) + (-8)(-4)
= 3 y^{2} - 4y - 24y + 32
= 3 y^{2} - 28y + 32

Question:1(iii) Multiply the binomials

\small (2.5l-0.5m) and \small (2.5l+0.5m)

Answer:

We need to multiply (2.5l - 0.5m) and (2.5l + 0..5m)
(2.5l - 0.5m) X (2.5l + 0..5m) = (2.5l)^{2} - (0.5m)^{2} using (a-b)(a+b) = (a)^{2} - (b)^{2}
= 6.25 l^{2} - 0.25 m^{2}

Question:1(iv) Multiply the binomials.

\small (a+3b) and \small (x+5)

Answer:

(a + 3b) X (x + 5) = (a)(x) + (a)(5) + (3b)(x) + (3b)(5)
= ax + 5a + 3bx + 15b

Question:1(v) Multiply the binomials.

\small (2pq+3q^2) and \small (3pq-2q^2)

Answer:

(2pq + 3q 2 ) X (3pq - 2q 2 ) = (2pq)(3pq) + (2pq)(-2q 2 ) + ( 3q 2 )(3pq) + (3q 2 )(-2q 2 )
= 6p 2 q 2 - 4pq 3 + 9pq 3 - 6q 4
= 6p 2 q 2 +5pq 3 - 6q 4

Question:1(vi) Multiply the binomials.

\small (\frac{3}{4}a^2+3b^2) and \small 4(a^2-\frac{2}{3}b^2)

Answer:

Multiplication can be done as follows

\small (\frac{3}{4}a^2+3b^2) X \small (4a^2-\frac{8}{3}b^2) = \frac{3a^{2}}{4} \times 4a^{2} + \frac{3a^{2}}{4} \times (-\frac{8b^{2}}{3}) + 3b^{2} \times 4a^{2} + 3b^{2} \times (-\frac{8b^{2}}{3})


= 3a^{4} - 2a^{2}b^{2} + 12a^{2}b^{2} - 8b^{4}

= 3a^{4} + 10a^{2}b^{2} - 8b^{4}

Question:2(i) Find the product.

\small (5-2x) \small (3+x)

Answer:

(5 - 2x) X (3 + x) = (5)(3) + (5)(x) +(-2x)(3) + (-2x)(x)
= 15 + 5x - 6x - 2 x^{2}
= 15 - x - 2 x^{2}

Question:2(ii) Find the product.

\small (x+7y)(7x-y)

Answer:

(x + 7y) X (7x - y) = (x)(7x) + (x)(-y) + (7y)(7x) + (7y)(-y)
= 7 x^{2} - xy + 49xy - 7 y^{2}
= 7 x^{2} + 48xy - 7 y^{2}

Question:2(iii) Find the product.

\small (a^2+b)(a+b^2)

Answer:

( a^{2} + b) X (a + b^{2} ) = ( a^{2} )(a) + ( a^{2} )( b^{2} ) + (b)(a) + (b)( b^{2} )
= a^{3 } + a^{2}b^{2} + ab + b^{3}

Question:2(iv) Find the product.

\small (p^2-q^2)(2p+q)

Answer:

following is the solution

( p^{2}- q^{2} ) X (2p + q) = (p^{2})(2p) + (p^{2})(q) + (-q^{2})(2p) + (-q^{2})(q)
2p^{3} + p^{2}q - 2q^{2}p - q^{3}

Question:3(i) Simplify.

\small (x^2-5)(x+5)+25

Answer:

this can be simplified as follows

( x^{2} -5) X (x + 5) + 25 = ( x^{2} )(x) + ( x^{2} )(5) + (-5)(x) + (-5)(5) + 25
= x^{3} + 5x^{2} - 5x -25 + 25
= x^{3} + 5x^{2} - 5x

Question:3(ii) Simplify .

(a^2+5)(b^3+3)+5

Answer:

This can be simplified as

( a^{2} + 5) X ( b^{3} + 3) + 5 = ( a^{2} )( b^{3} ) + ( a^{2} )(3) + (5)( b^{3} ) + (5)(3) + 5
= a^{2}b^{3} + 3a^{2} + 5b^{3} + 15+5
= a^{2}b^{3} + 3a^{2} + 5b^{3} + 20

Question:3(iii) Simplify.

(t+s^2)(t^2-s)

Answer:

simplifications can be

(t + s^{2} )( t^{2} - s) = (t)( t^{2} ) + (t)(-s) + ( s^{2} )( t^{2} ) + ( s^{2} )(-s)
= t^{3} - ts + s^{2}t^{2} - s^{3}

Question:3(iv) Simplify.

(a+b)(c-d)+(a-b)(c+d)+2 (ac+bd)

Answer:

(a + b) X ( c -d) + (a - b) X (c + d) + 2(ac + bd )
= (a)(c) + (a)(-d) + (b)(c) + (b)(-d) + (a)(c) + (a)(d) + (-b)(c) + (-b)(d) + 2(ac + bd )
= ac - ad + bc - bd + ac +ad -bc - bd + 2(ac + bd )
= 2(ac - bd ) + 2(ac +bd )
= 2ac - 2bd + 2ac + 2bd
= 4ac

Question:3(v) Simplify.

(x+y)(2x+y)+(x+2y)(x-y)

Answer:

(x + y) X ( 2x + y) + (x + 2y) X (x - y)
=(x)(2x) + (x)(y) + (y)(2x) + (y)(y) + (x)(x) + (x)(-y) + (2y)(x) + (2y)(-y)
= 2 x^{2} + xy + 2xy + y^{2} + x^{2} - xy + 2xy - 2 y^{2}
=3 x^{2} + 4xy - y^{2}

Question:3(vi) Simplify.

(x+y)(x^2-xy+y^2)

Answer:

simplification is done as follows

(x + y) X ( x^{2} -xy + y^{2} ) = x X ( x^{2} -xy + y^{2} ) + y ( x^{2} -xy + y^{2} )
= x^{3} -x^{2}y + xy^{2} + yx^{2} - xy^{2} + y^{3}
= x^{3}+ y^{3}

Question:3(vii) Simplify.

(1.5x-4y)(1.5x+4y+3)-4.5x+12y

Answer:

(1.5x - 4y) X (1.5x + 4y + 3) - 4.5x + 12y = (1.5x) X (1.5x + 4y + 3) -4y X (1.5x + 4y + 3) - 4.5x + 12y
= 2.25 x^{2} + 6xy + 4.5x - 6xy - 16 y^{2} - 12y -4.5x + 12 y
= 2.25 x^{2} - 16 y^{2}

Question:3(viii) Simplify.

(a+b+c)(a+b-c)

Answer:

(a + b + c) X (a + b - c) = a X (a + b - c) + b X (a + b - c) + c X (a + b - c)
= a^{2} + ab - ac + ab + b^{2} -bc + ac + bc - c^{2}
= a^{2} + b^{2} - c^{2} + 2ab

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities - Topic 9.11 Standard Identities

Question:1(i) Put -b in place of b in identity 1. Do you get identity 2?

Answer:

Identity 1 \Rightarrow (a+b)^{2} = a^{2} + 2ab + b^{2}
If we replace b with -b in identity 1
We get,
a^{2} + 2a(-b) + (-b)^{2} = a^{2} - 2ab + b^{2}
which is equal to
(a-b)^{2} which is identity 2
So, we get identity 2 by replacing b with -b in identity 1

NCERT Free Solutions for Class 8 Maths Chapter 9 Algebraic Expressions And Identities - Topic 9.11 Standard Identities

Question:1 Verify Identity (IV), for a=2,b=3,x=5 .

Answer:

Identity IV
(a + x)(b + x) = x^{2} + (a+b)x + ab
So, it is given that a = 2, b = 3 and x = 5
Lets put these value in identity IV
(2 + 5)(3 + 5) = 5^{2} + (2 + 3)5 +2 X 3
7 X 8 = 25 + 5 X 5 + 6
56 = 25 + 25 + 6
= 56
L.H.S. = R.H.S.
So, by this we can say that identity IV satisfy with given value of a,b and x

Question:2 Consider, the special case of Identity (IV) with a = b, what do you get? Is it related to Identity

Answer:

Identity IV is \Rightarrow (a +x)(b+x) = x^{2} + (a+b)x + ab
If a =b than

(a + x)(a + x) = x^{2} + (a+a)x + a\times a
(a+x)^{2} = x^{2} + 2ax + a^{2}
Which is identity I

Question:3 Consider, the special case of Identity (IV) with a=-c and b=-c What do you get? Is it related to Identity ?

Answer:

Identity IV is \Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab
If a = b = -c than,
(x - c)(x - c) = x^{2} + (-c + (-c))x + (-c) \times (-c)
(x-c)^{2} = x^{2} + -2cx + c^{2}
Which is identity II

Question:4 Consider the special case of Identity (IV) with b=-a . What do you get? Is it related to Identity?

Answer:

Identity IV is \Rightarrow (a+x )(b +x) = x^{2} + (a+b)x + ab
If b = -a than,

(x + a)(x - a) = x^{2} + (a +(-a))x + (-a) \times a
= x^{2} - a^{2}
Which is identity III

Class 8 algebraic expressions and identities NCERT solutions - exercise: 9.5

Question:1(i) Use a suitable identity to get each of the following products.

(x+3)(x+3)

Answer:

(x + 3) X (x +3) = (x +3)^{2}
So, we use identity I for this which is
(a+b)^{2} = a^{2} + 2ab + b^{2}
In this a=x and b = x
(x+3)^{2} = x^{2} + 2(x)(3)+ 3^{2}
= x^{2} + 6x+ 9

Question:1(ii) Use a suitable identity to get each of the following products in bracket.

(2y+5)(2y+5)

Answer:

(2y + 5) X ( 2y + 5) = (2y +5)^{2}
We use identity I for this which is
(a+b)^{2} = a^{2} + 2ab + b^{2}
IN this a = 2y and b = 5
(2y+5)^{2} = (2y)^{2} + 2(2y)(5) + 5^{2}
= (2y+5)^{2} = 4y^{2} + 20y + 25

Question:1(iii) Use a suitable identity to get each of the following products in bracket.

(2a-7)(2a-7)

Answer:

(2a -7) X (2a - 7) = (2a - 7)^{2}
We use identity II for this which is
(a-b)^{2} = a^{2} - 2ab + b^{2}
in this a = 2a and b = 7
(2a-7)^{2} = (2a)^{2} - 2(2a)(7) + 7^{2}
= 4a^{2} - 28a + 49

Question:1(iv) Use a suitable identity to get each of the following products in bracket.

(3a - \frac{1}{2}) (3a -\frac{1}{2} )

Answer:

(3a - \frac{1}{2}) \times (3a -\frac{1}{2} ) = ((3a - \frac{1}{2}))^{2}
We use identity II for this which is
(a-b)^{2} = a^{2} -2ab + b^{2}
in this a = 3a and b = -1/2
(3a-\frac{1}{2})^{2} = (3a)^{2} -2(3a)(\frac{1}{2}) + (\frac{1}{2})^{2}
= 9a^{2} -3a + \frac{1}{4}

Question:1(v) Use a suitable identity to get each of the following products in bracket.

(1.1m - 4)(1.1m+4)

Answer:

(1.1m - 4)(1.1m+4)
We use identity III for this which is
(a - b)(a + b) = a^{2} - b^{2}
In this a = 1.1m and b = 4
(1.1m - 4)(1.1m+4) = (1.1m)^{2} - (4)^{2}
= 1.21 m^{2} - 16

Question:1(vi) Use a suitable identity to get each of the following products in bracket.

(a^2+b^2)(-a^2+b^2)

Answer:

take the (-)ve sign common so our question becomes
- -(a^{2}+b^{2})(a^{2}-b^{2})
We use identity III for this which is
(a - b)(a + b) = a^{2} - b^{2}
In this a = a^{2} and b = b^{2}

-(a^{2}+b^{2})(a^{2}-b^{2}) = -((a^{2})^{2} -(b^{2})^{2}) = -a^{4} + b^{4}

Question:1(vii) Use a suitable identity to get each of the following.

(6x-7) (6x+7)

Answer:

(6x -7) X (6x - 7) = (6x-7)^{2}
We use identity III for this which is
(a - b)(a + b) = a^{2} - b^{2}
In this a = 6x and b = 7
(6x -7) X (6x - 7) = (6x)^{2} - (7)^{2} = 36x^{2} - 49

Question:1(viii) Use a suitable identity to get each of the following product.

(-a+c)(-a+c)

Answer:

take (-)ve sign common from both the brackets So, our question become
(a -c) X (a -c) = (a -c)^{2}
We use identity II for this which is
(a-b)^{2} =a^{2} -2ab + b^{2}
In this a = a and b = c
(a-c)^{2} =a^{2} -2ac + c^{2}

Question:1(ix) Use a suitable identity to get each of the following product.

(\frac{x}{2}+ \frac{3y}{4})(\frac{x}{2}+ \frac{3y}{4})

Answer:

(\frac{x}{2}+ \frac{3y}{4}) \times (\frac{x}{2}+ \frac{3y}{4}) = (\frac{x}{2}+ \frac{3y}{4})^{2}

We use identity I for this which is
(a+b)^{2} =a^{2}+2ab + b^{2}
In this a = \frac{x}{2} and b = \frac{3y}{4}

(\frac{x}{2}+ \frac{3y}{4})^{2} = (\frac{x}{2})^{2} + 2 (\frac{x}{2})(\frac{3y}{4}) + (\frac{3y}{4})^{2}
= \frac{x^{2}}{4} + \frac{3xy}{4} + \frac{9y^{2}}{16}

Question:1(x) Use a suitable identity to get each of the following products.

(7a-9b)(7a-9b)

Answer:

(7a-9b) \times (7a-9b) = (7a-9b)^{2}


We use identity II for this which is
(a-b)^{2} =a^{2}-2ab + b^{2}
In this a = 7a and b = 9b
(7a-9b)^{2} =(7a)^{2}-2(7a)(9b) + (9b)^{2}
= 49a^{2}-126ab + 81b^{2}

Question:2(i) Use the identity (x+a) (x+b) = x^2+(a+b)x+ab to find the following products.

(x+3)(x+7)

Answer:

We use identity (x+a) (x+b) = x^2+(a+b)x+ab
in this a = 3 and b = 7
(x+3)(x+7) = x^2+(3+7)x+3 \times 7
= x^2+10x+ 21

Question:2(ii) Use the identity (x+a)(x+b)=x^2+(a+b)x+ab to find the following products.

(4x+5)(4x+1)

Answer:

We use identity (x+a)(x+b)=x^2+(a+b)x+ab
In this a= 5 , b = 1 and x = 4x
(4x+5)(4x+1) = (4x)^2+(5+1)4x+(5)(1)
= 16x^2+24x+5

Question:2(iii) Use the identity (x+a)(x+b)= x^2+(a+b)x+ab to find the following products.

(4x-5)(4x-1)

Answer:

We use identity (x+a)(x+b)= x^2+(a+b)x+ab
in this x = 4x , a = -5 and b = -1
(4x-5)(4x-1) = (4x)^2+(-5-1)4x+(-5)(-1)
= 16x^2 - 24x+ 5

Question:1(iv) Use the identity (x+a)(x+b)=x^2+(a+b)x+ab to find the following products.

(4x+5)(4x-1)

Answer:

We use identity (x+a)(x+b)=x^2+(a+b)x+ab
In this a = 5 , b = -1 and x = 4x
(4x+5)(4x-1) = (4x)^2+(5+(-1))4x+(5)(-1)
= 16x^2+16x- 5

Question:2(v) Use the identity (x+a)(x+b)=x^2+(a+b)x+ab to find the following products.

(2x+5y)(2x+3y)

Answer:

We use identity (x+a)(x+b)=x^2+(a+b)x+ab
In this a = 5y , b = 3y and x = 2x
(2x+5y)(2x+3y) = (2x)^2+(5y+3y)(2x)+(5y)(3y)
= 4x^2+16xy + 15y^{2}

Question:2(vi) Use the identity (x+a)(x+b)=x^2+(a+b)x+ab to find the following products.

(2a^2+9)(2a^2+5)

Answer:

We use identity (x+a)(x+b)=x^2+(a+b)x+ab
In this a = 9 , b = 5 and x = 2a^{2}
(2a^{2}+9)(2a^{2}+5) = (2a^{2})^2+(9+5)2a^{2}+(9)(5)
= 4a^{4} + 28a^{2} + 45

Question:2(vii) Use the identity (x+a) (x+b)=x^2+(a+b)x+ab to find the following products.

(xyz-4) (xyz-2)

Answer:

We use identity (x+a)(x+b)=x^2+(a+b)x+ab
In this a = -4 , b = -2 and x = xyz
(xyz-4)(xyz-2) = (xyz)^2+((-4)+(-2))xyz+(-4)(-2)
= x^{2}y^{2}z^{2} -6xyz + 8

Question:3(i) Find the following squares by using the identities.

(b-7)^2

Answer:

We use identity
(a-b)^{2} = a^{2} - 2ab + b^{2}
In this a =b and b = 7
(b-7)^{2} = b^{2} - 2(b)(7) + 7^{2}
= b^{2} - 14b + 49

Question:3(ii) Find the following squares by using the identities.

(xy+3z)^2

Answer:

We use
(a+b)^{2} = a^{2} + 2ab + b^{2}

In this a = xy and b = 3z
(xy+3z)^{2} = (xy)^{2} + 2(xy)(3z) + (3z)^{2}
= x^{2}y^{2} + 6xyz+ 9z^{2}

Question:3(iii) Find the following squares by using the identities.

(6x^2-5y)^2

Answer:

We use
(a-b)^{2} = a^{2} - 2ab + b^{2}
In this a = 6x^{2} and b = 5y^{2}
(6x-5y)^{2} = (6x)^{2} - 2(6x)(5y) + (5y)^{2}
= 36x^{2} - 60xy + 25y^{2}

Question:3(iv) Find the following squares by using the identities.

(\frac{2}{3}m+\frac{3}{2}n)^2

Answer:

we use the identity
(a+b)^{2} = a^{2} + 2ab + b^{2}
In this a = \frac{2m}{3} and b = \frac{3n}{2}
(\frac{2m}{3} + \frac{3n}{2})^{2} = (\frac{2m}{3})^{2} + 2(\frac{2m}{3})( \frac{3n}{2}) + ( \frac{3n}{2})^{2}

= \frac{4m^{2}}{9} + 2mn + \frac{9n^{2}}{4}

Question:3(v) Find the following squares by using the identities.

(0.4p-0.5q)^2

Answer:

we use
(a-b)^{2} = a^{2} - 2ab + b^{2}
In this a = 0.4p and b =0.5q
(0.4p-0.5q)^{2} = (0.4p)^{2} - 2(0.4p)(0.5q) + (0.5q)^{2}
= 0.16p^{2} - 0.4pq + 0.25q^{2}

Question:3(vi) Find the following squares by using the identities.

(2xy+5y)^2

Answer:

we use the identity
(a+b)^{2} = a^{2} + 2ab + b^{2}
In this a = 2xy and b =5y
(2xy+5y)^{2} = (2xy)^{2} + 2(2xy)(5y) + (5y)^{2}
= 4x^{2}y^{2} + 20xy^{2} + 25y^{2}

Question:4(i) Simplify:

(a^2-b^2)^2

Answer:

we use
(a-b)^{2} = a^{2} - 2ab + b^{2}
In this a = a^{2} and b = b^{2}
(a^{2}-b^{2})^{2} = (a^{2})^{2} - 2(a^{2})(b^{2}) + (b^{2})^{2}
= a^{4} - 2a^{2}b^{2} + b^{4}

Question:4(ii) Simplify.

(2x+5)^2-(2x-5)^2

Answer:

we use
a^{2} - b^{2} = (a-b)(a+b)
In this a = (2x + 5) and b = (2x - 5)
(2x + 5)^{2} - (2x - 5)^{2} = ( (2x + 5)- (2x - 5))( (2x + 5)+ (2x - 5))
= ( 2x + 5- 2x + 5)( 2x + 5+ 2x - 5)
= (4x)(10)
=40x

or

remember that

(a+b)^2-(a-b)^2=4ab

here a= 2x, b= 5

4ab=4\times 2x \times 5=40x

Question:4(iii) Simplify.

(7m-8n)^2+(7m+8n)^2

Answer:

we use
(a-b)^{2} = a^{2} -2ab + b^{2} and (a+b)^{2} = a^{2} +2ab + b^{2}
In this a = 7m and b = 8n
(7m-8n)^{2} = (7m)^{2} -2(7m)(8n) + (8n)^{2}
= 49m^{2} -112mn + 64n^{2}
and
(7m+8n)^{2} = (7m)^{2} +2(7m)(8n) + (8n)^{2}
= 49m^{2} +112mn + 64n^{2}

So, (7m - 8n)^{2} + (7m + 8n)^{2} = 49m^{2} -112mn + 64n^{2} + 49m^{2} +112mn + 64n^{2}
= 2(49m^{2} + 64n^{2})

or

remember that

(a-b)^2+(a+b)^2=2(a^2+b^2)

Question: 4(iv) Simplify.

(4m+5n)^2+(5m+4n)^2

Answer:

we use
(a+b)^{2} = a^{2} +2ab + b^{2}
1 ) In this a = 4m and b = 5n

(4m+5n)^{2} = (4m)^{2} +2(4m)(5n) + (5n)^{2}
= 16m^{2} +40mn + 25n^{2}
2 ) in this a = 5m and b = 4n
(5m+4n)^{2} = (5m)^{2} +2(5m)(4n) + (4n)^{2}
= 25m^{2} +40mn + 16n^{2}

So, (4m + 5n)^{2} + (5m + 4n)^{2} = 16m^{2} +40mn + 25n^{2} + 25m^{2} +40mn + 16n^{2}
= 41m^{2} +80mn + 41n^{2}

Question: 4(v) Simplify.

(2.5p-1.5q)^2-(1.5p-2.5q)^2

Answer:

we use
a^{2}- b^{2} = (a-b)(a+b)
1 ) In this a = (2.5p- 1.5q) and b = (1.5p - 2.5q)
(2.5p- 1.5q)^{2}- (1.5p- 2.5q)^{2} = ( (2.5p- 1.5q)- (1.5p- 2.5q))( (2.5p- 1.5q)+ (1.5p- 2.5q))
= ( 2.5p- 1.5q- 1.5p + 2.5q)(2.5p- 1.5q+ 1.5p- 2.5q)
= 4(p + q ) (p - q)
= 4 (p^{2} - q^{2})

Question:4(vi) Simplify.

(ab+bc)^2-2ab^2c

Answer:

We use identity
(a+b)^{2} = a^{2} + 2ab + b^{2}
In this a = ab and b = bc
(ab+bc)^{2} = (ab)^{2} + 2(ab)(bc) + (bc)^{2}
= a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2}
Now, a^{2}b^{2} + 2ab^{2}c + b^{2}c^{2} - 2ab^{2}c
= a^{2}b^{2} + b^{2}c^{2}

Question:4(vii) Simplify.

(m^2 -n^2m)^2+2m^3n^2

Answer:

We use identity
(a-b)^{2} = a^{2} - 2ab + b^{2}
In this a = m^{2} and b = n^{2}m
(m^{2}-n^{2}m)^{2} = (m^{2})^{2} - 2(m^{2})(n^{2}m) + (n^{2}m)^{2}
= m^{4} - 2m^{3}n^{2} + n^{4}m^{2}
Now, m^{4} - 2m^{3}n^{2} + n^{4}m^{2} + 2m^{3}n^{2}
= m^{4} + n^{4}m^{2}

Question:5(i) Show that

(3x+7)^2-84x=(3x-7)^2

Answer:

L.H.S. = (3x+7)^2 - 84x = 9x^2 + 42x + 49 - 84x

= 9x^2 - 42 x +49

= (3x - 7)^2

= R.H.S.

Hence it is prooved

Question:5(ii) Show that

(9p-5q)^2+180pq=(9p+5q)^2

Answer:

L.H.S. = (9p-5q)^2+180pq = 81p^2 - 90pq + 25q^2 + 180pq (Using (a-b)^2 = a^2 - 2ab + b^2 )

= 81p^2 +90pq + 25q^2

= (9p + 5q)^2 \left ( (a+b)^2 = a^2 + 2ab + b^2 \right )

= R.H.S.

Question:5(iii) Show that.

(\frac{4}{3}m-\frac{3}{4}n)^2 +2mn=\frac{16}{9}m^2+\frac{9}{16}n^2

Answer:

First we will solve the LHS :

= (\frac{4}{3}m-\frac{3}{4}n)^2 +2mn = \frac{16}{9}m^2 - 2mn + \frac{9}{16}n^2 + 2mn

or = \frac{16}{9}m^2 + \frac{9}{16}n^2

= RHS

Question:5(iv) Show that.

(4pq+3q)^2-(4pq-3q)^2=48pq^2

Answer:

Opening both brackets we get,

(4pq+3q)^2-(4pq-3q)^2 = 16p^2q^2 + 24pq^2 + 9q^2 - (16p^2q^2 - 24pq^2 + 9q^2)

= 16p^2q^2 + 24pq^2 + 9q^2 - 16p^2q^2 + 24pq^2 - 9q^2)

= 48pq^2

= R.H.S.

Question:5(v) Show that

(a-b)(a+b) + (b-c) ( b +c)+(c-a)(c+a)=0

Answer:

Opening all brackets from the LHS, we get :

(a-b)(a+b) + (b-c) ( b +c)+(c-a)(c+a)\\\\ =\ a^2 +ab - ab- b^2 + b^2+bc - bc -c^2 + c^2 +ca - ac -a^2

= 0 = RHS

Question:6(i) Using identities, evaluate.

71^2

Answer:

We will use the identity:

(a + b)^2 = a^2 + 2ab + b^2

So, 71^2 = (70 +1)^2 = 70^2 + 2(70)(1) + 1^2

= 4900 + 140 + 1

= 5041

Question:6(ii) Using identities, evaluate.

99^2

Answer:

Here we will use the identity :

(a - b)^2 = a^2 - 2ab + b^2

So : 99^2 = (100 - 1) ^2 = 100^2 - 2(100)(1) + 1^2

or = 10000 - 200 + 1

= 9801

Question:6(iii) Using identities, evaluate.

102^2

Answer:

Here we will use the identity :

(a+b)^2 = a^2 +2ab + b^2

So :

102^2 = (100 + 2)^2 = 100^2 + 2(100)(2) + 2^2

or = 10000 + 400 + 4

= 10404

Question:6(iv) Using identities, evaluate.

998^2

Answer:

Here we will the identity :

998^2=(1000 - 2)^2 = 1000^2 - 2(1000)(2) + 2^2

or = 1000000 - 4000+ 4

or = 996004

Question:6(v) Using identities, evaluate.

5.2^2

Answer:

Here we will use :

(a + b)^2 = a^2 + 2ab + b^2

Thus

(5.2)^2 = (5.0 + 0.2)^2 = 5^2 + 2(5)(0.2) + (0.2)^2

or = 25 + 2 + 0.04

= 27.04

Question:6(vi) Using identities, evaluate.

297 \times 303

Answer:

This can be written as :

297\times303 = (300-3)\times(300+3)

using (a-b)(a+b)=a^2-b^2

or = 90000 - 9

= 89991

Question:6(vii) Using identities, evaluate.

78 \times 82

Answer:

This can be written in form of :

78\times82 = (80 - 2) \times(80+2)

or = 80^2 - 2^2 \because \left ( a-b \right )\left ( a+b \right ) = a^2 - b^2

or = 6400- 4 = 6396

Question:6(viii) Using identities, evaluate.

8.9^2

Answer:

Here we will use the identity :

(a - b)^2 = a^2 - 2ab + b^2

Thus :

8.9^2 = (9 - 0.1) ^2 = 9^2 - 2(9)(0.1) + 0.1^2

or = 81 - 1.8 + 0.01

or = 79.21

Question:6(ix) Using identities, evaluate.

10.5\times9.5

Answer:

This can be written as :

10.5\times9.5 = (10 +0.5)\times(10-0.5)

or = 10^2 - 0.5^2 \because (a+b)(a-b) = a^2 - b^2

or = 100 - 0.25

or = 99.75

Question:7(i) Using a^2-b^2=(a+b)(a-b) , find

51^2-49^2

Answer:

We know,

a^2-b^2=(a+b)(a-b)

Using this formula,

51^2-49^2 = (51 + 49)(51 - 49)

= (100)(2)

= 200

Question:7(ii) Using a^2-b^2=(a+b)(a-b) , find

(1.02)^2-(0.98)^2

Answer:

We know,

a^2-b^2=(a+b)(a-b)

Using this formula,

(1.02)^2-(0.98)^2 = (1.02 + 0.98)(1.02 - 0.98)
= (2.00)(0.04)

= 0.08

Question:7(iii) Using a^2-b^2=(a+b)(a-b) , find.

153^2-147^2

Answer:

We know,

a^2-b^2=(a+b)(a-b)

Using this formula,

153^2-147^2 = (153 - 147)(153 +147)

=(6) (300)

= 1800

Question:7(iv) Using a^2-b^2=(a+b )(a-b) , find

12.1^2-7.9^2

Answer:

We know,

a^2-b^2=(a+b)(a-b)

Using this formula,

(1.02)^2-(0.98)^2 = (1.02 + 0.98)(1.02 - 0.98)

= (2.00)(0.04)

= 0.08

Question:8(i) Using (x+a)(x+b)=x^2+(a+b)x+ab 103 \times 104

Answer:

We know,

(x+a)(x+b)=x^2+(a+b)x+ab

Using this formula,

103 \times 104 = (100 + 3)(100 + 4)

Here x =100, a = 3, b = 4

\therefore 103 \times 104 = 100^2+(3+4)100+(3\times4)

= 10000+1200+12

= 11212

Question:8(ii) Using (x+a)(x+b)=x^2+(a+b)x+ab , find

5.1\times 5.2

Answer:

We know,

(x+a)(x+b)=x^2+(a+b)x+ab

Using this formula,

5.1\times 5.2 = (5 + 0.1)(5 + 0.2)

Here x =5, a = 0.1, b = 0.2

\therefore 5.1\times 5.2 =5^2+(0.1 + 0.2)5+(0.1\times0.2)

= 25+1.5+0.02

= 26.52

Question:8(iii) Using (x+a)(x+b)=x^2+(a+b)x+ab , find

103 \times 98

Answer:

We know,

(x+a)(x+b)=x^2+(a+b)x+ab

Using this formula,

103 \times 98 = (100 + 3)(100 - 2) = (100 + 3){100 + (-2)}

Here x =100, a = 3, b = -2

\therefore 103 \times 98 = 100^2+(3 + (-2))100+(3\times(-2))

= 10000+100-6

= 10094

Question: 8(iv) Using (x+a)(x+b)=x^2+(a+b)x+ab , find

9.7 \times 9.8

Answer:

We know,

(x+a)(x+b)=x^2+(a+b)x+ab

Using this formula,

9.7 \times 9.8 = (10 - 0.3)(10 - 0.2) = {10 + (-0.3)}{10 + (-0.2)}

Here x =10, a = -0.3, b = -0.2

\therefore 9.7 \times 9.8 = 10^2+((-0.3) + (-0.2))10+((-0.3)\times(-0.2))

= 100-5+0.06

= 95.

NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities - Topics

  • What are Expressions?
  • Terms, Factors and Coefficients
  • Monomials, Binomials,f and Polynomials
  • Like and Unlike Terms
  • Addition and Subtraction of Algebraic Expressions
  • Multiplication of Algebraic Expressions: Introduction
  • Multiplying a Monomial by a Monomial
  • Multiplying two monomials
  • Multiplying three or more monomials
  • Multiplying a Monomial by a Polynomial
  • Multiplying a monomial by a binomial
  • Multiplying a monomial by a trinomial
  • Multiplying a Polynomial by a Polynomial
  • Multiplying a binomial by a binomial
  • Multiplying a binomial by a trinomial
  • What is an Identity?
  • Standard Identities
  • Applying Identities

NCERT Solutions for Class 8 Maths - Chapter Wise

NCERT Solutions for Class 8 - Subject Wise

Some Important Identities From NCERT Book for Class 8 Chapter 9 Algebraic Expressions And Identities

  • (a+b)^2=(a^2+2ab+b^2)

you can write

(a+b)^2=(a+b)(a+b)

Can be simplified as follows

(a+b)(a+b)=a\times a+a\times b+a\times b+b\times b

Now add each term

a\times a+a\times b+a\times b+b\times b=a^2+ab+ab+b^2

=a^2+2ab+b^2

(a+b)^2=(a^2+2ab+b^2)

  • (a-b)^2=(a^2-2ab+b^2)
  • (a-b)(a+b)=a^2-b^2
  • (x+a)(x+b)=x^{2}+(a+b) x+a b

You can form the above identities by yourself. These above identities have been used in many problems of NCERT solutions for class 8 maths chapter 9 algebraic expression and identities.

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. What are the important topics of NCERT syllabus chapter Algebraic Expressions and Identities ?

Addition and subtraction of the algebraic expression, multiplication of the algebraic expression, standard identities, and application of identities are important topics in this chapter.

2. Does CBSE class maths is tough ?

CBSE class 8 maths is not tough at all. It teaches a very basic and simple maths.

3. How many chapters are there in the CBSE class 8 maths ?

There are 16 chapters starting from rational number to playing with numbers in the CBSE class 8 maths.

4. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

5. Where can I find the complete solutions of NCERT for class 8 maths ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

6. Which is the official website of NCERT ?

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

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0.67\; J

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Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

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 9.89×10−3 kg

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12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

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K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

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33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

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twice that in 60 g carbon

Option 2)

6.023 × 1022

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half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

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more than 9

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