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

 

NCERT solutions for class 8 maths chapter 9 Algebraic Expressions and Identities- As you have learned in previous classes that expressions are formed from variables and constants. This chapter will introduce you to the application of algebraic terms and variables to solve various problems. NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities will give you a detailed explanation for every problem of this chapter. 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. You will find the questions related to these topics in CBSE NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities which will make your task easy while solving the problems. As it is a new concept, you may find difficulties while dealing with algebraic parts of mathematics but if you practice questions and go through solutions of NCERT for class 8 maths chapter 9 algebraic expressions and identities, you will find it very easy and one of the strongest parts in mathematics. You will get NCERT solutions from class 6 to 12 for science and maths by clicking on the above link. 

Topics of NCERT for class 8 maths chapter 9 - Algebraic expressions and identities:-

9.1 What are Expressions?

9.2 Terms, Factors and Coefficients

9.3 Monomials, Binomials and Polynomials

9.4 Like and Unlike Terms

9.5 Addition and Subtraction of Algebraic Expressions

9.6 Multiplication of Algebraic Expressions: Introduction

9.7 Multiplying a Monomial by a Monomial

9.7.1 Multiplying two monomials

9.7.2 Multiplying three or more monomials

9.8 Multiplying a Monomial by a Polynomial

9.8.1 Multiplying a monomial by a binomial

9.8.2 Multiplying a monomial by a trinomial

9.9Multiplying a Polynomial by a Polynomial

9.9.1 Multiplying a binomial by a binomial

9.9.2 Multiplying a binomial by a trinomial

9.10 What is an Identity?

9.11 Standard Identities

9.12 Applying Identities

NCERT Solutions to the exercises of Chapter 9: Algebraic Expressions and Identities 

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities topic 9.1 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:

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

x-4

Answer:

x-4 on the number line:

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

 2x+1

Answer:

2x+1 on the number  line:

 

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

 3x-2

Answer:

3x - 2 on the number line

Solutions of NCERT for class 8 maths chapter 9 algebraic expressions and identities 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\\

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities 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}

Solutions of NCERT 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 are5xyz^{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:2(i) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

 ab+bc+cd+da

Answer:

This polynomial does not fit in any of these three categories.

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

Solutions for NCERT class 8 maths chapter 9 algebraic expressions and identities 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.

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-Exercise: 9.2

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

 4,7p

Answer:

4\times 7p=28p

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

     -4p,7pq

Answer:

-4p\times 7pq\\=(-4\times 7)p\times pq\\=-28p^{2}q

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

               4p,0

Answer:

\\4p\times 0=0

Question:2(A)  Find the areas of rectangles with the following pairs of monomials as their lengths and breadths  respectively.

            (p,q)

Answer:

The question can be solved as follows

\\Area=length\times breadth\\ =(p\times q)\\ =pq

Question:2(B) Find the areas of rectangles with the following pairs of monomials as their lengths and breadth respectively.

          (10m,5n)

Answer:

the area is calculated as follows

\\Area=length\times breadth\\ =10m\times 5n\\ =50mn

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

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities 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 textbook 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

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-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:1(v) Carry out the multiplication of the expressions in each of the following pairs.

 pq+qr+rp, 0

Answer:

Using distributive law :

             (pq + qr + rp)(0) = pq(0) + qr(0) + rp(0) = 0

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)
                                       = 8x^{2} - 6x + 20x - 15
                                       = 8x^{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)
                             =   3y^{2} - 4y - 24y  + 32
                              = 3y^{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}
                                                  = 6.25l^{2}  - 0.25m^{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 + 3q2) X (3pq - 2q2)  =  (2pq)(3pq) + (2pq)(-2q2) + ( 3q2)(3pq) + (3q2)(-2q2)
                                            = 6p2q2 - 4pq3  + 9pq3 - 6q4
                                            = 6p2q2 +5pq3 - 6q4

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 - 2x^{2}
                            = 15  - x - 2x^{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)
                             = 7x^{2} - xy + 49xy - 7y^{2}
                             =  7x^{2} + 48xy - 7y^{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)
= 2x^{2} + xy + 2xy + y^{2} + x^{2} - xy + 2xy - 2y^{2}
=3x^{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
                                                                 =  2.25x^{2} + 6xy + 4.5x  - 6xy - 16y^{2} - 12y   -4.5x + 12 y
                                                                 = 2.25x^{2}    - 16y^{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 standerd 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

NCERT solutions for class 8 maths chapter 9 algebraic expressions and identities-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}
                                                                                           =  1.21m^{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: