NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities

Question

Here you will find NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities, which are essential for understanding the chapter's basic concepts. You can get a PDF of Chapter 9 Algebraic Expressions and Identities NCERT Solutions for Class 8 Maths that will make learning fun and help you finish the syllabus in less time. These NCERT Solutions will assist you in developing your own homework answers and will also make you aware of the difficulty of the questions. Class 8 NCERT Solutions will help you solve difficult problems in a given exercise. Chapter 9 NCERT Solutions will be extremely beneficial in exam preparation and understanding of concepts.

Topics in the chapter:

• What are Algebraic Expressions?
• Monomials, Binomials and Polynomials
• Like and Unlike Terms 
• Addition and Subtraction of Algebraic Expressions
• Multiplication of monomial by a monomial
• Multiplication of a monomial by a binomial or trinomial or polynomial
• Multiplying a Polynomial by a Polynomial
• What is an Identity Applying Identities

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Answer 1

NCERT Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities

1. Identify the terms, their coefficients for each of the following expressions.

(i) 5xyz² – 3zy

(ii) 1 + x + x²

(iii) 4x²y² – 4x²y²z² + z²

(iv) 3 – pq + qr – p

(v) (x/2) + (y/2) – xy

(vi) 0.3a – 0.6ab + 0.5b

Solution:

Sl. No.	Expression	Term	Coefficient
(i)	5xyz2 – 3zy	5xyz2 -3zy	5 -3
(ii)	1 + x + x2	1 x x2	1 1 1
(iii)	4x2y2 – 4x2y2z2 + z2	4x2y2 -4 x2y2z2 z2	4 -4 1
(iv)	3 – pq + qr – p	3 -pq qr -p	3 -1 1 -1
(v)	(x/2) + (y/2) – xy	x/2 Y/2 -xy	1/2 1/2 -1
(vi)	0.3a – 0.6ab + 0.5b	0.3a -0.6ab 0.5b	0.3 -0.6 0.5

2. Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories?

x + y, 1000, x + x² + x³ + x⁴ , 7 + y + 5x, 2y – 3y² , 2y – 3y² + 4y³ , 5x – 4y + 3xy, 4z – 15z² , ab + bc + cd + da, pqr, p²q + pq² , 2p + 2q

Solution:

Let us first define the classifications of these 3 polynomials:

Monomials, Contain only one term.

Binomials, Contain only two terms.

Trinomials, Contain only three terms.

x + y	two terms	Binomial
1000	one term	Monomial
x + x2 + x3 + x4	four terms	Polynomial, and it does not fit in listed three categories
2y – 3y2	two terms	Binomial
2y – 3y2 + 4y3	three terms	Trinomial
5x – 4y + 3xy	three terms	Trinomial
4z – 15z2	two terms	Binomial
ab + bc + cd + da	four terms	Polynomial, and it does not fit in listed three categories
pqr	one term	Monomial
p2q + pq2	two terms	Binomial
2p + 2q	two terms	Binomial
7 + y + 5x	three terms	Trinomial

3.  Add the following.

(i) ab – bc, bc – ca, ca – ab

(ii) a – b + ab, b – c + bc, c – a + ac

(iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²

(iv) l² + m², m² + n², n² + l², 2lm + 2mn + 2nl

Solution:

i) (ab – bc) + (bc – ca) + (ca-ab)

= ab – bc + bc – ca + ca – ab

= ab – ab – bc + bc – ca + ca

= 0

ii) (a – b + ab) + (b – c + bc) + (c – a + ac)

= a – b + ab + b – c + bc + c – a + ac

= a – a +b – b +c – c + ab + bc + ca

= 0 + 0 + 0 + ab + bc + ca

= ab + bc + ca

iii) 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²

= (2p²q² – 3pq + 4) + (5 + 7pq – 3p²q²)

= 2p²q² – 3p²q² – 3pq + 7pq + 4 + 5

= – p²q² + 4pq + 9

iv)(l² + m²) + (m² + n²) + (n² + l²) + (2lm + 2mn + 2nl)

= l² + l² + m² + m² + n² + n² + 2lm + 2mn + 2nl

= 2l² + 2m² + 2n² + 2lm + 2mn + 2nl

4. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3

(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz

(c) Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from 18 – 3p – 11q + 5pq – 2pq² + 5p²q

Solution:

(a) (12a – 9ab + 5b – 3) – (4a – 7ab + 3b + 12)

= 12a – 9ab + 5b – 3 – 4a + 7ab – 3b – 12

= 12a – 4a -9ab + 7ab +5b – 3b -3 -12

= 8a – 2ab + 2b – 15

(b) (5xy – 2yz – 2zx + 10xyz) – (3xy + 5yz – 7zx)

= 5xy – 2yz – 2zx + 10xyz – 3xy – 5yz + 7zx

= 5xy – 3xy – 2yz – 5yz – 2zx + 7zx + 10xyz

= 2xy – 7yz + 5zx + 10xyz

(c) (18 – 3p – 11q + 5pq – 2pq² + 5p²q) – (4p²q – 3pq + 5pq² – 8p + 7q – 10)

= 18 – 3p – 11q + 5pq – 2pq² + 5p²q – 4p²q + 3pq – 5pq² + 8p – 7q + 10

= 18+10 -3p+8p -11q – 7q + 5 pq + 3pq - 2pq² – 5pq^2 + 5 p² q – 4p² q

= 28 + 5p – 18q + 8pq – 7pq² + p²q

Answer 2

5. Find the product of the following pairs of monomials.

(i) 4, 7p

(ii) – 4p, 7p

(iii) – 4p, 7pq

(iv)  4p³, – 3p

(v) 4p, 0

Solution:

(i) 4 , 7 p =  4 × 7 × p = 28p

(ii) – 4p × 7p = (-4 × 7 ) × (p × p )= -28p²

(iii) – 4p × 7pq =(-4 × 7 ) (p × pq) =  -28p²q

(iv) 4p³ × – 3p = (4 × -3 ) (p³ × p ) =  -12p⁴

(v) 4p ×  0 = 0

6. Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively.

(p, q) ; (10m, 5n) ; (20x² , 5y²) ; (4x, 3x²) ; (3mn, 4np)

Solution:

Area of rectangle = Length x breadth. So, it is multiplication of two monomials.

The results can be written in square units.

(i) p × q = pq

(ii)10m ×  5n = 50mn

(iii) 20x² ×  5y² =  100x²y²

(iv) 4x × 3x² = 12x³

(v) 3mn ×  4np = 12mn²p

7. Complete the table of Products.

Solution:

Completed Table

8. Obtain the volume of rectangular boxes with the following length, breadth and height respectively.

(i) 5a, 3a², 7a⁴

(ii) 2p, 4q, 8r

(iii) xy, 2x²y, 2xy²

(iv) a, 2b, 3c

Solution:

Volume of rectangle = length x  breadth x  height. To evaluate volume of rectangular boxes, multiply all the monomials.

(i) 5a x 3a² x 7a⁴ = (5 × 3 × 7) (a × a² × a⁴ ) = 105a⁷

(ii) 2p x 4q x 8r = (2 × 4 × 8 ) (p × q × r ) = 64pqr

(iii) y × 2x²y × 2xy² =(1 × 2 × 2 )( x × x² × x × y × y × y² ) =  4x⁴y⁴

(iv) a x  2b x 3c = (1 × 2 × 3 ) (a × b × c) = 6abc

9. Obtain the product of

(i) xy,  yz, zx

(ii) a, – a² , a³

(iii) 2, 4y, 8y² , 16y³

(iv) a, 2b, 3c, 6abc

(v) m, – mn, mnp

Solution:

(i) xy × yz × zx = x² y² z²

(ii) a × – a²  × a³ = – a⁶

(iii) 2 × 4y × 8y² × 16y³ = 1024 y⁶

(iv) a × 2b × 3c × 6abc = 36a² b² c²

(v) m × – mn × mnp = –m³ n² p

10. Carry out the multiplication of the expressions in each of the following pairs.

(i) 4p, q + r

(ii) ab, a – b

(iii) a + b, 7a²b²

(iv) a² – 9, 4a

(v) pq + qr + rp, 0

Solution:

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

(ii) ab(a – b) = a² b – a b²

(iii) (a + b) (7a²b²) = 7a³b² + 7a²b³

(iv) (a² – 9)(4a) = 4a³ – 36a

(v) (pq + qr + rp) × 0 = 0 ( Anything multiplied by zero is zero )

11. Complete the table.

S.No.	First Expression	Second Expression	Product
(i)	a	b + c + d	–
(ii)	x + y – 5	5xy	–
(iii)	p	6p2 – 7p + 5	–
(iv)	4p2q2	p2 – q2	–
(v)	a + b + c	abc	–

Solution:

S.No.	First expression	Second expression	Product
(i)	a	b + c + d	a(b+c+d) = a×b + a×c + a×d = ab + ac + ad
(ii)	x + y – 5	5xy	5 xy (x + y – 5) = 5 xy x x + 5 xy x y – 5 xy x 5 = 5 x2y + 5 xy2 – 25xy
(iii)	p	6p2 – 7p + 5	p (6 p 2-7 p +5) = p× 6 p2 – p× 7 p + p×5 = 6 p3 – 7 p2 + 5 p
(iv)	4 p2 q2	P2 – q2	4p2 q2 * (p2 – q2 ) =4 p4 q2– 4p2 q4
(v)	a + b + c	abc	abc(a + b + c) = abc × a + abc × b + abc × c = a2bc + ab2c + abc2

Answer 3

12. Find the product.

(i) a² x (2a²²) x (4a²⁶)

(ii) (2/3 xy) × (-9/10 x²y²)

(iii) (-10/3 pq³/) × (6/5 p³q)

(iv) (x) × (x²) × (x³) × (x⁴)

Solution:

(i) a² x (2a²²) x (4a²⁶)

= (2 × 4) ( a² × a²² × a²⁶ )

= 8 × a^{2 + 22 + 26} 

= 8a⁵⁰

(ii) (2xy/3) ×(-9x²y²/10)

= (2/3 × -9/10 ) ( x × x² × y × y² )

= (-3/5 x³y³)

(iii) (-10pq³/3) ×(6p³q/5)

= ( -10/3 × 6/5 ) (p × p³× q³ × q)

= (-4p⁴q⁴)

(iv)  ( x) x (x²) x (x³) x (x⁴)

= x ^{1 + 2 + 3 + 4} 

=  x¹⁰

13. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =1/2

(b) Simplify a (a²+ a + 1) + 5 and find its value for (i) a = 0, (ii) a = 1 (iii) a = – 1.

Solution:

a) 3x (4x – 5) + 3

=3x ( 4x) – 3x( 5) +3

=12x² – 15x + 3

(i) Putting x=3 in the equation we gets 12x² – 15x + 3 =12(3²) – 15 (3) +3

= 108 – 45 + 3

= 66

(ii) Putting x=1/2 in the equation we get

12x² – 15x + 3 = 12 (1/2)² – 15 (1/2) + 3

= 12 (1/4) – 15/2 +3

= 3 – 15/2 + 3

= 6- 15/2

= (12- 15 ) /2

= -3/2

b) a(a² + a + 1) + 5

= a x a² + a x a + a x 1 + 5 =a³+a²+a+ 5

(i) putting a=0 in the equation we get 0³+0²+0+5=5

(ii) putting a=1 in the equation we get 1³ + 1² + 1+5 = 1 + 1 + 1+5 = 8

(iii) Putting a = -1 in the equation we get (-1)³+(-1)² + (-1)+5 = -1 + 1 – 1+5 = 4

14. (a) Add: p ( p – q), q ( q – r) and r ( r – p) 

(b) Add: 2x (z – x – y) and 2y (z – y – x) 

(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l ) 

(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c)  from 4c ( – a + b + c )

Solution:

a) p ( p – q) + q ( q – r) + r ( r – p)

= (p² – pq) + (q² – qr) + (r² – pr)

= p² + q² + r² – pq – qr – pr

b) 2x (z – x – y) + 2y (z – y – x)

= (2xz – 2x² – 2xy) + (2yz – 2y² – 2xy)

= 2xz – 4xy + 2yz – 2x² – 2y²

c) 4l ( 10 n – 3 m + 2 l ) – 3l (l – 4 m + 5 n)

= (40ln – 12lm + 8l²) – (3l² – 12lm + 15ln)

= 40ln – 12lm + 8l² – 3l² +12lm -15 ln

= 25 ln + 5l²

d) 4c ( – a + b + c ) – (3a (a + b + c ) – 2 b (a – b + c))

= (-4ac + 4bc + 4c²) – (3a² + 3ab + 3ac – ( 2ab – 2b² + 2bc ))

=-4ac + 4bc + 4c² – (3a² + 3ab + 3ac – 2ab + 2b² – 2bc)

= -4ac + 4bc + 4c² – 3a² – 3ab – 3ac +2ab – 2b² + 2bc

= -7ac + 6bc + 4c² – 3a² – ab – 2b²

15. Multiply the binomials.

(i) (2x + 5) and (4x – 3)

(ii) (y – 8) and (3y – 4)

(iii) (2.5l – 0.5m) and (2.5l + 0.5m)

(iv) (a + 3b) and (x + 5)

(v) (2pq + 3q²) and (3pq – 2q²)

(vi) (3/4 a² + 3b²) and 4( a² – 2/3 b²)

Solution:

(i) (2x + 5)(4x – 3)

= 2x x 4x – 2x x 3 + 5 x 4x – 5 x 3

= 8x² – 6x + 20x -15

= 8x² + 14x -15

(ii) ( y – 8)(3y – 4)

= y x 3y – 4y – 8 x 3y + 32

= 3y² – 4y – 24y + 32

= 3y² – 28y + 32

(iii) (2.5l – 0.5m)(2.5l + 0.5m)

= 2.5l x 2.5 l + 2.5l x 0.5m – 0.5m x 2.5l – 0.5m x 0.5m

= 6.25l² + 1.25 lm – 1.25 lm – 0.25 m²

= 6.25l²– 0.25 m²

(iv) (a + 3b) (x + 5)

= ax + 5a + 3bx + 15b

(v) (2pq + 3q²) (3pq – 2q²)

= 2pq x 3pq – 2pq x 2q² + 3q² x 3pq – 3q² x 2q²

= 6p²q² – 4pq³ + 9pq³ – 6q⁴

= 6p²q² + 5pq³ – 6q⁴

(vi) (3/4 a² + 3b² ) and 4( a² – 2/3 b² )

= (3/4 a² + 3b² ) x 4( a² – 2/3 b² )

= (3/4 a² + 3b² ) x (4a² – 8/3 b² )

= 3/4 a² x (4a² – 8/3 b² ) + 3b² x (4a² – 8/3 b² )

= 3/4 a² x 4a² -3/4 a² x 8/3 b² + 3b² x 4a² – 3b² x 8/3 b²

= 3a⁴ – 2a² b² + 12 a²  b² – 8b⁴

= 3a⁴ + 10a²  b² – 8b⁴

16. Find the product.

(i) (5 – 2x) (3 + x)

(ii) (x + 7y) (7x – y)

(iii) (a²+ b) (a + b²)

(iv) (p² – q²) (2p + q)

Solution:

(i) (5 – 2x) (3 + x)

= 5 (3 + x) – 2x (3 + x)

=15 + 5x – 6x – 2x²

= 15 – x -2 x ²

(ii) (x + 7y) (7x – y)

= x(7x-y) + 7y ( 7x-y)

=7x² – xy + 49xy – 7y²

= 7x² – 7y² + 48xy

(iii) (a²+ b) (a + b²)

= a²  (a + b²) + b(a + b²)

= a³ + a²b² + ab + b³

= a³ + b³ + a²b² + ab

(iv) (p²– q²) (2p + q)

= p² (2p + q) – q² (2p + q)

=2p³ + p²q – 2pq² – q³

= 2p³ – q³ + p²q – 2pq²

Answer 4

17. Simplify.

(i) (x²– 5) (x + 5) + 25

(ii) (a²+ 5) (b³+ 3) + 5

(iii)(t + s²)(t² – s)

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

(v) (x + y)(2x + y) + (x + 2y)(x – y)

(vi) (x + y)(x²– xy + y²)

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

(viii) (a + b + c)(a + b – c)

Solution:

(i) (x²– 5) (x + 5) + 25

= x³ + 5x² – 5x – 25 + 25

= x³ + 5x² – 5x

(ii) (a²+ 5) (b³+ 3) + 5

= a²b³ + 3a² + 5b³ + 15 + 5

= a²b³ + 5b³ + 3a² + 20

(iii) (t + s²)(t² – s)

= t (t² – s) + s²(t² – s)

= t³ – st + s²t² – s³

= t³ – s³ – st + s²t²

(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

= (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)

=(ac – ad + bc – bd) + (ac + ad – bc – bd) + (2ac + 2bd)

= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd

= 4ac

(v) (x + y)(2x + y) + (x + 2y)(x – y)

= 2x² + xy + 2xy + y² + x² – xy + 2xy – 2y²

= 3x² + 4xy – y²

(vi) (x + y)(x²– xy + y²)

= x³ – x²y + xy² + x²y – xy² + y³

= x³ + y³

(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y

= 2.25x² + 6xy + 4.5x – 6xy – 16y² – 12y – 4.5x + 12y = 2.25x² – 16y²

(viii) (a + b + c)(a + b – c)

= a² + ab – ac + ab + b² – bc + ac + bc – c²

= a² + b² – c² + 2ab  

18. Use a suitable identity to get each of the following products.

(i) (x + 3) (x + 3)

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

(iii) (2a – 7) (2a – 7)

(iv) (3a – 1/2)(3a – 1/2)

(v) (1.1m – 0.4) (1.1m + 0.4)

(vi) (a²+ b²) (- a²+ b²)

(vii) (6x – 7) (6x + 7)

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

(ix) (1/2x + 3/4y) (1/2x + 3/4y)

(x) (7a – 9b) (7a – 9b)

Solution:

(i) (x + 3) (x + 3) = (x + 3)²

= x² + 6x + 9

Using (a+b) ² = a² + b² + 2ab

(ii) (2y + 5) (2y + 5) = (2y + 5)²

= 4y² + 20y + 25

Using (a+b) ² = a² + b² + 2ab

(iii) (2a – 7) (2a – 7) = (2a – 7)²

= 4a² – 28a + 49

Using (a-b) ² = a² + b² – 2ab

(iv) (3a – 1/2)(3a – 1/2) = (3a – 1/2)²

=  9a² -3a+(1/4)

Using (a-b) ²  = a² + b² – 2ab

(v)   (1.1m – 0.4) (1.1m + 0.4)

= 1.21m² – 0.16

Using (a – b)(a + b) = a² – b²

(vi) (a²+ b²) (– a²+ b²)

= (b² + a² ) (b² – a²)

= -a⁴ + b⁴

Using (a – b)(a + b) = a² – b²

(vii) (6x – 7) (6x + 7)

= 36x² – 49

Using (a – b)(a + b) = a² – b²

(viii) (– a + c) (– a + c) = (– a + c)²

= c² + a² – 2ac

Using (a-b) ² = a² + b² – 2ab

= (x²/4) + (9y²/16) + (3xy/4)

Using (a+b) ² = a² + b² + 2ab

(x) (7a – 9b) (7a – 9b) = (7a – 9b)²

= 49a² – 126ab + 81b²

Using (a-b) ² = a² + b² – 2ab

19. Use the identity (x + a) (x + b) = x² + (a + b) x + ab to find the following products.

(i) (x + 3) (x + 7)

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

(iii) (4x – 5) (4x – 1)

(iv) (4x + 5) (4x – 1)

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

(vi) (2a² + 9) (2a² + 5)

(vii) (xyz – 4) (xyz – 2)

Solution:

(i) (x + 3) (x + 7)

= x² + (3+7)x + 21

= x² + 10x + 21

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

= 16x² + 4x + 20x + 5

= 16x² + 24x + 5

(iii) (4x – 5) (4x – 1)

= 16x² – 4x – 20x + 5

= 16x² – 24x + 5

(iv) (4x + 5) (4x – 1)

= 16x² + (5-1)4x – 5

= 16x² +16x – 5

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

= 4x² + (5y + 3y)2x + 15y²

= 4x² + 16xy + 15y²

(vi) (2a²+ 9) (2a²+ 5)

= 4a⁴ + (9+5)2a² + 45

= 4a⁴ + 28a² + 45

(vii) (xyz – 4) (xyz – 2)

= x²y²z² + (-4 -2)xyz + 8

= x²y²z² – 6xyz + 8

Answer 5

20. Find the following squares by using the identities.

(i) (b – 7)²

(ii) (xy + 3z)²

(iii) (6x² – 5y)²

(iv) [(2m/3) + (3n/2)]²

(v) (0.4p – 0.5q)²

(vi) (2xy + 5y)²

Solution:

Using identities:

(a – b) ² = a² + b² – 2ab (a + b) ² = a² + b² + 2ab

(i) (b – 7)²= b² – 14b + 49

(ii) (xy + 3z)² = x²y² + 6xyz + 9z²

(iii) (6x² – 5y)^{2 =} 36x⁴ – 60x²y + 25y²

(iv) [(2m/3}) + (3n/2)]² = (4m²/9) +(9n²/4) + 2mn

(v) (0.4p – 0.5q)² = 0.16p² – 0.4pq + 0.25q²

(vi) (2xy + 5y)² = 4x²y² + 20xy² + 25y²

21. Simplify.

(i) (a² – b²)²

(ii) (2x + 5)²  – (2x – 5)²

(iii) (7m – 8n)² + (7m + 8n)²

(iv) (4m + 5n)² + (5m + 4n)²

(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²

(vi) (ab + bc)²– 2ab²c

(vii) (m² – n²m)² + 2m³n²

Solution:

(i) (a²– b²)²

= a⁴ + b⁴ – 2a²b²

(ii) (2x + 5)²  – (2x – 5)²

= 4x² + 20x + 25 – (4x² – 20x + 25)

= 4x² + 20x + 25 – 4x² + 20x – 25

= 40x

(iii) (7m – 8n)² + (7m + 8n)²

= 49m² – 112mn + 64n² + 49m² + 112mn + 64n²

= 98m² + 128n²

(iv) (4m + 5n)² + (5m + 4n)²

= 16m² + 40mn + 25n² + 25m² + 40mn + 16n²

= 41m² + 80mn + 41n²

(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²

= 6.25p² – 7.5pq + 2.25q² – 2.25p² + 7.5pq – 6.25q²

= 4p² – 4q²

(vi) (ab + bc)²– 2ab²c

= a²b² + 2ab²c + b²c² – 2ab²c

= a²b² + b²c²

(vii) (m² – n²m)² + 2m³n²

= m⁴ – 2m³n² + m²n⁴ + 2m³n²

= m⁴ + m²n⁴

22. Show that.

(i) (3x + 7)² – 84x = (3x – 7)²

(ii) (9p – 5q)²+ 180pq = (9p + 5q)²

(iii) (4/3m – 3/4n)² + 2mn = 16/9 m² + 9/16 n²

(iv) (4pq + 3q)²– (4pq – 3q)² = 48pq²

(v) (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

Solution:

(i) LHS = (3x + 7)² – 84x

= 9x² + 42x + 49 – 84x

= 9x² – 42x + 49

= RHS

LHS = RHS

(ii)  LHS = (9p – 5q)²+ 180pq

= 81p² – 90pq + 25q² + 180pq

= 81p² + 90pq + 25q²

RHS = (9p + 5q)²

= 81p² + 90pq + 25q²

LHS = RHS

LHS = RHS

(iv)  LHS = (4pq + 3q)²– (4pq – 3q)²

= 16p²q² + 24pq² + 9q² – 16p²q² + 24pq² – 9q²

= 48pq²

= RHS

LHS = RHS

(v) LHS = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

= a² – b² + b² – c² + c² – a²

= 0

= RHS

LHS = RHS

Answer 6

23. Using identities, evaluate.

(i) 71²

(ii) 99²

(iii) 102²

(iv) 998²

(v) 5.2²

(vi) 297 x 303

(vii) 78 x 82

(viii) 8.9²

(ix) 10.5 x 9.5

Solution:

(i) 71²

= (70+1)²

= 70² + 140 + 1²

= 4900 + 140 +1

= 5041

(ii) 99²

= (100 -1)²

= 100² – 200 + 1²

= 10000 – 200 + 1

= 9801

(iii) 102²

= (100 + 2)²

= 100² + 400 + 2²

= 10000 + 400 + 4 = 10404

(iv) 998²

= (1000 – 2)²

= 1000² – 4000 + 2²

= 1000000 – 4000 + 4

= 996004

(v) 5.2²

= (5 + 0.2)²

= 5² + 2 + 0.2²

= 25 + 2 + 0.04 = 27.04

(vi) 297 x 303

= (300 – 3 )(300 + 3)

= 300² – 3²

= 90000 – 9

= 89991

(vii) 78 x 82

= (80 – 2)(80 + 2)

= 80² – 2²

= 6400 – 4

= 6396

(viii) 8.9²

= (9 – 0.1)²

= 9² – 1.8 + 0.1²

= 81 – 1.8 + 0.01

= 79.21

(ix) 10.5 x 9.5

= (10 + 0.5)(10 – 0.5)

= 10² – 0.5²

= 100 – 0.25

= 99.75

24. Using a² – b² = (a + b) (a – b), find

(i) 51² – 49²

(ii) (1.02)² – (0.98)²

(iii) 153² – 147²

(iv) 12.1² – 7.9²

Solution:

(i) 51²– 49²

= (51 + 49)(51 – 49) = 100 x 2 = 200

(ii) (1.02)²– (0.98)²

= (1.02 + 0.98)(1.02 – 0.98) = 2 x 0.04 = 0.08

(iii) 153² – 147²

= (153 + 147)(153 – 147) = 300 x 6 = 1800

(iv) 12.1² – 7.9²

= (12.1 + 7.9)(12.1 – 7.9) = 20 x 4.2= 84

25. Using (x + a) (x + b) = x² + (a + b) x + ab, find

(i) 103 x 104

(ii) 5.1 x 5.2

(iii) 103 x 98

(iv) 9.7 x 9.8

Solution:

(i) 103 x 104

= (100 + 3)(100 + 4)

= 100² + (3 + 4)100 + 12

= 10000 + 700 + 12

= 10712

(ii) 5.1 x 5.2

= (5 + 0.1)(5 + 0.2)

= 5² + (0.1 + 0.2)5 + 0.1 x 0.2

= 25 + 1.5 + 0.02

= 26.52.

(iii) 103 x 98

= (100 + 3)(100 – 2)

= 100² + (3-2)100 – 6

= 10000 + 100 – 6

= 10094

(iv) 9.7 x 9.8

= (9 + 0.7 )(9 + 0.8)

= 9² + (0.7 + 0.8)9 + 0.56

= 81 + 13.5 + 0.56

= 95.06