NCERT Solutions for Class 8 Maths Chapter 9: Algebraic Expressions and Identities
Orchid The International School presents comprehensive NCERT Solutions for Class 8 Maths Chapter 9 on Algebraic Expressions and Identities. Crafted by seasoned educators, these solutions offer a meticulous breakdown of each question within the NCERT textbook, aiding students in both homework assignments and exam preparation.
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Students can access the NCERT Solutions for Class 8 Maths Chapter 9: Algebraic Expressions and Identities. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 9.1
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
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
(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
(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- 2pq^2 – 5pq^2 + 5 p^2 q – 4p^2 q
= 28 + 5p – 18q + 8pq – 7pq2 + p2q
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
|
Sl. No. |
Expression |
Term |
Coefficient |
|
i) |
5xyz2 – 3zy |
Term: 5xyz2 Term: -3zy |
5 -3 |
|
ii) |
1 + x + x2 |
Term: 1 Term: x Term: x2 |
1 1 1 |
|
iii) |
4x2y2 – 4x2y2z2 + z2 |
Term: 4x2y2 Term: -4 x2y2z2 Term : z2 |
4 -4 1 |
|
iv) |
3 – pq + qr – p |
Term : 3 -pq qr -p |
3 -1 1 -1 |
|
v) |
(x/2) + (y/2) – xy |
Term : x/2 Y/2 -xy |
½ 1/2 -1 |
|
vi) |
0.3a – 0.6ab + 0.5b |
Term : 0.3a -0.6ab 0.5b |
0.3 -0.6 0.5 |
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
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 the 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 the listed three categories |
|
pqr |
one term |
Monomial |
|
p2q + pq2 |
two terms |
Binomial |
|
2p + 2q |
two terms |
Binomial |
|
7 + y + 5x |
three terms |
Trinomial |
Exercise 9.2
Find the product of the following pairs of monomials.
(i) 4, 7p
(ii) – 4p, 7p
(iii) – 4p, 7pq
(iv) 4p3, – 3p
(v) 4p, 0
(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
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)
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
Complete the following table of products:
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
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
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
(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
Exercise 9.3
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
(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 )
Complete the table.
|
|
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 |
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)
i) a2 x (2a22) x (4a26)
= (2 × 4) ( a2 × a22 × a26 )
= 8 × a2 + 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
(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.
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
(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 )
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
Exercise 9.4
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)
(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
Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a2+ b) (a + b2)
(iv) (p2 – q2) (2p + q)
(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
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)
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
Exercise 9.5
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)
(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
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)
(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
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
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
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
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
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
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
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
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
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
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
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
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
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Frequently Asked Questions
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