NCERT Solutions for Class 8 Math Chapter 9 – Algebraic Expressions and Identities

Page No 140:

Question 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 + qrrp

(v)

(vi) 0.3a − 0.6ab + 0.5b

Answer:

The terms and the respective coefficients of the given expressions are as follows.

Terms

Coefficients

(i)

5xyz2

− 3zy

5

− 3

(ii)

1

x

x2

1

1

1

(iii)

4x2y2

− 4x2y2z2

z2

4

− 4

1

(iv)

3

pq

qr

rp

3

−1

1

−1

(v)

xy

− 1

(vi)

0.3a

− 0.6ab

0.5b

0.3

− 0.6

0.5

Question 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

Answer:

The given expressions are classified as

Monomials: 1000, pqr

Binomials: x + y, 2y − 3y2, 4z − 15z2, p2q + pq2, 2p + 2q

Trinomials: 7 + y + 5x, 2y − 3y2 + 4y3, 5x − 4y + 3xy

Polynomials that do not fit in any of these categories are

x + x2 + x3 + x4, ab + bc + cd + da

Question 3:

Add the following.

(i) abbc, bcca, caab

(ii) ab + ab, bc + bc, ca + ac

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

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

Answer:

The given expressions written in separate rows, with like terms one below the other and then the addition of these expressions are as follows.

(i)

Thus, the sum of the given expressions is 0.

(ii)

Thus, the sum of the given expressions is ab + bc + ac.

(iii)

Thus, the sum of the given expressions is −p2q2 + 4pq + 9.

(iv)

Thus, the sum of the given expressions is 2(l2 + m2 + n2 + lm + mn + nl).

Question 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

Answer:

The given expressions in separate rows, with like terms one below the other and then the subtraction of these expressions is as follows.

(a)

(b)

(c)

Page No 143:

Question 1:

Find the product of the following pairs of monomials.

(i) 4, 7p (ii) − 4p, 7p (iii) − 4p, 7pq

(iv) 4p3, − 3p (v) 4p, 0

Answer:

The product will be as follows.

(i) 4 × 7p = 4 × 7 × p = 28p

(ii) − 4p × 7p = − 4 × p × 7 × p = (− 4 × 7) × (p × p) = − 28 p2

(iii) − 4p × 7pq = − 4 × p × 7 × p × q = (− 4 × 7) × (p × p × q) = − 28p2q

(iv) 4p3 × − 3p = 4 × (− 3) × p × p × p × p = − 12 p4

(v) 4p × 0 = 4 × p × 0 = 0

Question 2:

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)

Answer:

We know that,

Area of rectangle = Length × Breadth

Area of 1st rectangle = p × q = pq

Area of 2nd rectangle = 10m × 5n = 10 × 5 × m × n = 50 mn

Area of 3rd rectangle = 20x2 × 5y2 = 20 × 5 × x2 × y2 = 100 x2y2

Area of 4th rectangle = 4x × 3x2 = 4 × 3 × x × x2 = 12x3

Area of 5th rectangle = 3mn × 4np = 3 × 4 × m × n × n × p = 12mn2p

Page No 144:

Question 3:

Complete the table of products.

2x

− 5y

3x2

− 4xy

7x2y

− 9x2y2

2x

4x2

− 5y

− 15x2y

3x2

− 4xy

7x2y

− 9x2y2

Answer:

The table can be completed as follows.

2x

− 5y

3x2

− 4xy

7x2y

− 9x2y2

2x

4x2

− 10xy

6x3

− 8x2y

14x3y

− 18x3y2

− 5y

− 10xy

25 y2

− 15x2y

20xy2

− 35x2y2

45x2y3

3x2

6x3

− 15x2y

9x4

− 12x3y

21x4y

− 27x4y2

− 4xy

− 8x2y

20xy2

− 12x3y

16x2y2

− 28x3y2

36x3y3

7x2y

14x3y

− 35x2y2

21x4y

− 28x3y2

49x4y2

− 63x4y3

− 9x2y2

− 18x3y2

45 x2y3

− 27x4y2

36x3y3

− 63x4y3

81x4y4

Question 4:

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

Answer:

We know that,

Volume = Length × Breadth × Height

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

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

(iii) Volume = xy × 2x2y × 2xy2 = 2 × 2 × xy ×x2y × xy2 = 4x4y4

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

Question 5:

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

Answer:

(i) xy × yz × zx = x2y2z2

(ii) a × (− a2) × a3 = − a6

(iii) 2 × 4y × 8y2 × 16y3 = 2 × 4 × 8 × 16 × y × y2 × y3 = 1024 y6

(iv) a × 2b × 3c × 6abc = 2 × 3 × 6 × a × b × c × abc = 36a2b2c2

(v) m × (− mn) × mnp = − m3n2p

Page No 146:

Question 1:

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

(i) 4p, q + r (ii) ab, ab (iii) a + b, 7a2b2

(iv) a2 − 9, 4a (v) pq + qr + rp, 0

Answer:

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

(ii) (ab) × (ab) = (ab × a) + [ab × (− b)] = a2bab2

(iii) (a + b) × (7a2 b2) = (a × 7a2b2) + (b × 7a2b2) = 7a3b2 + 7a2b3

(iv) (a2 − 9) × (4a) = (a2 × 4a) + (− 9) × (4a) = 4a3 − 36a

(v) (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

5 xy

(iii)

p

6p2 − 7p + 5

(iv)

4p2q2

p2 q2

(v)

a + b + c

abc

Answer:

The table can be completed as follows.

First expressionSecond Expression

Product

(i)

a

b + c + d

ab + ac + ad

(ii)

x + y − 5

5 xy

5x2y + 5xy2 − 25xy

(iii)

p

6p2 − 7p + 5

6p3 − 7p2 + 5p

(iv)

4p2q2

p2 q2

4p4q2 − 4p2q4

(v)

a + b + c

abc

a2bc + ab2c + abc2

Question 3:

Find the product.

(i) (a2) × (2a22) × (4a26)

(ii)

(iii)

(iv) x × x2 × x3 × x4

Answer:

(i) (a2) × (2a22) × (4a26) = 2 × 4 ×a2 × a22 × a26 = 8a50

(ii)

(iii)

(iv) x × x2 × x3 × x4 = x10

Question 4:

(a) Simplify 3x (4x −5) + 3 and find its values for (i) x = 3, (ii) .

(b) a (a2 + a + 1) + 5 and find its values for (i) a = 0, (ii) a = 1, (iii) a = − 1.

Answer:

(a) 3x (4x − 5) + 3 = 12x2 − 15x + 3

(i) For x = 3, 12x2 − 15x + 3 = 12 (3)2 − 15(3) + 3

= 108 − 45 + 3

= 66

(ii) For

(b)a (a2 + a + 1) + 5 = a3 + a2 + a + 5

(i) For a = 0, a3 + a2 + a + 5 = 0 + 0 + 0 + 5 = 5

(ii) For a = 1, a3 + a2 + a + 5 = (1)3 + (1)2 + 1 + 5

= 1 + 1 + 1 + 5 = 8

(iii) For a = −1, a3 + a2 + a + 5 = (−1)3 + (−1)2 + (−1) + 5

= − 1 + 1 − 1 + 5 = 4

Question 5:

(a) Add: p (pq), q (q ­­­r) and r (r ­− p)

(b) Add: 2x (zxy) and 2y (zyx)

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

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

Answer:

(a) First expression = p (pq) = p2pq

Second expression = q (q ­­­r) = q2qr

Third expression = r (r ­− p) = r2pr

Adding the three expressions, we obtain

Therefore, the sum of the given expressions is p2 + q2 + r2pqqrrp.

(b) First expression = 2x (zxy) = 2xz − 2x2 − 2xy

Second expression = 2y (zyx) = 2yz − 2y2 − 2yx

Adding the two expressions, we obtain

Therefore, the sum of the given expressions is − 2x2 − 2y2 − 4xy + 2yz + 2zx.

(c) 3l (l − 4m + 5n) = 3l2 − 12lm + 15ln

4l (10n − 3m + 2l) = 40ln − 12lm + 8l2

Subtracting these expressions, we obtain

Therefore, the result is 5l2 + 25ln.

(d) 3a (a + b + c) − 2b (ab + c) = 3a2 +3ab + 3ac − 2ba + 2b2 − 2bc

= 3a2 + 2b2 + ab + 3ac − 2bc

4c (− a + b + c) = − 4ac + 4bc + 4c2

Subtracting these expressions, we obtain

Therefore, the result is −3a2 −2b2 + 4c2ab + 6bc − 7ac.

Page No 148:

Question 1:

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)

Answer:

(i) (2x + 5) × (4x − 3) = 2x × (4x − 3) + 5 × (4x − 3)

= 8x2 − 6x + 20x − 15

= 8x2 + 14x −15 (By adding like terms)

(ii) (y − 8) × (3y − 4) = y × (3y − 4) − 8 × (3y − 4)

= 3y2 − 4y − 24y + 32

= 3y2 − 28y + 32 (By adding like terms)

(iii) (2.5l − 0.5m) × (2.5l + 0.5m) = 2.5l × (2.5l + 0.5m) − 0.5m (2.5l + 0.5m)

= 6.25l2 + 1.25lm − 1.25lm − 0.25m2

= 6.25l2 − 0.25m2

(iv) (a + 3b) × (x + 5) = a × (x + 5) + 3b × (x + 5)

= ax + 5a + 3bx + 15b

(v) (2pq + 3q2) × (3pq − 2q2) = 2pq × (3pq − 2q2) + 3q2 × (3pq − 2q2)

= 6p2q2 − 4pq3 + 9pq3 − 6q4

= 6p2q2 + 5pq3 − 6q4

(vi)

Question 2:

Find the product.

(i) (5 − 2x) (3 + x) (ii) (x + 7y) (7xy)

(iii) (a2 + b) (a + b2) (iv) (p2q2) (2p + q)

Answer:

(i) (5 − 2x) (3 + x) = 5 (3 + x) − 2x (3 + x)

= 15 + 5x − 6x − 2x2

= 15 − x − 2x2

(ii) (x + 7y) (7xy) = x (7xy) + 7y (7xy)

= 7x2xy + 49xy − 7y2

= 7x2 + 48xy − 7y2

(iii) (a2 + b) (a + b2) = a2 (a + b2) + b (a + b2)

= a3 + a2b2 + ab + b3

(iv) (p2q2) (2p + q) = p2 (2p + q) − q2 (2p + q)

= 2p3 + p2q − 2pq2q3

Question 3:

Simplify.

(i) (x2 − 5) (x + 5) + 25

(ii) (a2 + 5) (b3 + 3) + 5

(iii) (t + s2) (t2 − s)

(iv) (a + b) (cd) + (ab) (c + d) + 2 (ac + bd)

(v) (x + y) (2x + y) + (x + 2y) (xy)

(vi) (x + y) (x2xy + y2)

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

(viii) (a + b + c) (a + bc)

Answer:

(i) (x2 − 5) (x + 5) + 25

= x2 (x + 5) − 5 (x + 5) + 25

= x3 + 5x2 − 5x − 25 + 25

= x3 + 5x2 − 5x

(ii) (a2 + 5) (b3 + 3) + 5

= a2 (b3 + 3) + 5 (b3 + 3) + 5

= a2b3 + 3a2 + 5b3 + 15 + 5

= a2b3 + 3a2 + 5b3 + 20

(iii) (t + s2) (t2 − s)

= t (t2 s) + s2 (t2 − s)

= t3st + s2t2 s3

(iv) (a + b) (cd) + (ab) (c + d) + 2 (ac + bd)

= a (cd) + b (cd) + a (c + d) − b (c + d) + 2 (ac + bd)

= acad + bcbd + ac + adbcbd + 2ac + 2bd

= (ac + ac + 2ac) + (adad) + (bcbc) + (2bdbdbd)

= 4ac

(v) (x + y) (2x + y) + (x + 2y) (xy)

= x (2x + y) + y (2x + y) + x (xy) + 2y (xy)

= 2x2 + xy + 2xy + y2 + x2xy + 2xy − 2y2

= (2x2 + x2) + (y2 − 2y2) + (xy + 2xyxy + 2xy)

= 3x2y2 + 4xy

(vi) (x + y) (x2xy + y2)

= x (x2xy + y2) + y (x2xy + y2)

= x3x2y + xy2 + x2yxy2 + y3

= x3 + y3 + (xy2xy2) + (x2yx2y)

= x3 + y3

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

= 1.5x (1.5x + 4y + 3) − 4y (1.5x + 4y + 3) − 4.5x + 12y

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

= 2.25 x2 + (6xy − 6xy) + (4.5x − 4.5x) − 16y2 + (12y − 12y)

= 2.25x2 − 16y2

(viii) (a + b + c) (a + bc)

= a (a + bc) + b (a + bc) + c (a + bc)

= a2 + abac + ab + b2bc + ca + bcc2

= a2 + b2c2 + (ab + ab) + (bcbc) + (caca)

= a2 + b2c2 + 2ab

Page No 151:

Question 1:

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)

(v) (1.1m − 0.4) (1.1 m + 0.4) (vi) (a2 + b2) (− a2 + b2)

(vii) (6x − 7) (6x + 7) (viii) (− a + c) (− a + c)

(ix) (x) (7a − 9b) (7a − 9b)

Answer:

The products will be as follows.

(i) (x + 3) (x + 3) = (x + 3)2

= (x)2 + 2(x) (3) + (3)2 [(a + b)2 = a2 + 2ab + b2]

= x2 + 6x + 9

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

= (2y)2 + 2(2y) (5) + (5)2 [(a + b)2 = a2 + 2ab + b2]

= 4y2 + 20y + 25

(iii) (2a ­− 7) (2a − 7) = (2a − 7)2

= (2a)2 − 2(2a) (7) + (7)2 [(ab)2 = a2 − 2ab + b2]

= 4a2 − 28a + 49

(iv)

[(ab)2 = a2 − 2ab + b2]

(v) (1.1m − 0.4) (1.1 m + 0.4)

= (1.1m)2 − (0.4)2 [(a + b) (ab) = a2b2]

= 1.21m2 − 0.16

(vi) (a2 + b2) (− a2 + b2) = (b2 + a2) (b2a2)

= (b2)2 − (a2)2 [(a + b) (ab) = a2b2]

= b4a4

(vii) (6x − 7) (6x + 7) = (6x)2 − (7)2 [(a + b) (ab) = a2b2]

= 36x2 − 49

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

= (− a)2 + 2(− a) (c) + (c)2 [(a + b)2 = a2 + 2ab + b2]

= a2 − 2ac + c2

(ix)

[(a + b)2 = a2 + 2ab + b2]

(x) (7a − 9b) (7a − 9b) = (7a − 9b)2

= (7a)2 − 2(7a)(9b) + (9b)2 [(a b)2 = a2 − 2ab + b2]

= 49a2 − 126ab + 81b2

Question 2:

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)

Answer:

The products will be as follows.

(i) (x + 3) (x + 7) = x2 + (3 + 7) x + (3) (7)

= x2 + 10x + 21

(ii) (4x + 5) (4x + 1) = (4x)2 + (5 + 1) (4x) + (5) (1)

= 16x2 + 24x + 5

(iii)

= 16x2 − 24x + 5

(iv)

= 16x2 + 16x − 5

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

= 4x2 + 16xy + 15y2

(vi) (2a2 +9) (2a2 + 5) = (2a2)2 + (9 + 5) (2a2) + (9) (5)

= 4a4 + 28a2 + 45

(vii) (xyz − 4) (xyz − 2)

=

= x2y2z2 − 6xyz + 8

Question 3:

Find the following squares by suing the identities.

(i) (b − 7)2 (ii) (xy + 3z)2 (iii) (6x2 − 5y)2

(iv) (v) (0.4p − 0.5q)2 (vi) (2xy + 5y)2

Answer:

(i) (b − 7)2 = (b)2 − 2(b) (7) + (7)2 [(ab)2 = a2 − 2ab + b2]

= b2 − 14b + 49

(ii) (xy + 3z)2 = (xy)2 + 2(xy) (3z) + (3z)2 [(a + b)2 = a2 + 2ab + b2]

= x2y2 + 6xyz + 9z2

(iii) (6x2 − 5y)2 = (6x2)2 − 2(6x2) (5y) + (5y)2 [(ab)2 = a2 − 2ab + b2]

= 36x4 − 60x2y + 25y2

(iv) [(a + b)2 = a2 + 2ab + b2]

(v) (0.4p − 0.5q)2 = (0.4p)2 − 2 (0.4p) (0.5q) + (0.5q)2

[(ab)2 = a2 − 2ab + b2]

= 0.16p2 − 0.4pq + 0.25q2

(vi) (2xy + 5y)2 = (2xy)2 + 2(2xy) (5y) + (5y)2

[(a + b)2 = a2 + 2ab + b2]

= 4x2y2 + 20xy2 + 25y2

Question 4:

Simplify.

(i) (a2b2)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) (m2n2m)2 + 2m3n2

Answer:

(i) (a2b2)2 = (a2)2 − 2(a2) (b2) + (b2)2 [(ab)2 = a2 − 2ab + b2 ]

= a4 − 2a2b2 + b4

(ii) (2x +5)2 − (2x − 5)2 = (2x)2 + 2(2x) (5) + (5)2 − [(2x)2 − 2(2x) (5) + (5)2]

[(ab)2 = a2 − 2ab + b2]

[(a + b)2 = a2 + 2ab + b2]

= 4x2 + 20x + 25 − [4x2 − 20x + 25]

= 4x2 + 20x + 25 − 4x2 + 20x − 25 = 40x

(iii) (7m − 8n)2 + (7m + 8n)2

= (7m)2 − 2(7m) (8n) + (8n)2 + (7m)2 + 2(7m) (8n) + (8n)2

[(ab)2 = a2 − 2ab + b2 and (a + b)2 = a2 + 2ab + b2]

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

= 98m2 + 128n2

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

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

[ (a + b)2 = a2 + 2ab + b2]

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

= 41m2 + 80mn + 41n2

(v) (2.5p − 1.5q)2 − (1.5p − 2.5q)2

= (2.5p)2 − 2(2.5p) (1.5q) + (1.5q)2 − [(1.5p)2 − 2(1.5p)(2.5q) + (2.5q)2]

[(ab)2 = a2 − 2ab + b2 ]

= 6.25p2 − 7.5pq + 2.25q2 − [2.25p2 − 7.5pq + 6.25q2]

= 6.25p2 − 7.5pq + 2.25q2 − 2.25p2 + 7.5pq − 6.25q2]

= 4p2 − 4q2

(vi) (ab + bc)2 − 2ab2c

= (ab)2 + 2(ab)(bc) + (bc)2 − 2ab2c [(a + b)2 = a2 + 2ab + b2 ]

= a2b2 + 2ab2c + b2c2 − 2ab2c

= a2b2 + b2c2

(vii) (m2n2m)2 + 2m3n2

= (m2)2 − 2(m2) (n2m) + (n2m)2 + 2m3n2 [(ab)2 = a2 − 2ab + b2 ]

= m4 − 2m3n2 + n4m2 + 2m3n2

= m4 + n4m2

Question 5:

Show that

(i) (3x + 7)2 − 84x = (3x − 7)2 (ii) (9p − 5q)2 + 180pq = (9p + 5q)2

(iii)

(iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2

(v) (ab) (a + b) + (bc) (b + c) + (ca) (c + a) = 0

Answer:

(i) L.H.S = (3x + 7)2 − 84x

= (3x)2 + 2(3x)(7) + (7)2 − 84x

= 9x2 + 42x + 49 − 84x

= 9x2 − 42x + 49

R.H.S = (3x − 7)2 = (3x)2 − 2(3x)(7) +(7)2

= 9x2 − 42x + 49

L.H.S = R.H.S

(ii) L.H.S = (9p − 5q)2 + 180pq

= (9p)2 − 2(9p)(5q) + (5q)2 − 180pq

= 81p2 − 90pq + 25q2 + 180pq

= 81p2 + 90pq + 25q2

R.H.S = (9p + 5q)2

= (9p)2 + 2(9p)(5q) + (5q)2

= 81p2 + 90pq + 25q2

L.H.S = R.H.S

(iii) L.H.S =

(iv) L.H.S = (4pq + 3q)2 − (4pq − 3q)2

= (4pq)2 + 2(4pq)(3q) + (3q)2 − [(4pq)2 − 2(4pq) (3q) + (3q)2]

= 16p2q2 + 24pq2 + 9q2 − [16p2q2 − 24pq2 + 9q2]

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

= 48pq2 = R.H.S

(v) L.H.S = (ab) (a + b) + (bc) (b + c) + (ca) (c + a)

= (a2b2) + (b2c2) + (c2a2) = 0 = R.H.S.

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Question 6:

Using identities, evaluate.

(i) 712 (ii) 992 (iii) 1022 (iv) 9982

(v) (5.2)2 (vi) 297 × 303 (vii) 78 × 82

(viii) 8.92 (ix) 1.05 × 9.5

Answer:

(i) 712 = (70 + 1)2

= (70)2 + 2(70) (1) + (1)2 [(a + b)2 = a2 + 2ab + b2 ]

= 4900 + 140 + 1 = 5041

(ii) 992 = (100 − 1)2

= (100)2 − 2(100) (1) + (1)2 [(ab)2 = a2 − 2ab + b2 ]

= 10000 − 200 + 1 = 9801

(iii) 1022 = (100 + 2)2

= (100)2 + 2(100)(2) + (2)2 [(a + b)2 = a2 + 2ab + b2 ]

= 10000 + 400 + 4 = 10404

(iv) 9982 = (1000 − 2)2

= (1000)2 − 2(1000)(2) + (2)2 [(ab)2 = a2 − 2ab + b2 ]

= 1000000 − 4000 + 4 = 996004

(v) (5.2)2 = (5.0 + 0.2)2

= (5.0)2 + 2(5.0) (0.2) + (0.2)2 [(a + b)2 = a2 + 2ab + b2 ]

= 25 + 2 + 0.04 = 27.04

(vi) 297 × 303 = (300 − 3) × (300 + 3)

= (300)2 − (3)2 [(a + b) (ab) = a2b2]

= 90000 − 9 = 89991

(vii) 78 × 82 = (80 − 2) (80 + 2)

= (80)2 − (2)2 [(a + b) (ab) = a2b2]

= 6400 − 4 = 6396

(viii) 8.92 = (9.0 − 0.1)2

= (9.0)2 − 2(9.0) (0.1) + (0.1)2 [(ab)2 = a2 − 2ab + b2 ]

= 81 − 1.8 + 0.01 = 79.21

(ix) 1.05 × 9.5 = 1.05 × 0.95 × 10

= (1 + 0.05) (1− 0.05) ×10

= [(1)2 − (0.05)2] × 10

= [1 − 0.0025] × 10 [(a + b) (ab) = a2b2]

= 0.9975 × 10 = 9.975

Question 7:

Using a2 b2 = (a + b) (ab), find

(i) 512 − 492 (ii) (1.02)2 − (0.98)2 (iii) 1532 − 1472

(iv) 12.12 − 7.92

Answer:

(i) 512 − 492 = (51 + 49) (51 − 49)

= (100) (2) = 200

(ii) (1.02)2 − (0.98)2 = (1.02 + 0.98) (1.02 ­− 0.98)

= (2) (0.04) = 0.08

(iii) 1532 − 1472 = (153 + 147) (153 − 147)

= (300) (6) = 1800

(iv) 12.12 − 7.92 = (12.1 + 7.9) (12.1 − 7.9)

= (20.0) (4.2) = 84

Question 8:

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

(i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8

Answer:

(i) 103 × 104 = (100 + 3) (100 + 4)

= (100)2 + (3 + 4) (100) + (3) (4)

= 10000 + 700 + 12 = 10712

(ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2)

= (5)2 + (0.1 + 0.2) (5) + (0.1) (0.2)

= 25 + 1.5 + 0.02 = 26.52

(iii) 103 × 98 = (100 + 3) (100 − 2)

= (100)2 + [3 + (− 2)] (100) + (3) (− 2)

= 10000 + 100 − 6

= 10094

(iv) 9.7 × 9.8 = (10 − 0.3) (10 − 0.2)

= (10)2 + [(− 0.3) + (− 0.2)] (10) + (− 0.3) (− 0.2)

= 100 + (− 0.5)10 + 0.06 = 100.06 − 5 = 95.06

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