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 + qr − rp

(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) 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

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 1^st rectangle = p × q = pq

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

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

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

Area of 5^th 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, a − b (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) × (a − b) = (ab × a) + [ab × (− b)] = a²b − ab²

(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 expression	Second 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 (p − q), q (q _­­­− r) and r (r ­− p)

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

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

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

Answer:

(a) First expression = p (p − q) = p² − pq

Second expression = q (q _­­­− r) = q² − qr

Third expression = r (r ­− p) = r² − pr

Adding the three expressions, we obtain

Therefore, the sum of the given expressions is p² + q² + r² − pq − qr − rp.

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

Second expression = 2y (z − y − x) = 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 (a − b + 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 + 4c² − ab + 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) (7x − y)

(iii) (a² + b) (a + b²) (iv) (p² − q²) (2p + q)

Answer:

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

= 15 + 5x − 6x − 2x²

= 15 − x − 2x²

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

= 7x² − xy + 49xy − 7y²

= 7x² + 48xy − 7y²

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

= a³ + a²b² + ab + b³

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

= 2p³ + p²q − 2pq² − q³

Question 3:

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)

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)

= t³ − st + s²t² − s³

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

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

= ac − ad + bc − bd + ac + ad − bc − bd + 2ac + 2bd

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

= 4ac

(v) (x + y) (2x + y) + (x + 2y) (x − y)

= x (2x + y) + y (2x + y) + x (x − y) + 2y (x − y)

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

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

= 3x² − y² + 4xy

(vi) (x + y) (x² − xy + y²)

= x (x² − xy + y²) + y (x² − xy + y²)

= x³ − x²y + xy² + x²y − xy² + y³

= x³ + y³ + (xy² − xy²) + (x²y − x²y)

= 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 + b − c)

= a (a + b − c) + b (a + b − c) + c (a + b − c)

= a² + ab − ac + ab + b² − bc + ca + bc − c²

= a² + b² − c² + (ab + ab) + (bc − bc) + (ca − ca)

= a² + b² − c² + 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 [(a − b)² = a² − 2ab + b²]

= 4a² − 28a + 49

(iv) 

 [(a − b)² = a² − 2ab + b²]

(v) (1.1m − 0.4) (1.1 m + 0.4)

= (1.1m)² − (0.4)² [(a + b) (a − b) = a² − b²]

= 1.21m² − 0.16

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

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

= b⁴ − a⁴

(vii) (6x − 7) (6x + 7) = (6x)² − (7)² [(a + b) (a − b) = a² − b²]

= 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 [(a − b)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 [(a − b)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

[(a − b)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) (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

Answer:

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

[(a − b)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

[(a − b)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]

[(a − b)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) (m² − n²m)² + 2m³n²

= (m²)² − 2(m²) (n²m) + (n²m)² + 2m³n² [(a − b)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) (a − b) (a + b) + (b − c) (b + c) + (c − a) (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 = (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a)

= (a² − b²) + (b² − c²) + (c² − a²) = 0 = R.H.S.

Page No 152:

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 [(a − b)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 [(a − b)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) (a − b) = a² − b²]

= 90000 − 9 = 89991

(vii) 78 × 82 = (80 − 2) (80 + 2)

= (80)² − (2)² [(a + b) (a − b) = a² − b²]

= 6400 − 4 = 6396

(viii) 8.9² = (9.0 − 0.1)²

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

= 0.9975 × 10 = 9.975

Question 7:

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

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

Read this too