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 − 15z2ab + bc + cd + dapqrp2q + pq2, 2p + 2q

Answer:

The given expressions are classified as
Monomials: 1000, pqr
Binomials: x + y, 2y − 3y2, 4z − 15z2p2q + 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 + x4ab + bc + cd + da

Question 3:

Add the following.
(i) ab − bcbc − caca − ab
(ii) a − b + abb − c + bcc − a + ac
(iii) 2p2q2 − 3pq + 4, 5 + 7pq − 3p2q2
(iv) l2 + m2m2 + n2n2 + 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 + mnlm + 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.
(pq); (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, yzzx (ii) a, − a2a3 (iii) 2, 4y, 8y2, 16y3
(iv) a, 2b, 3c, 6abc (v) m, − mnmnp

Answer:

(i) xy × yz × zx = x2y2z2
(ii) a × (− a2) × a3 = − a6
(iii) 2 × 4× 8y2 × 16y= 2 × 4 × 8 × 16 × × y2 × y3 = 1024 y6
(iv) × 2b × 3c × 6abc = 2 × 3 × 6 × a × b × c × abc = 36a2b2c2
(v) × (− mn) × mnp = − m3n2p

Page No 146:

Question 1:

Carry out the multiplication of the expressions in each of the following pairs.
(i) 4pq + r (ii) aba − 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
xy
(iii)
p
6p− 7p + 5
(iv)
4p2q2
p− 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
xy
5x2y + 5xy2 − 25xy
(iii)
p
6p− 7p + 5
6p− 7p2 + 5p
(iv)
4p2q2
p− 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 + 2bab + 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
a2b+ 3a2 + 5b3 + 15 + 5
= a2b+ 3a2 + 5b3 + 20
(iii) (t + s2) (t2 − s)
t (t− s) + s2 (t2 − s)
t3 − st + s2t− 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)− (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 [(− b)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) (4− 1) (iv) (4x + 5) (4− 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)
x+ 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)(ii) (xy + 3z)(iii) (6x2 − 5y)2
(iv)  (v) (0.4p − 0.5q)(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]
= 0.16p− 0.4pq + 0.25q2
(vi) (2xy + 5y)2 = (2xy)2 + 2(2xy) (5y) + (5y)2
[(a + b)= a2 + 2ab + b2]
= 4x2y2 + 20xy2 + 25y2

Question 4:

Simplify.
(i) (a2 − b2)(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 − 2ab2(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(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)(ii) (9p − 5q)2 + 180pq = (9p + 5q)2
(iii) 
(iv) (4pq + 3q)− (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
= 9x− 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)− (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 a− 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 (a) (b) = x2 + (a + bx + 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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