Chapter 6: Algebraic Expressions and Identities Exercise – 6.7
Question: 1
Find the following products:
(i) (x + 4) (x + 7)
(ii) (x – 11) (x + 4)
(iii) (x + 7) (x – 5)
(iv) (x – 3) (x – 2)
(v) (y2 – 4) (y2 – 3)
(vi) (x + 4/3)(x + 3/4)
(vii) (3x + 5) (3x + 11)
(viii) (2x2 – 3) (2x2 + 5)
(ix) (z2 + 2) (z2 – 3)
(x) (3x – 4y) (2x – 4y)
(xi) (3x2 – 4xy) (3x2 – 3xy)
(xii) (x + 1/5)(x + 5)
(xiii) (z + 3/4)(z + 4/3)
(xiv) (x2 + 4) (x2 + 9)
(xv) (y2 + 12) (y2 + 6)
(xvi) (y2 + 5/7) (y2 – 14/5)
(xvii) (p2 + 16)(p2 – 14)
Solution:
Question: 2
Evaluate the following:
(i) 102 x 106
(ii) 109 x 107
(iii) 35 x 37
(iv) 53 x 55
(v) 103 x 96
(vi) 34 x 36
(vii) 994 x 1006
Solution:
(i) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab
102 x 106
= (100 + 2) (100 + 6)
= 1002 + (2 + 6)100 + 2 x 6
=10000 + 800 +12 = 10812
(ii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab
109 x 107
= (100 + 9) (100 + 7)
= 1002 + (9 + 7)100 + 9 x 7
=10000 + 1600 + 63 = 11663
(iii) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab
35 x 37
= (30 + 5) (30 + 7)
= 302 + (5 + 7)30 + 5 x 7
= 900 + 360 + 35 = 1295
(iv) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab
53 x 55
= (50 + 3) (50 + 5)
= 502 + (3 + 5)50 + 3 x 5
= 2500 + 400 + 15 = 2915
(v) Here, we will use the identity (x + a) (x – b) = x2 + (a – b)x – ab
103 x 96
= (100 + 3) (100 – 4)
= 1002 + (3 – 4)100 – 3 x 4
= 10000 – 100 – 12 = 9888
(vi) Here, we will use the identity (x + a) (x + b) = x2 + (a + b)x + ab
34 x 36
= (30 + 4) (30 + 6)
= 302 + (4 + 6)30 + 4 x 6
= 900 + 300 + 24 = 1224
(vii) Here, we will use the identity (x – a) (x + b) = x2 + (b – a)x – ab
994 x 1006
= (1000 – 6) x (1000 + 6)
= 10002 + (6 – 6) x 1000 – 6 x 6
= 1000000 – 36 = 999964