Page 150 - ICSE Math 8

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2
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Area of (II) = (b) = (4) = 16 cm , Area of (IV) = (a – b) ¥ b = 6 ¥ 4 = 24 cm 2
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2
2
2
2
2
RHS of (1) = a – {(a – b) ¥ b + b + (a – b) ¥ b} = a – {ab – b + b + ab – b }
2
2
2
= a – (2ab – b ) = a – 2ab + b 2
Area of (RUSK) = 100 – (24 + 16 + 24) = 100 – 64 = 36 cm 2
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2
\ RHS = a – 2ab + b = 36 cm 2
\ LHS = RHS
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2
Hence, the identity (a – b) = a – 2ab + b is verified.
Paste along IT, TN, ON, OG and GK. Do not paste other portions of II, III and IV.

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Identity 3: (a + b) (a – b) = a – b 2
In other words,
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(Sum of the two terms) ¥ (Difference of the two terms) = (First term) – (Second term) 2
Proof: (a + b)(a – b) = a(a – b) + b(a – b) (Distributive property of multiplication over addition)
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2
= a – ab + ab – b (Commutative property ab = ba)
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= a – b 2
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\ (a + b)(a – b) = a – b 2

Example 1: Find the following squares by using the identities:
-6
Ê
2
2
n
(a) (6x – 5y) (b) Ê 2 m + 3 ˆ 2 (c) (0.4p – 0.5q) (d) Á Ê Ë 7 p 3 + q 3 ˆ -6 p 3 - q 3 ˆ ˜ ¯
˜ Á
˜
Á
¯ Ë 7
Ë
2 ¯
3
2
2
2
2
2
2
Solution: (a) (6x – 5y) = (6x) – 2(6x)(5y) + (5y) {Using (a – b) = a – 2ab + b }
2
= 36x – 60xy + 25y 2
Ê 2 3 ˆ 2 Ê 2 ˆ 2 Ê 2 ˆ Ê 3 ˆ Ê 3 ˆ 2
2
2
2
n
n
n +
(b) Á 3 m + 2 ¯ = Á Ë 3 ¯ ˜ + 2 Á 3 ¯ Ë 2 ¯ Á Ë 2 ¯ {Using (a + b) = a + 2ab + b }
m
m
˜ Á
˜
˜
˜
Ë
Ë
4 9
2
= m + 2mn + n 2
9 4
2
2
2
2
2
2
(c) (0.4p – 0.5q) = (0.4p) – 2(0.4p)(0.5q) + (0.5q) {Using (a – b) = a – 2ab + b }
2
= 0.16p – 0.4pq + 0.25q 2
Ê -6 ˆ -6 ˆ Ê -6 ˆ 2
Ê
2
2
3 2
(d) Á p 3 + q 3 ˜ Á p 3 - q 3 ˜ = Á p 3 ˜ – (q ) {Using (a + b)(a – b) = a – b }
¯
¯
¯ Ë 7
Ë 7
Ë 7
36 6 6
= 49 p −− q
1 1 1
4
2
Example 2: If x + = 2, find the values of: (a) x + (b) x +
x x 2 x 4
1 Ê ˆ 1 2
2
Solution: (a) x + = 2 fi x + ˜ = 2 (Squaring both sides)
Á
x Ë x¯
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