Page 152 - ICSE Math 8

Ê q ˆ Ê q ˆ
1
Example 5: Find the product p - 5 - ˜ Á p + 5 + ˜ 1 .
Á
¯
¯ Ë
Ë
Ê q ˆ Ê q ˆ Ï Ê q ˆ ¸Ï Ê q ˆ ¸¸
1
1
Solution: Á p - 5 - ˜ Á p + 5 + ˜ = Ì p - Á Ë 5 + ˜ ˝Ì p + Á 5 + ˜ 1 ˝
1
¯
¯ Ë
Ë
¯
¯
Ë
˛
˛Ó
Ó
2
ˆ
2
2
2
= (p) – Á Ê q + 1 ˜ {Using (a – b)(a + b) = a – b }
Ë 5 ¯
Ï Ê qˆ 2 Ê qˆ ¸ 2 2 2
2
Ì
1 + ˝
= p – Á ˜ + Á ˜ () 1 2 {Using (a + b) = a + 2ab + b }
2
5
5
Ó Ë ¯ Ë ¯ ˛
Ê q 2 2 q ˆ q 2 2 q
2
= p – Á Ë 25 + 5 + 1 = p 2 - 25 - 5 -1
˜
¯
2
2
2
2
2
Example 6: Prove that 2x + 2y + 2z – 2xy – 2yz – 2zx = (x – y) + (y – z) + (z – x) 2
2
2
2
Solution: LHS = 2x + 2y + 2z – 2xy – 2yz – 2zx
2
2
2
2
2
2
= x + x + y + y + z + z – 2xy – 2yz – 2zx
2
2
2
2
2
2
= (x + y – 2xy) + (y + z – 2yz) + (z + x – 2zx)
2
2
= (x – y) + (y – z) + (z – x) 2
= RHS
We can also prove this by solving RHS.
2
Example 7: Using a suitable identity, evaluate: (a) (98) (b) (102) 2
2
Solution: (a) (98) = (100 – 2) 2
2
2
2
2
2
= (100) – 2(100)(2) + (2) {Using (a – b) = a – 2ab + b }
= 10,000 – 400 + 4 = 9,604
2
(b) (102) = (100 + 2) 2
2
2
2
2
2
= (100) + 2(100)(2) + (2) {Using (a + b) = a + 2ab + b }
= 10,000 + 400 + 4 = 10,404
2
2
Example 8: Simplify the following using (p + q)(p – q) = p – q :
2
(a) 102 ¥ 98 (b) 46 – 44 2
2
2
Solution: (a) 102 ¥ 98 = (100 + 2) ¥ (100 – 2) = (100) – (2) = 10,000 – 4 = 9,996
2
2
(b) 46 – 44 = (46 + 44)(46 – 44) = 90 ¥ 2 = 180
2
2
Example 9: Find the value of a if 5a = 30 – 25 .
2
2
2
2
Solution: 5a = 30 – 25 = (30 + 25)(30 – 25) {Using (p – q ) = (p + q)(p – q)}
5a 55 5¥
fi 5a = 55 ¥ 5 fi =
fi a = 55 5 5
EXERCISE 13.1
1. Evaluate the following. 2 2
Ê
yˆ
x
2
2
2 2
2
2
(a) (y + 3) (b) (3x – 4y) (c) a + b¯ ˆ 2 ˜ (d) (x y – xy ) (e) (2.3x + 3.5y) (f) Ê 3 - 2¯
˜
Á
Á
Ë
Ë
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