Page 104 - Start Up Mathematics_8 (Non CCE)

Ê ˆ 1 2
fi 16 + x - x¯ ˜ = 20
Á
Ë
Ê x - x¯ ˆ 1 ˜ 2 = 20 – 16 = 4 = (2) 2
Á
Ë
fi x – 1 = ± 2
x
2
2
Example 27: If 2x + 3y = 10 and xy = 5, find the value of 4x + 9y .
Solution: 2x + 3y = 10
2
2
fi (2x + 3y) = (10) (Squaring both sides)
2
2
2
2
2
fi (2x) + 2(2x)(3y) + (3y) = 100 {Using (p + q) = p + 2pq + q }
2
2
fi 4x + 12xy + 9y = 100
2
2
fi (4x + 9y ) + 12xy = 100
2
2
fi (4x + 9y ) + 12(5) = 100 (Putting the value of xy)
2
2
fi 4x + 9y + 60 = 100
2
2
fi 4x + 9y = 100 – 60 = 40
Ê q ˆ Ê q ˆ
1
Example 28: Find the product p - 5 - ˜ Á p + 5 + ˜ 1 ¯
Á
¯ Ë
Ë
ˆ ¸ Ï
ˆ ¸
Ï
ˆ Ê
ˆ
Solution: Ê p - q - ˜ Á p + q + ˜ = Ì p - Á Ê Ë q + ˜ ˝ + Ì p + Á Ê Ë q + ˜ 1 ˝
1
1
1
Á
¯
¯
Ë
¯
¯ Ë
5
5
5
5
˛ ÓÓ
˛
Ó
2
ˆ
2
2
2
= (p) – Á Ê Ë q + 1 ˜ {Using (p – q)(p + q) = p – q }
¯
5
Ï Ê qˆ 2 Ê qˆ ¸ 2 2 2
2
Ì
1 + ˝
= p – Á ˜ + Á ˜ () 1 2 {Using (p + q) = p + 2pq + q }
2
5
5
Ó Ë ¯ Ë ¯ ˛
ˆ
2
= p – Ê Á Ë q 2 + 2 5 q + 1 = p - q 2 - 2 5 q - 1
2
˜
25
25
¯
2
2
2
2
2
Example 29: 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 taking RHS.
2
Example 30: 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 (p – q) = p – 2pq + q }
= 10,000 – 400 + 4 = 9,604
2
(b) (102) = (100 + 2) 2
2
2
2
2
2
= (100) + 2(100)(2) + (2) {Using (p + q) = p + 2pq + q }
= 10,000 + 400 + 4 = 10,404
96