Page 82 - Maths Skills - 8

80 Maths

Example 2: Find the following products.

2 2

(i) xy xy (ii) (5x – 3y)(5x – 3y)

3 3
Solution: (i) We have,

x
= 2 xy 2 xy = 2 xy 2 = 2 2 2 y y [ (a – b) = a – 2ab + b ]
x

2
2
2
2

2

3 3 3 3 3
4
= x 4 xy y
2
2
9 3
(ii) We have,
(5x – 3y)(5x – 3y) = (5x – 3y) = (5x) – 2 × 5x × 3y + (3y) [ (a – b) = a – 2ab + b ]
2
2
2
2
2
2
= 25x – 30xy + 9y 2
2
Example 3: Find the following products.
(i) (x + y )(x – y ) (ii) (5x + 7y)(5x – 7y)
2
2
2
2
Solution: (i) (x + y )(x – y ) = (x ) – (y )
2 2
2
2
2
2 2
2
= x – y 4 [ (a + b)(a – b) = a – b ]
4
2
2
(ii) (5x + 7y)(5x – 7y) = (5x) – (7y) 2 [ (a + b)(a – b) = a – b ]
2
2
2
= 25x – 49y 2
2
Example 4: Using the identities, evaluate the following.
(i) 39 (ii) 52 2
2
Solution: We have
(i) 39 = (40 – 1) 2 (ii) 52 = (50 + 2) 2
2
2
= (40) – 2 × 40 × 1 + (1) 2 = (50) + 2 × 50 × 2 + (2) 2
2
2
= 1600 – 80 + 1 = 1521 = 2500 + 200 + 4 = 2704
1
Example 5: If x + = 7, find the value of:
x
(i) x + 1 (ii) x + 1
4
2
x 2 x 4
1 (ii) Now, we have, x + 1 = 47
2
Solution: (i) Given, x = 7 x 2

x
1 2 On squaring both sides again,
On squaring both sides, we get x = 7 2

x 1 2
2

1 2 we get x 2 = 47 2
1
⇒ x + 2 × x × = 49 x
2
x
x 1 1 2
2
2 2
2
⇒ x + 2 + 1 = 49 ⇒ (x ) + 2 × x × x 2 x = 2209
2

x 2 1
4
⇒ x + 1 = 49 – 2 = 47 ⇒ x + 2 + x 4 = 2209
2
x 2 ⇒ x + 1 = 2209 – 2 = 2207
4
\ x + 1 = 47 x 4
2
x 2 1
\ x + = 2207
4
x 4