Question

Question 1
Use a suitable identity to get each of the following products.




(a) (p - 11) (p + 11)
(b) (2y + 5) (2y - 5)
(c) (12a - 9) (12a +9)

​

Answer 1

$\red\bigstar$Answer:

1. p² - 121
2. 4y² - 25
3. 144a² - 81

$\pink\bigstar$Given:

• To find the product of :

1. ( p - 11 )( p + 11 )
2. ( 2y + 5 )( 2y - 5 )
3. ( 12a - 9 )( 12a + 9 )

$\blue\bigstar$To find:

1. ( p - 11 )( p + 11 )
2. ( 2y + 5 )( 2y - 5 )
3. ( 12a - 9 )( 12a + 9 )

$\green\bigstar$ Solution:

Here, we can use the following formula :

• a² - b² = (a + b) ( a - b)

So,

(1)

(p - 11) (p + 11)

= (p)² - (11)²

[ By using a² - b² = (a + b) ( a - b) ]

= p² - 121

(2)

(2y + 5)(2y - 5)

= (2y)² - (5)²

[ By using a² - b² = (a + b) ( a - b) ]

= 4y² - 25

(3)

( 12a - 9 )( 12a + 9 )

= (12a)² - (9)²

[ By using a² - b² = (a + b) ( a - b) ]

= 144a² - 81

$\red\bigstar$ Concepts Used:

• a² - b² = (a + b) ( a - b)

• Squaring of numbers and variables

$\pink\bigstar$ Additional Information:

• (a + b)² = a ² + b² + 2ab

• (a - b)² = a² + b² - 2ab

• a² - b² = ( a + b )( a-b )

• a² + b² = (a + b)² - 2ab

• a³ + b³ = (a + b)(a² - ab + b²)

• a³ - b³ = (a - b)(a² + ab + b²) = (a - b)³ + 3ab(a- b)

• 2(a² + b²) = (a+ b)² + (a- b)²

• (a + b)² - (a - b)² = 4ab

• (a + b + c)² = a² + b² +c² + 2ab + 2bc + 2ca

• (a + b - c)² = a² + b² + c² + 2ab – 2bc – 2ca