Question

a square + b square + c square upon B square -ac

Answer 1

How this identity of (a + b)2 = a2 + b2 + 2ab is obtained: 

Taking LHS of the identity: 
(a + b)2 

This can also be written as: 
= (a + b) (a + b) 

Multiply as we do multiplication of two binomials and we get: 
= a(a + b) + b(a + b) 
= a2 + ab + ab + b2 

Add like terms and we get: 
= a2 + 2ab + b2 

Rearrange the terms and we get: 
= a2 + b2 + 2ab 

Hence, in this way we obtain the identity i.e. (a + b)2 = a2 + b2 + 2ab 


Following are few applications of identity first: 

Example 1: Solve (4p + 5q)2 
Solution: This proceeds as: 
Given polynomial (4p + 5q)2 represents identity first i.e. (a + b)2 
Where a = 4p and b = 5q 

Now apply values of a and b on the identity i.e. (a + b)2 = a2 + b2 + 2ab and we get: 
(4p + 5q)2 = (4p)2 + (5q)2 + 2(4p)(5q) 

Expand the exponential forms and we get: 
= 16p2 + 25q2 + 2(4p)(5q) 

Solve multiplication process and we get: 
= 16p2 + 25q2 + 40pq 

Hence, (4p + 5q)2 = 16p2 + 25q2 + 40pq 


Example 2: Solve (8x + 4y)2 
Solution: This proceeds as: 
Given polynomial (8x + 4y)2 represents identity first i.e. (a + b)2 
Where a = 8x and b = 4y 

Now apply values of a and b on the identity i.e. (a + b)2 = a2 + b2 + 2ab and we get: 
(8x + 4y)2 = (8x)2 + (4y)2 + 2(8x)(4y) 

Expand the exponential forms and we get: 
= 64x2 + 16y2 + 2(8x)(4y) 

Solve multiplication process and we get: 
= 64x2 + 16y2 + 64xy 

Hence, (8x + 4y)2 = 64x2 + 16y2 + 64xy