Question

show that b) (3p-q)square+12pq =(3p+q)square

Answer 1

Answer:

Step-by-step explanation:

To prove: $(3p-q)^{2} + 12pq = (3p+q)$

Identity used:

1. $(a+b)^{2} = a^{2}+2ab+b^{2}$

2. $(a-b)^{2} =a^{2} - 2ab + b^{2}$

Solution:

LHS=

$(3p-q)^{2} +12pq\\= (3p)^{2} - 6pq + q^{2} + 12pq\\\\= (3p)^{2} + 6pq + q^{2}\\= (3p+q)^{2}$

= RHS

Hence proved.