Question

this is to solve...
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Answer 1

Answer:

Proved below.

Step-by-step explanation:

Given,

      2a - b + c = 0

⇒ 2a - b + c = 0

⇒ 2a + c = b

Square on both sides :

⇒ ( 2a + c )^2 = b^2

⇒ 4a^2 + c^2 + 2( 2a )c = b^2           [ using ( a + b )^2 = a^2 + b^2 + 2ab ]

⇒ 4a^2 + c^2 + 4ac = b^2

⇒ 4a^2 + c^2 + 4ac - b^2 = 0

⇒ 4a^2 - b^2 + c^2 + 4ac = 0

      Proved.