Question

prove: (a+b)^2= (a-b)^2+4ab

Answer 1

LHS
$= {(a + b)}^{2} \\ = {a }^{2} + {b}^{2} + 2ab$
RHS
$= {(a - b)}^{2} + 4ab \\ = {a }^{2} + {b}^{2} - 2ab + 4ab \\ = {a}^{2} + {b}^{2} + 2ab$
LHS=RHS
Hence proved

Answer 2

SOLUTION...

(a+b)^2 = a^2 + b^2 + 2ab................... eq (1.)

(a-b)^2 = a^2 + b^2 - 2ab ...................... eq (2.)

subtracting eq (2.) from eq (1.)
we get,
(a+b)^2 - (a-b)^2 = 2ab + 2ab
(a+b)^2 = 4ab + (a-b)^2
(a+b)^2 = (a-b)^2 + 4ab.
HENCE, PROVED.

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