Math, asked by khushi1787, 11 months ago

prove; (2a+3b)^2+ (2a-3b)^2 =8a^2+18b^2

Answers

Answered by krimusa7524
13
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Answered by abhi569
20

⇔ ( 2a + 3b )^2 + ( 2a - 3b )^2


Using identities : ( a + b )^2 = a^2 + b^2 + 2ab

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


⇒ ( 2a + 3b )^2 + ( 2a - 3b )^2


⇒ { ( 2a )^2 + ( 3b )^2 + 2( 2a × 3b ) } + { ( 2a )^2 + ( 3b )^2 - 2( 2a × 3b ) }


⇒ { 4a^2 + 9b^2 + 12ab } + { 4a^2 + 9b^2 - 12ab }


⇒ 4a^2 +  9b^2 + 12ab + 4a^2 + 9b^2 - 12ab


⇒ 4a^2 + 4a^2 + 9b^2 + 9b^2 + 12ab - 12ab


⇒ 8a^2 + 18b^2



Hence, proved that  ( 2a+3b )^2+ (  2a-3b )^2 =8a^2 + 18b^2

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