Question

1. (v)(3m-4n) whole square =(3m-4n) whole square + 48mn
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Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO PROVE

$\sf{ {(3m + 4n)}^{2} \: = {(3m - 4n)}^{2} + 48mn\: }$

FORMULA TO BE IMPLEMENTED

We are aware of the identity that

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

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

CALCULATION

$\sf{ {(3m + 4n)}^{2} \: \: }$

$\sf{ = {(3m)}^{2} + 2 \times 3m \times 4n + {(4n)}^{2} \: \: } \: \: (by \: identity \: 1)$

$\sf{ = {9m}^{2} + 24mn + {16n}^{2} \: \: }$

$\sf{ = {9m}^{2} - 24mn + {16n}^{2} + 48mn\: \: }$

$\sf{ = {(3m)}^{2} - 2 \times 3m \times 4n + {(4n)}^{2} \: + 48mn\: } \:$

$= \sf{ {(3m - 4n)}^{2} \: + 48mn \: } \: \: \: \: (\: by \: identity \: 2)$

Hence proved

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