Question

show that (2x + 3y)^2 = (2x + 3y)^2 + 24xy​

Answer 1

Aɴꜱᴡᴇʀ

Thus LHS≠RHS(just use common sense both sides have the same number and (2x+3y)² and if the rhs is added with a 24xy how will they be the same)

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Gɪᴠᴇɴ

$\large \tt(2 x + 3y {)}^{2} = (2x + 3y {)}^{2} + 24xy$

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Sᴛᴇᴘꜱ

$\tt(2 x + 3y {)}^{2} = (2x + 3y {)}^{2} + 24xy \\ \\ \large \tt \: with \: (x + y {)}^{2} = {x}^{2} +2xy + {y}^{2} \\ \\ \tt first \: lhs \\ \\ \tt(2x {)}^{2} + (3y {)}^{2} + 2(2x)(3y) \\ \\ \tt \dashrightarrow 4 {x}^{2} + 9 {y}^{2} + 12xy \\ \\ \tt{}next \: rhs \\ \\ \tt{}as \: they \: both \: are \: the \: same \\ \\ \tt{}4 {x}^{2} + 9 {y}^{2} + 12xy + 24xy \\ \\ \tt \dashrightarrow4 {x}^{2} + 9 {y}^{2} + 36xy$

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$\huge{\mathfrak{\purple{hope\; it \;helps}}}$

Answer 2

$\large \tt(2 x + 3y {)}^{2} = (2x + 3y {)}^{2} + 24xy$

$\tt(2 x + 3y {)}^{2} = (2x + 3y {)}^{2} + 24xy \\ \\ \large \tt \: with \: (x + y {)}^{2} = {x}^{2} +2xy + {y}^{2} \\ \\ \tt first \: lhs \\ \\ \tt(2x {)}^{2} + (3y {)}^{2} + 2(2x)(3y) \\ \\ \tt \dashrightarrow 4 {x}^{2} + 9 {y}^{2} + 12xy \\ \\ \tt{}next \: rhs \\ \\ \tt{}as \: they \: both \: are \: the \: same \\ \\ \tt{}4 {x}^{2} + 9 {y}^{2} + 12xy + 24xy \\ \\ \tt \dashrightarrow4 {x}^{2} + 9 {y}^{2} + 36xy$

$\huge{\mathfrak{\red{hope\; it \;helps}}}$