Question

Prove: [4m/3 -3/4n]^2 2mn

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Answer 1

Given

LHS

$\bf \: \{ \dfrac{4m}{3} - \dfrac{3n}{4}\}^{2} + 2mn$

RHS

$\bf \: \dfrac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16}$

To Prove

LHS=RHS

Proof

LHS

• Open bracket using identity
• $\boxed{ \rm \: (a - b) ^{2} = {a}^{2} + {b}^{2} - 2ab }$

$\rm \: \dfrac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16} - 2 \times \dfrac{4m}{3} \times \dfrac{3n}{4} + 2mn \\ \\ \implies \rm \frac{16 {m}^{2} }{9} + \frac{9 {n}^{2} }{16} - 2mn + 2mn \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \rm \: \frac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \underline{\boxed{ \huge \bf \: LHS = \frac{16 {m}^{2} }{ 9 } + \frac{9 {n}^{2} }{16} }}$

$\underline{ \boxed{ \huge \rm \: RHS = \bf \: \dfrac{16 {m}^{2} }{9} + \dfrac{9 {n}^{2} }{16} }}$

LHS=RHS

HENCE PROVED