Question

Show that: \[\left( \frac{4m}{3} - \frac{3n}{4} \right)^2 + 2mn = \frac{16 m^2}{9} + \frac{9 n^2}{16}\]

Answer 1

\[\text { LHS } = \left( \frac{4m}{3} - \frac{3n}{4} \right)^2 + 2mn\]

\[ = \left( \frac{4m}{3} - \frac{3n}{4} \right)^2 + 2 \times \frac{4m}{3} \times \frac{3n}{4}\]

\[ = \left( \frac{4m}{3} \right)^2 + \left( \frac{3n}{4} \right)^2 \left[ \because \left( a - b \right)^2 + 2ab = a^2 + b^2 \right]\]

\[ = \frac{16 m^2}{9} + \frac{9 n^2}{16}\]

= RHS

Because LHS is equal to RHS, the given equation is verified.