Question

Solve the following equation and verify your answer:

\[\left( \frac{x + 1}{x + 2} \right)^2 = \frac{x + 2}{x + 4}\]

Answer 1

\[ \left( \frac{x + 1}{x + 2} \right)^2 = \frac{x + 2}{x + 4}\]

\[\text{ or }\frac{x^2 + 2x + 1}{x^2 + 4x + 4} = \frac{x + 2}{x + 4}\]

\[\text{ or }x^3 + 2 x^2 + x + 4 x^2 + 8x + 4 = x^3 + 4 x^2 + 4x + 2 x^2 + 8x + 8 [\text{ After cross multiplication }]\]

\[\text{ or }x^3 - x^3 + 6 x^2 - 6 x^2 + 9x - 12x = 8 - 4\]

\[\text{ or }- 3x = 4\]

\[\text{ or }x = \frac{4}{- 3} = \frac{- 4}{3}\]

\[\text{ Thus, }x = \frac{- 4}{3}\text{ is the solution of the given equation .} \]

\[\text{ Check: }\]

\[ \text{ Substituting }x = \frac{- 4}{3}\text{ in the given equation, we get: }\]

\[\text{ L . H . S . }= \left( \frac{\frac{- 4}{3} + 1}{\frac{- 4}{3} + 2} \right)^2 = \left( \frac{- 4 + 3}{- 4 + 6} \right)^2 = \frac{1}{4}\]

\[\text{ R . H . S . }= \frac{\frac{- 4}{3} + 2}{\frac{- 4}{3} + 4} = \frac{- 4 + 6}{- 4 + 12} = \frac{2}{8} = \frac{1}{4}\]

\[ \therefore\text{ L . H . S . = R . H . S . for }x = \frac{- 4}{3}\]