Solve quadratic equation ^@\dfrac {x} { x + 1 } + \dfrac { x + 1 } { x } = \dfrac { 25 } { 12 } ^@ (^@x \neq 0 ^@ and ^@ x \neq -1^@).


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Answer:

^@x = 3^@, or ^@x = -4^@

Step by Step Explanation:
  1. On adding two fractions on LHS,
    ^@\begin{align} &\dfrac { x^2 + (x + 1)^2 } { x(x + 1) } = \dfrac { 25 } { 12 } \\ \implies &\dfrac { x^2 + (x^2 + 2x + 1) } { x(x + 1) } = \dfrac { 25 } { 12 } \end{align}^@
  2. ^@ \begin{align} & 12 (2x^2 + 2x + 1) = 25 (x^2 + x) \\ \implies & 24x^2 + 24x + 12 = 25x^2 + 25x \\ \implies & 25x^2 + 25x - 24x^2 - 24x - 12 = 0 \\ \implies & x^2 + x - 12 = 0 \\ \implies & x^2 + 4 x - 3 x - 12 = 0 \\ \implies & x(x + 4) - 3(x + 4) = 0 \\ \implies & (x - 3) (x + 4) \\ \implies & x = 3 \space or \space -4 \end{align}^@
  3. Hence, ^@x = 3^@, or ^@x = -4^@.

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