Solve quadratic equation ^@x^2 + (\dfrac { p } { p + q } + \dfrac { p + q } { p })x + 1^@ ^@= 0 ^@ using factorization.


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Answer:

^@ x = - \dfrac { p } { p + q } \space or \space x = - \dfrac { p + q } { p }^@

Step by Step Explanation:
  1. ^@\begin{align} LHS = \\ & x^2 + \dfrac { p } { p + q }x + \dfrac { p + q } { p }x + 1 \\ \implies & x^2 + \dfrac { p } { p + q }x + \dfrac { p + q } { p }x + 1 \times \dfrac { p } { p + q } \times \dfrac { p + q } { p } \\ \implies & x(x + \dfrac { p } { p + q } ) + \dfrac { p + q } { p } ( x + \dfrac { p } { p + q } ) \\ \implies & (x + \dfrac { p } { p + q } ) ( x + \dfrac { p + q } { p } ) = 0 \\ \implies & x = - \dfrac { p } { p + q } \space or \space - \dfrac { p + q } { p } \end{align}^@
  2. Hence, ^@ x = - \dfrac { p } { p + q } \space or \space x = - \dfrac { p + q } { p }^@.

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