. So the required solutions are + − , , , , Example . Solve the equation The given equation can be written as This is an odd degree reciprocal equation of Type I. Thus − is a solution and hence x + is a factor.
Dividing the polynomial by the factor x + ,we get x + as a quotient. Solving this we get and as roots. Thus − , are the solutions of the given equation. Example .
Solve the following equation: x + = - - Theory of Equations This equation is Type I even degree reciprocal equation. Hence it can be rewritten as − + = Since x ≠ , we get x − + Let y = x + . Then, we get