3.8.2 Reciprocal Equations

. . Reciprocal Equations Let α be a solution of the equation. = .

... ( ) Then α ¹ (why?) and = . Substituting α for x in the left side of ( ), we get  −  +  +   +  −  + = Thus α is also a solution of ( ). Similarly we can see that if α is a solution of the equation = ...

( ) then α is also a solution of ( ). Equations ( ) and ( ) have a common property that, if we replace x by x in the equation and write it as a polynomial equation, then we get back the same equation. The immediate question that flares up in our mind is “Can we identify whether a given equation has this property or not just by seeing it?” Theorem . below answers this question.

Definition . A polynomial P x ( ) of degree n is said to be a reciprocal polynomial if one of the following conditions is true: (i) P x x P x ( ) = (ii) P x x P x ( ) = − . A polynomial P x ( ) of degree n is said to be a reciprocal polynomial of Type I if P x x P x ( ) = . is called a reciprocal equation of Type I.

A polynomial P x ( ) of degree n is said to be a reciprocal polynomial of Type II P x x P x ( ) = − . is called a reciprocal equation of Type II. Theory of Equations Theorem . A polynomial equation a x a x a x  = , ( a n ¹ is a reciprocal equation if, and only if, one of the following two statements is true: (i) a n = , a n − = , a n − =  (ii) a n = − , a n − = − , a n − = − ,  Proof Consider the polynomial equation P x ( ) = a x a x a x  = .

… ( ) Replacing x by x in ( ), we get P x   = a  = . … ( ) Multiplying both sides of ( ) by x n , we get x P x   = a x a x a x  = . … ( ) Now, ( ) is a reciprocal equation ⇔ P x x P x ( ) = ± ⇔ ( ) and ( ) are same . This is possible ⇔ a  Let the proportion be equal to λ .

Then, we get a = λ and a a n = λ . Multiplying these equations, we get λ = . So, we get two cases λ = 1and λ = − . Case (i) : λ = In this case, we have a  .

That is, the coefficients of ( ) from the beginning are equal to the coefficients from the end. Case (ii) : λ = − In this case, we have a  . That is, the coefficients of ( ) from the beginning are equal in magnitude to the coefficients from the end, but opposite in sign. Note Reciprocal equations of Type I correspond to those in which the coefficients from the beginning are equal to the coefficients from the end.

For instance, the equation is of type I. Reciprocal equations of Type II correspond to those in which the coefficients from the beginning are equal in magnitude to the coefficients from the end, but opposite in sign. For instance, the equation is of Type II. Remark (i) A reciprocal equation cannot have as a solution.

(ii) The coefficients and the solutions are not restricted to be real. (iii) The statement “ If P x ( ) = is a polynomial equation such that whenever α is a root, is also a root, then the polynomial equation P x ( ) = must be a reciprocal equation ” is not true. For instance is a polynomial equation whose roots are , , Note that x P x  ≠ ± P ( x ) and hence it is not a reciprocal equation. Reciprocal equations are classified as Type I and Type II according to a n r − = or a n r − = − , r = , , ,...

n . We state some results without proof : • For an odd degree reciprocal equation of Type I, x = − must be a solution. • For an odd degree reciprocal equation of Type II, x = 1must be a solution. • For an even degree reciprocal equation of Type II, the middle term must be .

Further x = 1and x = − are solutions. • For an even degree reciprocal equation, by taking x + or x − as y , we can obtain a polynomial equation of degree one half of the degree of the given equation ; solving this polynomial equation, we can get the roots of the given polynomial equation. As an illustration, let us consider the polynomial equation which is an even degree reciprocal equation of Type II. So and − are two solutions of the equation and hence x − is a factor of the polynomial.

Dividing the polynomial by the factor x − , we get as a factor. Dividing this factor by x and rearranging the terms we get  −  + . Setting u it becomes a quadratic polynomial as u u ) − which reduces to u u . Solving we obtain u = .

Taking u = gives x = and taking u = gives x = . 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