. . Roots in Progressions As already noted to solve higher degree polynomial equations, we need some information about the solutions of the equation or about the polynomial. “The roots are in arithmetic progression” and “the roots are in geometric progression” are some of such information.
Let us discuss an equation of this type. - - Example . Obtain the condition that the roots of x px qx are in A.P. Let the roots be in A.P.
Then, we can assume them in the form α α α , , Applying the Vieta’s formula ( d = − p = p ⇒ α = − p ⇒ α = − p . But, we note that α is a root of the given equation. Therefore, we get + + + q ⇒ pq Example . Find the condition that the roots of ax bx cx are in geometric progression.
Assume a b c d , , , ¹ Let the roots be in G.P. Then, we can assume them in the form α λ α αλ , , Applying the Vieta’s formula, we get å = α λ + + = − b … ( ) å = α + + = c … ( ) å = α = − d a . … ( ) Dividing ( ) by ( ), we get α = − c … ( ) Substituting ( ) in ( ), we get − = − d a ⇒ ac db Example . If the roots of x px qx are in H.P.
, prove that pqr q Assume p , q , r ≠ Let the roots be in H.P. Then, their reciprocals are in A.P. and roots of the equation p x q x rx qx px + + + ⇔ + = … ( ) Since the roots of ( ) are in A.P., we can assume them as α α α , , Applying the Vieta’s formula, we get å = ( d = − q r ⇒ α = − q r ⇒ α = − q . - - Theory of Equations But, we note that α is a root of ( ).
Therefore, we get q q q q q q pqr + + + = ⇒− ⇒ pqr q r . Example . It is known that the roots of the equation x are in arithmetic progression. Find its roots.
Let the roots be a d a a , , . Then the sum of the roots is a which is equal to from the given equation. Thus a = and hence a = . The product of the roots is a ad which is equal to − from the given equation.
Substituting the value of a , we get and hence d = ± . If we take d = we get − , , as roots and if we take d = − , we get , , − as roots (same roots given in reverse order) of the equation.