(These are the solutions from a student of mine, whose identity is private. I will call him, RI, Bengaluru.)
Solve as elegantly as possible:
Solution I (of RI, Bengaluru):
Find the necessary and sufficient conditions on the coefficients p, q, and r of the given cubic equation such that the roots of the cubic are in AP:
Solution II: (credit to RI, Bengaluru) :
Let the roots of the above cubic be .
By Viete’s relations: and and so that we get so that and from the second equation we get , that is, so that , and exploiting the third Viete’s relation we get , that is , which is the required necessary and sufficient condition, where .
Method II: For pedagogical purposes. The above solution to question 2 was quick and elegant because of the right choice of three quantities in AP as . Do you want to know how ugly and messy it can get if a standard assumption is made:
Let the roots of the cubic be . Let d be the common difference. So that the three roots in AP are , and
Then, applying Viete’s relations, we get so that and which changes to and the third viete’s relation gives us .
The second Viete’s relation is a quadratic in and the third Viete’s relation is a cubic in . This is how messy it can get…at least, you will agree RI’s judicious choice has rendered a clean, quick solution.