Question

Find the condition that the zeroes of the polynomial $ p(x) = {x^3} - p{x^2} + qx - r $ may be in arithmetic progression.

Answer 1

Hint: Since, roots of the a cubic polynomial are in A.P. there taking its roots are (a –d) , (a) and (a + d) and then forming three different equations using relation between roots and coefficients and then solving value of ‘a’ and ‘d’ from two of these and using values in third to get required relation or solution of the problem.
Relation between roots and coefficients for a cubic polynomial are given as:
 $
 Sum\,\,of\,the\,\,roots\left( {\alpha + \beta + \gamma } \right) = \dfrac{{ - coff.\,\,of\,\,{x^2}}}{{coff.\,\,of\,\,{x^3}}} \\
  Product\,\,of\,\,the\,\,roots\,\left( {\alpha .\beta .\gamma } \right) = \dfrac{{ - \,cons\tan t\,\,term}}{{coff.\,\,of\,\,{x^3}}} \\
  Sum\,\,product\,\,taken\,\,two\,at\,time\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) = \dfrac{{coff.\,\,of\,\,x}}{{coff.\,\,of\,\,{x^3}}} \\
  $

Complete step-by-step answer:
Given polynomial in question is $ p(x) = {x^3} - p{x^2} + qx - r $ .
Since, the degree of the given polynomial is $ 3 $ . Therefore, it will have $ 3 $ zeroes. But as it is given that zeroes of the polynomial are in A.P.
And we know that for three terms which are in A.P. always taken as (a – d), (a) and (a + d).
Now, from relation between roots and coefficient of polynomial we have
Sum of the roots = $ \dfrac{{ - (coff.\,\,of\,\,{x^2})}}{{(coff.\,\,of\,\,{x^3})}} $
 $
   \Rightarrow \left( {a - d} \right) + (a) + \left( {a + d} \right) = \dfrac{{ - ( - p)}}{{(1)}} \\
   \Rightarrow a - d + a + a + d = p \\
   \Rightarrow 3a = p \\
   \Rightarrow a = \dfrac{p}{3}....................(i) \;
  $
Product of the roots = $ \dfrac{{ - (cons\tan t\,\,term)}}{{\left( {coff.\,\,of\,\,{x^3}} \right)}} $
 $
  \left( {a - d} \right)(a)\left( {a + d} \right) = \dfrac{{ - ( - r)}}{1} \\
   \Rightarrow \left( {{a^2} - {d^2}} \right)(a) = r \;
  $
Substituting value of ‘a’ calculated in above equation. We have,
 $
  \left( {\dfrac{{{p^2}}}{9} - {d^2}} \right)\left( {\dfrac{p}{3}} \right) = r \\
   \Rightarrow \left( {\dfrac{{{p^2}}}{9} - {d^2}} \right) = \dfrac{{3r}}{p} \\
   \Rightarrow \dfrac{{{p^2}}}{9} - \dfrac{{3r}}{p} = {d^2} \\
   \Rightarrow {d^2} = \dfrac{{{p^2}}}{9} - \dfrac{{3r}}{p}...................(ii) \;
  $
Also, we know that sum product of zeroes of polynomial is given as
 $ \left( {a - d} \right)(a) + (a)\left( {a + d} \right) + \left( {a - d} \right)\left( {a + d} \right) = \dfrac{{\left( {coff.\,\,of\,\,x} \right)}}{{\left( {coff.\,\,of\,\,{x^3}} \right)}} $
 $
   \Rightarrow a\left( {a - d + a + d} \right) + \left( {{a^2} - {d^2}} \right) = \dfrac{q}{1} \\
   \Rightarrow 2{a^2} + \left( {{a^2} - {d^2}} \right) = q \\
   \Rightarrow 3{a^2} - {d^2} = q \;
  $
Substituting the value of $ {d^2} $ calculated in (ii) we have:
 $
  3{\left( {\dfrac{p}{3}} \right)^2} - \left( {\dfrac{{{p^2}}}{9} - \dfrac{{3r}}{p}} \right) = q \\
   \Rightarrow 3\left( {\dfrac{{{p^2}}}{9}} \right) - \dfrac{{{p^2}}}{9} + \dfrac{{3r}}{p} = q \\
   \Rightarrow \dfrac{{{p^2}}}{3} - \dfrac{{{p^2}}}{9} + \dfrac{{3r}}{p} = q \\
   \Rightarrow \dfrac{{3{p^2} - {p^2}}}{9} + \dfrac{{3r}}{p} = q \\
   \Rightarrow \dfrac{{2{p^2}}}{9} + \dfrac{{3r}}{p} = q \;
  $
From above we see that required relation between p, q and r when zeroes of the $ p(x) = {x^3} - p{x^2} + qx - r $ are in A.P. is: $ \dfrac{{2{p^2}}}{9} + \dfrac{{3r}}{p} = q $ .

Note: We know that in A.P. three numbers may be taken as ‘a’, ‘a + d’ and a + 2d’. But we know that for a problem when only three terms of A.P. are to be taken and there is a sum of roots either given or condition we can use. In this type of problems we can take three A.P. terms are (a –d), (a) and (a + d) as these make calculation easier to get the required solution.