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If alpha,beta,gamma be the roots of x^3+qx+r=0,then find the equation whose roots are Beta/gamma+gamma/beta,gamma/alpha+alpha/gamma,alpha/beta+beta /alpha

If alpha,beta,gamma be the roots of x^3+qx+r=0,then find the equation whose roots are 
Beta/gamma+gamma/beta,gamma/alpha+alpha/gamma,alpha/beta+beta /alpha

Grade:12th pass

1 Answers

Pavan Kumar
32 Points
5 years ago
Heyy, thank you for asking the question 
                       i suppose you know what to do if roots are shifted by or multiplied by 1 or 2 towards right such type of problems
     now we know from the equation  αβγ= -r
                                                         αβ+βγ+αγ=q
                                                          α+β+γ  =0
                                                  (α+β+γ)2 = α2+β2+γ2+ 2αβ+2βγ+2αγ
                                                    0=α2+β2+γ2+2q
                                                       α2+β2+γ2= -2q
        now consider new roots  β/γ+γ/β=β2+γ2/βγ
                                                               = (  -2q -   α2)/(-r/α)
                                                               =    ( 2qα +   α3)/r
                                     but we have  α3 + qα+r =0  substituting this in above equaiton we get
                                                              =    (qα – r)/r
                                                              =   q/r*α  – 1
                                 so from the last result we came to know that roots of the new equation are q/r times of previous roots and a shift of -1 so the final equation of new roots are
      
                                           (r/q)3 (x+1)3+q* r/q*(x+1)+r=0
now lets check for a root       (r/q)3 ( q/r*α  – 1+1)3+q* r/q*( q/r*α  – 1+1)+r=0
                                                (r/q)3 (q/r*α )3+q* r/q*( q/r*α )+r=0
                                                 α 3+q*α+r=0     [proved]
  Hope you got the solution.....

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