The four roots of x⁴-15x³+70x²-120x+m=0 are in geometric sequence. Determine m and solve the equation.

Let the roots be a, k*a, k^2*a, and k^3*a. Then

  (x-a)*(x-k*a)*(x-k^2*a)*(x-k^3*a) =
 
   x^4 - a*(k+1)*(k^2+1)*x^3 +
         a*2*k*(k^4+k^3+2k^2+k+1)*x^2 -
         a^3*k^3*(k+1)*(k^2+1)*x + a^4*k^6

Thus,

   a*(k+1)*(k^2+1) = 15                   

   a*2*k*(k^4+k^3+2k^2+k+1) = 70         

   a^3*k^3*(k+1)*(k^2+1) = 120           

   a^4*k^6 = m                          

Therefore,

   15*a^2*k^3 = a^2*k^3*[a*(k+1)*(k^2+1)] = 120

          or

   a^2*k^3 = 8

Hence,

   m = a^4*k^6 = 64

The roots of 

   x^4 - 15x^3 + 70x^2 -120x + 64 = 0

are 1, 2, 4, and 8.

Posted by Bractals on 2006-11-24 06:56:50