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Further Geometric Numbers (Posted on 2006-11-24) Difficulty: 2 of 5
The four roots of x4-15x3+70x2-120x+m=0 are in geometric sequence. Determine m and solve the equation.

See The Solution Submitted by K Sengupta    
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Solution Solution Comment 1 of 1

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
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