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 Further Geometric Numbers (Posted on 2006-11-24)
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 No Rating

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 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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