Q.354. Roots of a polynomial which are in G.P.

Question 354.
Solve the equation x^3 - 19x^2 + 114x - 216 = 0 given that the roots are in G.P.

Answer 354.
Let the roots in G.P. be
a/r, a, ar
=>
a/r + a + ar = 19 and
(a/r) * a * (ar) = 216
=> a^3 = 216 => a = 6
Plugging in the first eqn.,
6/r + 6 + 6r = 19
=> 6r^2 - 13r + 6 = 0
=> 6r^2 - 4r - 9r + 6 = 0
=> 2r (3r - 2) - 3 (3r - 2) = 0
=> (2r - 3) (3r - 2) = 0
=> r = 3/2 or 2/3
=> the roots are
a/r = 6/(3/2) = 4
a = 6 and ar = 6 *(3/2) = 9
Answer: 4, 6 and 9.

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