Pre Calc
posted by Connor .
Determine K and solve the equation x^3kx^2+3x+54=0, if one of its zeros is triple of another.

let the roots be a, 3a and b
then
(xa)(x3a)(xb) = x^3  kx^2 + 3x + 54
looking at the last term 3a^2b = 54
a^2b = 18 , the only square factors of 18 are 1 and 9
case 1: a=1, b = 18
then
(x1)(x3)(x+18) would be the expression
this give us x^3 + 14x^2  69x + 54
which would not match up the x terms
case 2: a=3, b=2
then
(x3)(x9)(x+2) would be the expression for
x^3  10x^2 + 3x + 54
this matches if k = 10
and the roots are 3,9 and 2
(there should be an easier way to do this)
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