Posted by **Connor** on Tuesday, August 31, 2010 at 7:29pm.

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

- Pre Calc -
**Reiny**, Tuesday, August 31, 2010 at 8:15pm
let the roots be a, 3a and b

then

(x-a)(x-3a)(x-b) = 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

(x-1)(x-3)(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

(x-3)(x-9)(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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