If the roots of the cubic equation x^3-2x^2+kx-27=0 are in geometric progression, determine the value of k?

i) Let the 3 roots of the given equation be, {p, q, r}

ii) Then by properties, p + q + r = 2 ------ (1)
pq + qr + rp = k ------ (2)
pqr = 27 -------- (3)

ii) As roots are in GP, pr = q^2
Substituting this in equation (3) and solving, q = 3; as well pr = q^2 = 9

iii) Substituting this in (1), p + r = -1 ---- (4)

iv) From (2) we have, k = q(p + r) + pr = 3*(-1) + 9 = -3 + 9 = 6

Thus k = 6

Let a/r, a and ar be the roots of the cubic equation x³ - 2x² + kx - 27 = 0

(a/r) + a + ar = -(-2) …… [1]
(a/r)a + a(ar) + (ar)(a/r) = k …… [2]
(a/r)(a)(ar) = -(-27) …… [3]

From [3] :
a³ = 27
a = 3

From [1] :
(3/r) + 3 + 3r = 2
3[(1/r) + 1 + r] = 2
(1/r) + 1 + r = 2/3 …… [4]

From [2]:
(a²/r) + a²r + a² = k
a²[(1/r) + r + 1] = k
k = a²[(1/r) + r + 1] …… [5]

Substitute a = 3 and [4] into [5] :
k = 3² * (2/3)
k = 6

If the 3 roots in GP are a/r, a and ar then
last term = -(product of roots) = -27, so a = 3
(x – 3/r)(x – 3r) = x^2 – (3r + 3/r)x + 9
(x – 3)[ x^2 – (3r + 3/r)x + 9] = x^3 - 2x^2 + kx – 27
We need the coefficient of x, 9(r + 1 + 1/r) = k ………….(1)
From the coefficient of x^2 by comparison
-3r – 3 - 3/r = -2,
k = 3*3(r + 1 + 1/r) = 6

x³ - 2x² + kx - 27 = 0 → supose that the 3 rots are x₁, x₂, x₃ ← where: x₁ ≠ 0 and x₂ ≠ 0 and x₃ ≠ 0

(x - x₁).(x - x₂).(x - x₃) = 0 → x₁, x₂, x₃ in geometric progression → x₂ = q.x₁

(x - x₁).(x - q.x₁).(x - x₃) = 0 → x₁, x₂, x₃ in geometric progression → x₃ = q.x₂

(x - x₁).(x - q.x₁).(x - q.x₂) = 0 → recall: x₂ = q.x₁

(x - x₁).(x - q.x₁).(x - q².x₁) = 0 → you expand

(x² - q.x.x₁ - x.x₁ + q.x₁²).(x - q².x₁) = 0

x³ - q².x².x₁ - q.x².x₁ + q³.x.x₁² - x².x₁ + q².x.x₁² + q.x.x₁² - q³.x₁³ = 0 → you order

x³ - q².x².x₁ - q.x².x₁ - x².x₁ + q³.x.x₁² + q².x.x₁² + q.x.x₁² - q³.x₁³ = 0 → you group

x³ - (q².x².x₁ + q.x².x₁ + x².x₁) + (q³.x.x₁² + q².x.x₁² + q.x.x₁²) - q³.x₁³ = 0 → you factorize

x³ - x².(q².x₁ + q.x₁ + x₁) + x.(q³.x₁² + q².x₁² + q.x₁²) - q³.x₁³ = 0 → you compare to: x³ - 2x² + kx - 27 = 0


(1) : - q³.x₁³ = - 27

(1) : q³.x₁³ = 27

(1) : (q.x₁)³ = 3³

(1) : q.x₁ = 3


(2) : q³.x₁² + q².x₁² + q.x₁² = k

(2) : q. x₁².(q² + q + 1) = k

(2) : q² + q + 1 = k/q.x₁²


(3) : q².x₁ + q.x₁ + x₁ = 2

(3) : x₁.(q² + q + 1) = 2

(3) : q² + q + 1 = 2/x₁ → recall (2) : q² + q + 1 = k/q.x₁²

(3) : k/q.x₁² = 2/x₁ → you can simplify by x₁ because: x₁ ≠ 0

(3) : k/q.x₁ = 2

(3) : k = 2.q.x₁ → recall (3) : q.x₁ = 3

(3) : k = 6