Question

Solve the equation 3x³ – 26x² + 52x – 24 = 0 if its roots form a geometric progression

Answer 1

Let the roots form a G.P. be `"a"/lambda, "a", "a"lambda`

Product of the roots, `"a"/lambda xx "a" xx "a"lambda = 24/3` = 8

a³ = 8

a = 2

Sum of the rots, `"a"/lambda + "a" + "a"lambda = 26/3`

`"a"(1/lambda + 1 + lambda) = 26/3`

`2((1 + lambda + lambda^2)/lambda) = 26/3`

3 + 3λ + 3λ² = 13λ

3λ² + 3λ – 13λ + 3 = 0

3λ² – 10λ + 3 = 0

(λ – 3)(3λ – 1) = 0

λ = 6 or λ = `1/3`

If λ = 3,  

`"a"/lambda = 2/3`

aλ = 2(3) = 6

If λ = `1/3`

`"a"/lambda = 2/(1/3)` = 

aλ = `2(1/3) = 2/3`

Roots are `"a"/lambda, "a", "a"lambda`

If λ = 3, roots are `2/3, 2, 6`

If λ = `1/3`, roots are `6, 2, 2/3`