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Question
Solve the equation 3x3 – 16x2 + 23x – 6 = 0 if the product of two roots is 1
Solution
Let the roots be α, β, γ
Given αβ = 1, β = `1/alpha`
Sum of the roots `alpha + 1/alpha + γ = (-beta)/alpha = 16/3` ........(1)
Product of the roots `alpha * 1/alpha * γ = (- "d")/(alpha) = 6/3` = 2
⇒ γ = 2
Put in (1)
`alpha + 1/alpha + 2 = 16/3`
⇒ `alpha + 1/alpha = 16/3 - 2`
= `(16 - 6)/3`
= `10/3`
`(alpha^2 + 1)/alpha = 10/3`
⇒ `3alpha^2 + 3 = 10alpha`
⇒ 3α2 – 10α + 3 = 0
3α2 – 9α – α + 3 = 0
3α(α – 3) – 1(α – 3) = 0
(3α – 1)(α – 3) = 0
α = 3 or α = `1/3`
If α = 3, β = `1/3`, γ = 2
or
α = `1/3`, β = 3, γ = 2
⇒ [α, β, γ] = `(1/3, 3, 2)`
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