Question

solve 81x^3+18x^2-36x+8=0 whose roots are in harmonic progession ​

Answer 1

Answer:

2/9, 2/3 and -2/3

Explanation:

[If a, b, c are in H.P. then 1/a,1/b, 1/c are in the A.P. we have b = 2ac/(a + c)

Let α, β, γ be the roots of the equation. 

αβ + βλ + αλ = -4/9

αβλ = -8/81

Since α, β, γ are in H.P

β = 2αλ/α + γ

αβ + βγ = 2αλ

 ∴ (1) 3αγ = -4/9 ⇒ αλ = -4/27 

 ∴ (2) αλβ = -8/81 

  -4/27β = -8/81 ⇒ β = -8/81 × -27/4 = 2/3

Now, α + λ = 2αγ/β 

⇒ α + λ = -8/27 × 3/2 

∴ α + λ = -4/9 --- (3) 

(α -  λ)² = (α + λ)² - 4αλ

        = 16/81 + 16/27

        = 64/27

∴ α - λ = 8/9 --- (4) 

Solving (3) and (4) we get, α = 2/9 and λ = -2/3

Therefore, the roots are 2/9, 2/3 and -2/3