Q.225. Relation between roots of a polynomial

Question 225.
The equation x³ + px² + qx + r = 0 (where p, q, r are non zero) has roots α, β, γ
such that 1/α , 1/β, 1/γ are consecutive terms in an arithmetic sequence,
show that β = -3r / q.
The equation x³ - 26x² + 216x - 576 = 0, has roots α, β, γ such that 1/α , 1/β, 1/γ are consecutive terms in an arithmetic sequence. Find the values of α, β, γ.

Answer 225.
If 1/α , 1/β, 1/γ are in A.P., then
1/α + 1/γ = 2/β,
=> βγ + αβ = 2γα
=> βγ + αβ + γα = 3γα ... ( 1 )

Also,
x^3 + px^2 + qx + r = (x - α) (x - β) (x - γ)
=>
x^3 + px^2 + qx + r = x^3 - (α + β + γ)x^2 + (βγ + αβ + γα)x - αβγ
=> βγ + αβ + γα = q ... ( 2 )
and αβγ = - r ... ( 3 )

From ( 1 ) an ( 2 ),
3γα = q
=> 3αβγ = qβ

Plugging αβγ = - r from ( 3 ),
- 3r = qβ
=> β = - 3r/q.

x^3 - 26x^2 + 216x - 576 = 0
Now, β = -3r / q
=> β = - 3(-576)/216 = 8

Coefficient of x^2 is
α + β + γ = 26
=> α + γ = 26 - β = 26 - 8 = 18 ... ( i )
and αβγ = r
=> αγ = r/β = 576/8 = 72 ... ( ii )
 Solving ( i ) and ( ii ),
α = 6, β = 8 and γ = 12
OR
α = 12, β = 8 and γ = 6.

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