Let equation x3-px2-qx-q-0 where p- q- - R - 0 has 3 real roots - - - in H-P- then 9p-2q has the value equal to

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Question

Let equation $x^{3} + p x^{2} + q x - q = 0$ where $p, q, \in R - {0}$ has 3 real roots $α, β, λ$ in $H . P .$ , then
$(9 p + 2 q)$ has the value equal to

9

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- 27

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Solution

The correct option is B $- 27$
$x^{3} + p x^{2} + q x - q = 0$
$α, β, λ$ are its roots
$∴ α + β + λ = - p ⟶ 1$
$α β + β λ + λ α = q ⟶ 2$
$α β λ = q ⟶ 3$
Now, $α, β, λ,$ are in H.P
$∴ β = \frac{2 α λ}{α + λ}$
$α β + β λ = 2 α λ$
$α β + β λ + λ α = 3 α λ$
$q = 3 α λ$ (from equation 2)
$q = 3 \frac{q}{β}$ (from equation 3)
$β = 3$
Put value of $β$ in equation 2
$3 α + 3 λ + α λ = q$
$3 (α + λ) + α λ = q$
use equation 2 and 3 we get
$3 (- p - β) + \frac{q}{β} = q$
$3 (- p - 3) + \frac{q}{3} = q$
$- 3 p - 9 = \frac{2 q}{3}$
$- 9 p - 27 = 2 q$
$9 p + 2 q = - 27$

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Let equation x3+px2+qx−q=0 where p,q,∈R−{0} has 3 real roots α,β,λ in H.P., then (9p+2q) has the value equal to

Let equation $x^{3} + p x^{2} + q x - q = 0$ where $p, q, \in R - {0}$ has 3 real roots $α, β, λ$ in $H . P .$ , then
$(9 p + 2 q)$ has the value equal to