First slide

Theory of equations

Question

The roots $α, β$ and $γ$ of an equation $x^{3} - 3 a x^{2} + 3 b x - c = 0$ are in H.P. Then,

Moderate

A

$β = \frac{1}{a}$
B

$β = b$
C

$β = \frac{b}{c}$
D

$β = \frac{c}{b}$

Solution

Given that the roots of the equation
$x^{3} - 3 a x^{2} + 3 b x - c = 0$ are in H.P. Therefore, the roots of the reciprocal equation i.e. $c y^{3} - 3 b y^{2} + 3 a y - 1 = 0$ are in A.P.
i.e. $\frac{1}{α}, \frac{1}{β}, \frac{1}{γ}$ are in A.P.
$\frac{2}{β} = \frac{1}{α} + \frac{1}{γ}$
$\Rightarrow \frac{3}{β} = \frac{1}{α} + \frac{1}{β} + \frac{1}{γ} \Rightarrow \frac{3}{β} = \frac{3 b}{c} \Rightarrow β = \frac{c}{b} [∵ \frac{1}{α} + \frac{1}{β} + \frac{1}{γ} = \frac{3 b}{c}]$

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