If the roots of the equation, $x^{3} + 3 a x^{2} + 3 b x + c = 0$ are in H.P. then

b^{2} = c (3 a b - c)

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2 b^{3} = c (3 a b - c)

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2 b^{3} = c^{2} (3 a b - c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 b^{2} = c^{2} (3 a b - c)

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Solution

The correct option is D $2 b^{3} = c (3 a b - c)$
Let the roots be $p, q, r$
$p, q, r$ are in H.P.
So, $q = \frac{2 p r}{p + r}$ ,
$q r + p q = 2 p r,$
$p q + q r + p r = 3 p r .$ ...(i)
Using the theory of polynomials,
$p q r = - c$ ...(ii)

$p q + q r + p r = 3 b$ ...(iii)

(i), (ii) and (iii) $\Rightarrow 3 b = \frac{- 3 c}{q}$
$\Rightarrow q = \frac{- c}{b} .$
But $q$ is a root of the equation. Hence, we substitute the value of $q$ in the equation.

$\frac{- c^{3}}{b^{3}} + \frac{3 a c^{2}}{b^{2}} - 3 c + c = 0,$

$\Rightarrow - c^{3} + 3 a b c^{2} = 2 c b^{3},$

$\Rightarrow 2 b^{3} = c (3 a b - c) .$

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