If the roots of the equation $x^{3} - p x^{2} + q x - r = 0$ are in AP, then

2 p^{3} = 9 p q - 27 r

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2 q^{3} = 9 p q - 27 r

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p^{3} = 9 p q - 27 r

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2 p^{3} = 9 p q + 27 r

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Solution

The correct option is A $2 p^{3} = 9 p q - 27 r$
Let the roots of the given equation be $(a - d), a, (a + d)$
Then,
$(a - d) + a + (a + d) = \frac{- (- p)}{1}$
$\Rightarrow a = \frac{p}{3}$
Since, $a$ is a root of the given equation
$∴ a^{3} - p a^{2} + q a - r = 0$
$\Rightarrow \frac{p^{3}}{27} - \frac{p^{3}}{9} + \frac{q p}{3} - r = 0$
$\Rightarrow 2 p^{3} - 9 p q + 27 r = 0$

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Similar questions

x^{3} - p x^{2} + q x - r = 0

9 p q^{2}

(- 3 p q^{2} + 4 p q) - (7 p q - 9 p q^{2}) .

o A

L i^{2 +}

n_{2} = 2

n = 1

(3 r^{2}) \times (9 r^{2})^{3 / 2} \div (27 r^{- 3})^{1 / 3}

r

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