First slide

Theory of equations

Question

lf the roots of the equation $x^{3} - p x^{2} + q x - r = 0$ are in H.P., then

Moderate

A

$27 r^{2} + 9 p q r + 2 q^{3} = 0$
B

$27 r^{2} - 9 p q r + 2 q^{3} = 0$
C

$2 r^{2} - 9 p q r + 27 q^{3} = 0$
D

$27 r^{2} - 9 p q r - 2 q^{3} = 0$

Solution

The equation whose roots are reciprocals of the roots of the given equation is given by
$\frac{1}{x^{3}} - \frac{p}{x^{2}} + \frac{q}{x} - r = 0 \Rightarrow r x^{3} - q x^{2} + p x - 1 = 0$
Since the roots of the given equation are in H.P. Therefore, the roots of equation (i) are in A.P. Let its roots be $a - d,$ $a$ and $a + d$ . Then
Then,
$(a - d) + a + (a + d) = - (- \frac{1}{r}) \Rightarrow 3 a = \frac{q}{r} \Rightarrow a = \frac{9}{3 r}$
Since a is a root of equation { i).
$\begin{array}{l} ∴ r a^{3} - q a^{2} + p a - 1 = 0 \\ \Rightarrow r {(\frac{q}{3 r})}^{3} - q {(\frac{q}{3 r})}^{2} + p (\begin{array}{c} q \\ 3 r \end{array}) - 1 = 0 \\ \Rightarrow \frac{q^{3}}{27 r^{2}} - \frac{q^{3}}{9 r^{2}} + \frac{p q}{3 r} - 1 = 0 \\ \Rightarrow q^{3} - 3 q^{3} + 9 p q r - 27 r^{2} = 0 \Rightarrow 27 r^{2} - 9 p q r + 2 q^{3} = 0 \end{array}$

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