First slide

Theory of equations

Question

If the roots of the equation $x^{3} + {px}^{2} + qx - 1 = 0$ form an increasing G.P., where p and q are real, then

Moderate

A

p + q = 0
B

$p \in (- 3, \infty)$
C

one of the roots is unity
D

one root is smaller than 1 and one root is greater than 1

Solution

Let the roots be a/r, a, ar, where a > 0, r > 1. Now,
$\begin{array}{l} a / r + a + ar = - p (1) \\ a (a / r) + a (ar) + (ar) (a / r) = q (2) \\ (a / r) (a) (ar) = 1 (3) \end{array}$
$\begin{array}{ll} \Rightarrow a^{3} = 1 \\ \Rightarrow a = 1 \end{array}$
Hence, (3) is correct. From (1), putting a = 1, we get
$- p - 3 > 0 (∵ r + \frac{1}{r} > 2) \Rightarrow p < - 3$
Hence, (2) is not correct. Also,
$1 / r + 1 + r = - p$ (4)
From (2), putting a = 1, we get
$1 / r + r + 1 = q$ (5)
From (4) and (5), we have
$- p = q \Rightarrow p + q = 0$
Hence, (1) is correct. Now, as r > 1
$\begin{matrix} a / r = 1 / r < 1 \\ and ar = r > 1 \end{matrix}$
Hence, (4) is correct.

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