Question

The condition that $ x^3 - px^2 + qx - r = 0 $ may have two of its roots equal to each other, but opposite in sign is

• A. $ r = pq $
• B. $ r = 2p^3 + pq $
• C. $ r = p^2q $
• D. $None \,of\, these$

Answer 1

Answer: (A)

The given cubic equation is
$x^{3}-px^{2}+qx-r=0$
Given condition, two roots are equal, but opposite in sign
ie, $(\alpha, -\alpha, \beta)$
Sum of roots $=\alpha-\alpha+\beta=p$
$\beta=P$
$‘\beta’$ is the roots of given cubic equation, so it satisfies
$p^{3}-p^{3}+pq-r=0$
$pq=r$