The condition that x3 - px2 + qx - r = 0 may have two of its roots equal to each other, but opposite in sign is

Question

The condition that x3 - px2 + qx - r = 0 may have two of its roots equal to each other, but opposite in sign is (A)  r = pq  (B)  r = 2p3 + pq  (C)  r = p2q  (D) None of these

Tardigrade Admin · Accepted Answer

The given cubic equation is x3-px2+qx-r=0 Given condition, two roots are equal, but opposite in sign ie, (α, -α, β) Sum of roots =α-α+β=p β=P ‘β’ is the roots of given cubic equation, so it satisfies p3-p3+pq-r=0 pq=r

Q. The condition that $x^{3} - p x^{2} + q x - r = 0$ may have two of its roots equal to each other, but opposite in sign is

$r = pq$

$r = 2 p^{3} + pq$

$r = p^{2} q$

$N o n e o f t h ese$

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

Q. The condition that x3−px2+qx−r=0 may have two of its roots equal to each other, but opposite in sign is

r=pq

r=2p3+pq

r=p2q

Noneofthese