Question

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

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

Answer 1

The correct option is C r = pq

Given,

$x^3-{px}^2+qx-r=0$

say the cubic equation has roots $a,-a,b$

then

$a+-a+b=(-1)(-p)⟶\left(1\right)$

$a(-a)+a\left(b\right)+(-a)b=+q⟶\left(2\right)$

$a(-a)b=(-1)(-r)⟶\left(3\right)$

$\therefore$ From $\left(1\right)$ we have,

$b=p⟶\left(4\right)$

From $\left(2\right)$ we have

${-a}^2=q⟶\left(5\right)$

From $\left(3\right)$ we have

${-a}^{2}b=r⟶\left(6\right)$

Using $\left(4\right)$ & $\left(5\right)$ we can write $\left(6\right)$ as

$+qp=r$ ($\because$ ${-a}^2=q,b=p$)

$\Rightarrow r=pq$

Option C.