Question

The condition that ${\mathrm{x}}^3-{\mathrm{px}}^2+\mathrm{qx}-\mathrm{r}=0$may have two of its roots equal in magnitude but of opposite sign, is

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

Answer 1

The correct option is A

$r=pq$

Finding the roots of the given equation:

Given, ${\mathrm{x}}^3-{\mathrm{px}}^2+\mathrm{qx}-\mathrm{r}=0$

Let $\alpha ,–\alpha ,\beta$ be the three roots for the given equation.

Then sum of roots $=-\frac{coefficientof{x}^2}{coefficientof{x}^3}$

$\alpha +(-\alpha )+\beta =p$

$\Rightarrow$ $\beta =p$

Since $\beta$ is the root of the given equation, substituting it in the given equation,

$$\begin{gathered} \therefore p^3–p^3+qp–r=0 \\ \Rightarrow pq=r \end{gathered}$$

Hence, option (A) is the correct answer.