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

Question

Dr. Abubacker · Accepted Answer

(a)
Let the rots of the equation be $\alpha,\ \beta,\ \gamma$. Also, $\alpha = - \beta$ [given]
$	herefore\ \ \alpha + \beta + \gamma = p\  \Rightarrow \  - \beta + \beta + \gamma = p$
$\Rightarrow \ \ \gamma = p$ ...(i)
Now, since $\gamma$ is a root of the equation.
$	herefore$ It satisfies the given equation
$\Rightarrow \ \ \gamma^{3} - p\gamma^{2} + q\gamma - r = 0$
$\Rightarrow p^{3} - pp^{2} + pq - r = 0$ [from Eq. (i)]
$\Rightarrow \ \ \ r = pq$

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

Explanation

Reviewed by Dr. Abubacker

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