Question

Let $\alpha +i\beta :\alpha ,\beta \in R$ , be a root of the equation $x^3+qx+r=0$ ; $q,r\in R$. Find a real cubic equation, independent of $\alpha$ and $\beta$ , whose one root is $2\alpha$

• A. $x^3+rx-r=0$
• B. $x^3+rx-q=0$
• C. $x^3+qx-r=0$
• D. $x^3+2qx-r=0$

Answer 1

Answer: (C)

$x^3+qx-r=0$

Given equation $x^3+qx+r=0$
Given, $\alpha +i\beta$ is a root of the equation.
$\Rightarrow \alpha -i\beta$ will be the other root.
Let the third root be $\gamma$

Now, sum of roots $\alpha +i\beta +\alpha -i\beta +\gamma =0$ ....(As coefficient of $x^2$ is $0$)
$\Rightarrow 2\alpha =-\gamma$ .....(1)
Product of roots $({\alpha }^2+{\beta }^2)\gamma =-r$
$\Rightarrow ({\alpha }^2+{\beta }^2)=-\frac{r}{\gamma }$ .....(2)
Also, sum of any two roots taken at a time
${\alpha }^2+{\beta }^2+(\alpha +i\beta )\gamma +(\alpha -i\beta )\gamma =q$
$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \gamma =q$
Put the values from (1) and (2), we get
$-\frac{r}{\gamma }-{\gamma }^2=q$
$\Rightarrow -{\gamma }^3-q\gamma -r=0$
$\Rightarrow (2\alpha {)}^3+q(2\alpha )=r$
So, the equation with $2\alpha$ as a root is
$x^3+qx=r$

i.e. $x^3+qx-r=0$