Question

Let $\alpha + i\beta :\alpha ,\beta {\text{ }} \in {\text{ }}\Re, $be a root of the equation ${x^3} + qx + r = 0;q,r \in \Re$. Find a real cubic equation, independent of $\alpha $ and $\beta $, whose one root is $2\alpha $.
$
  A{\text{ }}{{\text{x}}^3} + rx - r = 0 \\
  B{\text{ }}{{\text{x}}^3} + rx - q = 0 \\
  C{\text{ }}{{\text{x}}^3} + qx - r = 0 \\
  D{\text{ }}{{\text{x}}^3} + 2qx - r = 0 \\
 $

Answer 1

Hint- Here we will proceed by assuming the third root be $\gamma $. Then we will find the sum of roots and product of roots to form the required equation.

Complete step-by-step answer:
As we are given that ${x^3} + qx + r = 0$
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 $.
As the coefficient of ${x^2}$is 0-
Now we will find the sum of roots will be-
$\alpha + i\beta + \alpha - i\beta + \gamma = 0$
$ \Rightarrow 2\alpha = - \gamma $ ………………. (1)
And Product of roots $\left( {{\alpha ^2} + {\beta ^2}} \right)\gamma = - r$
 $ \Rightarrow \left( {{\alpha ^2} + {\beta ^2}} \right) = - \dfrac{r}{\gamma }$ ……….. (2)
Also the sum of any two roots taken at a time-
${\alpha ^2} + {\beta ^2} + \left( {\alpha + i\beta } \right)\gamma + \left( {\alpha - i\beta } \right)\gamma = q$
$ \Rightarrow {\alpha ^2} + {\beta ^2} + 2\alpha \gamma = q$
Now put the value of equation 1 and equation 2,
We get-
$ - \dfrac{r}{\gamma } - {\gamma ^2} = q$
$ \Rightarrow - {\gamma ^3} - q\gamma - r = 0$
Or ${\left( {2\alpha } \right)^3} + q\left( {2\alpha } \right) = r$
So the equation with $2\alpha $as a root is-
$
  {x^3} + qx = r \\
  i.e.{\text{ }}{{\text{x}}^3} + qx - r = 0 \\
 $
Hence option C is correct.
Note- In order to solve this type of question, we must understand that we have to calculate the sum of roots and product of roots so that when we take both sum and product of roots together, we will get the required equation.