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# 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 $2\alpha$.$A{\text{ }}{{\text{x}}^3} + rx - r = 0 \\ B{\text{ }}{{\text{x}}^3} + rx - q = 0 \\ C{\text{ }}{{\text{x}}^3} + qx - r = 0 \\ D{\text{ }}{{\text{x}}^3} + 2qx - r = 0 \\$  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.

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.
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