Question
A quadratic equation whose roots are $${\left( {\frac{\gamma }{\alpha }} \right)^2}$$ and $${\left( {\frac{\beta }{\alpha }} \right)^2},$$ where $$\alpha ,\beta ,\gamma $$ are the roots of $${x^3} + 27 = 0,$$ is
A.
$${x^2} - x + 1 = 0$$
B.
$${x^2} + 3x + 9 = 0$$
C.
$${x^2} + x + 1 = 0$$
D.
$${x^2} - 3x + 9 = 0$$
Answer :
$${x^2} + x + 1 = 0$$
Solution :
$$\eqalign{
& {x^3} + 27 = 0 \cr
& \Rightarrow \,\,x = {\left( { - 27} \right)^{\frac{1}{3}}} = - 3, - 3\omega , - 3{\omega ^2} \cr
& \therefore \,\,\frac{\gamma }{\alpha } = {\omega ^2},\frac{\beta }{\alpha } = \omega \,\,\,{\text{or, }}\frac{\gamma }{\alpha } = \frac{1}{\omega },\frac{\beta }{\alpha } = \omega \,\,{\text{or, }}\frac{\gamma }{\alpha } = \frac{1}{{{\omega ^2}}},\frac{\beta }{\alpha } = \frac{1}{\omega }. \cr} $$
In all the cases, $$\frac{\gamma }{\alpha },\frac{\beta }{\alpha }$$ are $$\omega \,\,{\text{or }}{\omega ^2}.$$
∴ the equation is $${x^2} - \left( {\omega + {\omega ^2}} \right)x + \omega \cdot {\omega ^2} = 0.$$