Question

If $\omega $ is an imaginary cube root of unity. Then the equation whose roots are $2\omega + 3{\omega ^2}$ and $2{\omega ^2} + 3\omega $ is

Answer 1

Hint: Using the formula, the sum of roots $ = - \dfrac{{Middle\,term}}{{First\,term}}$ $ = - \dfrac{b}{a}$ , find b in terms of a.
Now, after that, use, the product of roots $ = \dfrac{{Last\,term}}{{First\,term}}$ $ = \dfrac{c}{a}$ , find c in the terms of a.
Now, using the two relations b and c, form an equation using $a{x^2} + bx + c = 0$ .

Complete step-by-step answer:
The given roots of the equation, which is to be formed, are $2\omega + 3{\omega ^2}$ and $2{\omega ^2} + 3\omega $.
Now, in a linear quadratic equation, the sum of roots $ = - \dfrac{{Middle\,term}}{{First\,term}}$ $ = - \dfrac{b}{a}$ .
 $\therefore - \dfrac{b}{a} = \left( {2\omega + 3{\omega ^2}} \right) + \left( {2{\omega ^2} + 3\omega } \right)$
 $
  \therefore - \dfrac{b}{a} = 3{\omega ^2} + 3\omega + 2{\omega ^2} + 2\omega \\
  \therefore - \dfrac{b}{a} = 3\left( {{\omega ^2} + \omega } \right) + 2\left( {{\omega ^2} + \omega } \right) \\
 $
But, ${\omega ^2} + \omega = - 1$
 $
  \therefore - \dfrac{b}{a} = 3\left( { - 1} \right) + 2\left( { - 1} \right) \\
  \therefore - \dfrac{b}{a} = - 1\left( {3 + 2} \right) \\
  \therefore \dfrac{b}{a} = 5 \\
 $
 $\therefore b = 5a$ … (1)
Similarly, in a linear quadratic equation, the product of roots $ = \dfrac{{Last\,term}}{{First\,term}}$ $ = \dfrac{c}{a}$ .
 $
  \therefore \dfrac{c}{a} = \left( {2\omega + 3{\omega ^2}} \right)\left( {3\omega + 2{\omega ^2}} \right) \\
  \therefore \dfrac{c}{a} = \omega \left( {2 + 3\omega } \right) \times \omega \left( {3 + 2\omega } \right) \\
  \therefore \dfrac{c}{a} = {\omega ^2}\left( {2 + 3\omega } \right)\left( {3 + 2\omega } \right) \\
  \therefore \dfrac{c}{a} = {\omega ^2}\left( {6{\omega ^2} + 4\omega + 9\omega + 6} \right) \\
  \therefore \dfrac{c}{a} = {\omega ^2}\left( {6{\omega ^2} + 13\omega + 6} \right) \\
  \therefore \dfrac{c}{a} = 6{\omega ^4} + 13{\omega ^3} + 6{\omega ^2} \\
  \therefore \dfrac{c}{a} = 6{\omega ^4} + 6{\omega ^2} + 13{\omega ^3} \\
  \therefore \dfrac{c}{a} = 6{\omega ^2}\left( {{\omega ^2} + 1} \right) + 13{\omega ^3} \\
 $
But, ${\omega ^2} + 1 = - \omega $
 $
  \therefore \dfrac{c}{a} = 6{\omega ^2}\left( { - \omega } \right) + 13{\omega ^3} \\
  \therefore \dfrac{c}{a} = 13{\omega ^3} - 6{\omega ^3} \\
 $
Also, ${\omega ^3} = 1$
 $
  \therefore \dfrac{c}{a} = 13 - 6 \\
  \therefore \dfrac{c}{a} = 7 \\
 $
 $\therefore c = 7a$ … (2)
Now, the quadratic equation can be formed as $a{x^2} + bx + c = 0$ .
Substituting, $b = 5a$ and $c = 7a$ in the above equation.
 $
  \therefore a{x^2} + 5ax + 7a = 0 \\
  \therefore a\left( {{x^2} + 5x + 7} \right) = 0 \\
  \therefore {x^2} + 5x + 7 = 0 \\
 $
Thus, the quadratic equation is ${x^2} + 5x + 7 = 0$ .

Note: There are 3 cube roots of unity, which are 1, $\dfrac{{ - 1 + \sqrt 3 i}}{2}$ and $\dfrac{{ - 1 - \sqrt 3 i}}{2}$ .
We also write $\dfrac{{ - 1 + \sqrt 3 i}}{2}$ as $\omega $ and $\dfrac{{ - 1 - \sqrt 3 i}}{2}$ as ${\omega ^2}$ .
The properties of cube roots of unity are as follows:
1. $1 + \omega + {\omega ^2} = 0$
2. ${\omega ^3} = 1$
3. ${\omega ^{3n}} = 1,{\omega ^{3n + 1}} = \omega ,{\omega ^{3n + 2}} = {\omega ^2}$
4. $\overline \omega = {\omega ^2},{\left( {\overline \omega } \right)^2} = \omega $ .