Question

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

• A. $x^2+5x+7=0$
• B. $x^2+5x-7=0$
• C. $x^2-5x+7=0$
• D. $x^2-5x-7=0$

Answer 1

The correct option is A $x^2+5x+7=0$

The sum of the roots is
$=2w+3w^2+2w^2+3w$
$=3(w^2+w)+2(w^2+w)$
$=3(-1)+2(-1)$
$=-5$
$=\frac{-b}{a}$

Hence
$b=5a$ .............................(i) (where $a$ is the coefficient of $x^2$ and $b$ is the coefficient of $x$.)
Similarly product of roots is
$=\frac{c}{a}$
$=(2w+3w^2)(3w+2w^2)$
$=w^2(3w+2)(2w+3)$
$=w^2(6w^2+4w+9w+6)$
$=w^2(6w^2+13w+6)$
$=6w^4+13w^3+6w^2$
$=6w^2(w^2+1)+13$
$=6w^2(-w)+13$

$=13-6$

$=7$

Hence
$c=7a$ where 'c' denotes the constant term

Thus the equation becomes
$a{x}^2+bx+c=0$
$\to a{x}^2+5ax+7a=0$
Or
$x^2+5x+7=0$