Question

Let $\alpha$ and $\beta$ be the roots of the equation $x^2+x+1=0$. The equation whose roots are ${\alpha }^{19},{\beta }^7$ is

• A. $x^2-x-1=0$
• B. $x^2-x+1=0$
• C. $x^2+x-1=0$
• D. $x^2+x+1=0$

Answer 1

Answer: (D)

$x^2+x+1=0$

$x^2+x+1=0$

$\Rightarrow (x-\omega )(x-{\omega }^2)=0$

$\Rightarrow x=\omega ,{\omega }^2$

$\therefore \alpha =\omega ,\beta ={\omega }^2$ $(\because \omega ,{\omega }^2$ are cube roots of unity $)$

Hence, ${\alpha }^3={\omega }^3=1$

${\beta }^3=[{\omega }^3{]}^2=1$

$\alpha \beta ={\omega }^3=1$

$\therefore {\alpha }^{19}={\left({\alpha }^3\right)}^6\alpha =1^6\alpha =\alpha =\omega$ and ${\beta }^7={\beta }^6.\beta =1^2.\beta =\beta ={\omega }^2$

$\Rightarrow {\alpha }^{19}+{\beta }^7=\omega +{\omega }^2=-1$

$\Rightarrow {\alpha }^{19}{\beta }^7=\omega .{\omega }^2={\omega }^3=1$

Hence equation whose roots are ${\alpha }^{19},{\beta }^7$ is

$x^2-({\alpha }^{19}+{\beta }^7)x+{\alpha }^{19}{\beta }^7=0$

$\Rightarrow x^2+x+1$