Question

If $\alpha$ is an imaginary root of $x^5-1=0$ then the equation whose roots are $\alpha +{\alpha }^4$ and ${\alpha }^2+{\alpha }^3$ 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: (B)

$x^2+x-1=0$

$x^5-1=0$

$\Rightarrow {\alpha }^5-1=0$

$\Rightarrow (\alpha -1)({\alpha }^4+{\alpha }^3+{\alpha }^2+\alpha +1)=0$

$\Rightarrow ({\alpha }^4+{\alpha }^3+{\alpha }^2+\alpha +1)=0$ , $\alpha$ is non-real

For the required equation,

Sum of roots $=\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4=-1$

Product of roots $=(\alpha +{\alpha }^4)({\alpha }^2+{\alpha }^3)$

$={\alpha }^3+{\alpha }^4+{\alpha }^6+{\alpha }^7$

$={\alpha }^3+{\alpha }^4+\alpha +{\alpha }^2$

$=-1$

Thus, the quadratic equation formed is $x^2+x-1=0$.