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

Hint: In this problem, we have to find the quadratic equation whose roots are $\alpha + {\alpha ^4}$ and ${\alpha ^2} + {\alpha ^3}$. We know that the sum of roots of quadratic equation $a{x^2} + bx + c$ is $ - \dfrac{b}{a}$ and the product of roots is $\dfrac{c}{a}$. We will use this information to find the required equation.


Complete step-by-step solution: In this problem, it is given that $\alpha $ is an imaginary root of the equation ${x^5} - 1 = 0$. That is, $\alpha $ satisfies the equation ${x^5} - 1 = 0$. Therefore, we can write ${\alpha ^5} - 1 = 0 \cdots \cdots \left( 1 \right)$
We know that the factorization of ${x^n} - 1$ is given by ${x^n} - 1 = \left( {x - 1} \right)\left( {1 + x + {x^2} + ... + {x^{n - 2}} + {x^{n - 1}}} \right)$. We will use this information in equation $\left( 1 \right)$. Therefore, we get
${\alpha ^5} - 1 = 0$
$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 + \alpha + {\alpha ^2} + {\alpha ^3} + {\alpha ^4}} \right) = 0$
$ \Rightarrow \alpha - 1 = 0$ or $1 + \alpha + {\alpha ^2} + {\alpha ^3} + {\alpha ^4} = 0\quad \left[ {\because ab = 0 \Rightarrow a = 0} \right.$ or $\left. {b = 0} \right]$
$ \Rightarrow \alpha = 1$ or $1 + \alpha + {\alpha ^2} + {\alpha ^3} + {\alpha ^4} = 0$
It is given that $\alpha $ is an imaginary root. So, we can say that $\alpha \ne 1$ because $\alpha = 1$ is the real root. Therefore, we have only $1 + \alpha + {\alpha ^2} + {\alpha ^3} + {\alpha ^4} = 0 \cdots \cdots \left( 2 \right)$. Now from the equation $\left( 2 \right)$, we can write
$\alpha + {\alpha ^2} + {\alpha ^3} + {\alpha ^4} = - 1$
$ \Rightarrow \left( {\alpha + {\alpha ^4}} \right) + \left( {{\alpha ^2} + {\alpha ^3}} \right) = - 1 \cdots \cdots \left( 3 \right)$
From equation $\left( 3 \right)$, we can say that the sum of two roots $\left( {\alpha + {\alpha ^4}} \right)$ and $\left( {{\alpha ^2} + {\alpha ^3}} \right)$ is $ - 1$.
Let us multiply by $\alpha $ on both sides of the equation $\left( 1 \right)$. Therefore, we get ${\alpha ^6} = \alpha $ and ${\alpha ^7} = {\alpha ^2}$.
Let us find the product of two roots $\left( {\alpha + {\alpha ^4}} \right)$ and $\left( {{\alpha ^2} + {\alpha ^3}} \right)$ . Therefore, we get
$\left( {\alpha + {\alpha ^4}} \right)\left( {{\alpha ^2} + {\alpha ^3}} \right)$
$ = {\alpha ^3} + {\alpha ^4} + {\alpha ^6} + {\alpha ^7}$
$ = {\alpha ^3} + {\alpha ^4} + \alpha + {\alpha ^2}\quad \left[ {\because {\alpha ^6} = \alpha ,{\alpha ^7} = {\alpha ^2}} \right]$
$ = \alpha + {\alpha ^2} + {\alpha ^3} + {\alpha ^4}$
$ = - 1$
Note that here we write $\alpha + {\alpha ^2} + {\alpha ^3} + {\alpha ^4} = - 1$ from equation $\left( 2 \right)$.
Now we have the following information:
$\left( 1 \right)$ The sum of two roots $\left( {\alpha + {\alpha ^4}} \right)$ and $\left( {{\alpha ^2} + {\alpha ^3}} \right)$ is $ - 1$.
$\left( 2 \right)$ The product of two roots $\left( {\alpha + {\alpha ^4}} \right)$ and $\left( {{\alpha ^2} + {\alpha ^3}} \right)$ is $ - 1$.
If the sum and product of two roots is known then we can say that the quadratic equation will be
${x^2} - $ (sum of roots) $x + $ (product of roots) $ = 0$
Therefore, the required equation is
${x^2} - \left( { - 1} \right)x + \left( { - 1} \right) = 0$
$ \Rightarrow {x^2} + x - 1 = 0$
Therefore, if $\alpha $ is an imaginary root of ${x^5} - 1 = 0$ then the equation is ${x^2} + x - 1 = 0$ whose roots are $\alpha + {\alpha ^4}$ and ${\alpha ^2} + {\alpha ^3}$.
Therefore, option B is true.


Note: The standard form of quadratic equation is $a{x^2} + bx + c = 0$ where $a \ne 0$. Every quadratic equation has exactly two roots. For the roots of quadratic equation, there are three possible cases:
$\left( 1 \right)$ Roots are real and distinct $\left( 2 \right)$ Roots are real and equal $\left( 3 \right)$ Roots are imaginary numbers.