Question

If $a = \cos \left( {\dfrac{{2\pi }}{7}} \right) + i\sin \left( {\dfrac{{2\pi }}{7}} \right)$ , then the quadratic equation whose roots are $\alpha = a + {a^2} + {a^4}$ and $\beta = {a^3} + {a^5} + {a^6}$, is
A. ${x^2} - x + 2 = 0$
B. ${x^2} + 2x + 2 = 0$
C. ${x^2} + x + 2 = 0$
D. ${x^2} + x - 2 = 0$

Answer 1

Hint: The general form of quadratic equation where the roots are know is given by ${x^2} - (\alpha + \beta )x + \alpha \beta = 0$. As in the given question, we know the values of the roots of the equation, first, we sum those two roots and then the product of the roots. The vales will be substituted in the general formula.

Formula Used:
The Sum of nth term of G.P. $S_n=\dfrac{a(r^{n}-1)}{r-1}$
For the quadratic equation: $ax^{2} + bx+ c=0$,
Sum of roots $\alpha+\beta=-\dfrac{b}{a}$
Product of Roots $\alpha\beta=\dfrac{c}{a}$

Complete step by step Solution:
Consider the given data; $a = \cos \left( {\dfrac{{2\pi }}{7}} \right) + i\sin \left( {\dfrac{{2\pi }}{7}} \right)$. The roots of the equations are $\alpha = a + {a^2} + {a^4}$ and $\alpha = {a^3} + {a^5} + {a^6}$.
First, we determine the sum of the roots of the equation
$\alpha + \beta = a + {a^2} + {a^4} + {a^3} + {a^5} + {a^6}$
On rearranging the terms we have
$\alpha + \beta = a + {a^2} + {a^3} + {a^4} + {a^5} + {a^6}$
The above expression is in the form of G.P, the sum of the above terms will be
$\alpha + \beta = \dfrac{{a({a^6} - 1)}}{{a - 1}}$
$\alpha + \beta = \dfrac{{{a^7} - a}}{{a - 1}}$
Now we will determine the value of \[{a^7}\], given that $a = \cos \left( {\dfrac{{2\pi }}{7}} \right) + i\sin \left( {\dfrac{{2\pi }}{7}} \right)$, this can be written as ${e^{i\dfrac{{2\pi }}{7}}}$ and ${a^7} = {\left( {{e^{i\dfrac{{2\pi }}{7}}}} \right)^7} = {e^{i2\pi }}$
The value of ${e^{i2\pi }}$ is $1$.
So the sum of the roots of the equation will be
$\alpha + \beta = \dfrac{{1 - a}}{{a - 1}}$
Taking minus as a common we get
$\alpha + \beta = \dfrac{{ - (a - 1)}}{{a - 1}}$
By simplifying we get
$\alpha + \beta = - 1$
Now we will find the value of the product of the roots of the equation.
$\alpha \beta = (a + {a^2} + {a^4})({a^3} + {a^5} + {a^6})$
On multiplying
$\alpha \beta = 2{a^7} + {a^4} + {a^5} + {a^6} + {a^7} + {a^8} + {a^9} + {a^{10}}$
As we know that ${a^7} = 1$, so we have
$\alpha \beta = 2 + {a^4} + {a^5} + {a^6} + {a^7} + {a^8} + {a^9} + {a^{10}}$
Other terms are in the form of geometric series. On applying the formula we get
$\alpha \beta = 2 + \dfrac{{{a^4}({a^7} - 1)}}{{a - 1}}$
As we know that ${a^7} = 1$, so we have
$\alpha \beta = 2 + \dfrac{{{a^4}(1 - 1)}}{{a - 1}}$
By simplifying we get
$\alpha \beta = 2$
Therefore, the quadratic equation is ${x^2} + x + 2 = 0$.

Hence, the correct option is C.

Note:The complex number $\cos \theta + i\sin \theta $ can be written as ${e^{i\theta }}$. The value of ${e^{i2\pi }} = \cos 2\pi + i\sin 2\pi = 1$. Remember the general formula $\cos n\pi = {( - 1)^n}$ and $\sin n\pi = 0$. This is only applicable that n should be a natural number, suppose if it is in the form of a fraction then we can’t use this formula.