
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
161.7k+ views
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.
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.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
