Question

If $a=\cos \left(\frac{2\pi }{7}\right)+i\sin \left(\frac{2\pi }{7}\right)$, 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

The correct option is C

$x^2+x+2=0$

Explanation for the correct option:

Step 1. Find the quadratic equation:

Given, $a=\cos \left(\frac{2\pi }{7}\right)+i\sin \left(\frac{2\pi }{7}\right)$

$\Rightarrow$ $a^7=1$

Step 2. Find Sum of the roots:

$\begin{array}{rcl}ɑ+\beta & =& a+a^2+a^4+a^3+a^5+a^6\end{array}$

It is a GP, where, $a=a,r=a$

$\Rightarrow$$\begin{array}{rcl}\alpha +\beta & =& \frac{a(a^6–1)}{a–1}\end{array}$ $\left[\mathbf{\because }{\mathit{S}}_{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{a}\left(r^n-1\right)}{\mathbf{r}\mathbf{-}\mathbf{1}}\right]$

$\Rightarrow$$\begin{array}{rcl}\alpha +\beta & =& \frac{a^7–a}{a–1}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\alpha +\beta & =& \frac{1–a}{a–1}\end{array}$

$\therefore ɑ+\beta =–1$

Step 3. Find Product of the roots:

$\begin{array}{rcl}ɑ\beta & =& (a+a^2+a^4)(a^3+a^5+a^6)\\ & =& 2a^7+a^4+a^5+a^6+a^7+a^8+a^9+a^{10}\\ & =& 3a^7+a^4+a^5+a^6+a^1+a^2+a^3\\ & =& 3-1=2\end{array}$

$\therefore ɑ\beta =2$

Thus, the required quadratic equation is ${\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$ $\left[\mathbf{\because }{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\left(\alpha +\beta \right)\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{\alpha }\mathit{\beta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\right]$

Hence, Option ‘C’ is Correct.