Roots - real or complex ?

Quadratic equation whose roots are "a" and "b" is $x^{2}- (a + b)x + ab = 0$ Let $\alpha = a$ , thus :

$x^{2} - ( a + a^{2} +a^{3} + a^{4} + a^{5} + a^{6})x + (a + a^{2} + a^{4})(a^{3} + a^{5} + a^{6})=0$

$\implies x^{2} - ( 1 + a + a^{2} +a^{3} + a^{4} + a^{5} + a^{6} - 1)x + (a + a^{2} + a^{4})(\frac{1}{a^{4}} + \frac{1}{a^{2}} + \frac{1}{a})=0$

(since $a^{7} = 1)$

$\implies x^{2} + x + 3 + a + a^{2} + a^{3} + \frac{1}{a^{2}} + \frac{1}{a^{3}} + \frac{1}{a}=0$

$\implies x^{2} + x +3 + a + a^{2} + a^{3} + a^{4} + a^{5} + a^{6}=0$

$\implies x^{2} + x + 1 + a + a^{2} + a^{3} + a^{4} + a^{5} + a^{6} + 2=0$

$\implies x^{2} + x + 2=0$

Discriminant = $1 - 4.2 = -7$

The seventh roots of unity is given by Euler identity as

$z = e^{\frac{2n\pi}{7}i} = \cos{\frac{2n\pi}{7}} +\sin{\frac{2n\pi}{7}}i$

where $n = 0,1,2...6$ .

This is shown in Argand diagram below.

Therefore, if $\alpha = e^{\frac{2n\pi}{7}}$ is an non-real root, then $\alpha^2$ , $\alpha^3$ , $\alpha^4$ , $\alpha^5$ and $\alpha^6$ are also the non-real roots. (Note that we can choose any non-real root to be $\alpha$ , it will not change the above fact.).

Therefore the six non-real roots (see the Argand diagram) are:

Let $z_1=\alpha+\alpha^2+\alpha^4$ and $z_2=\alpha^3+\alpha^5+\alpha^6$ .

From the Argand diagram we note that:

• $z_1+z_2+1 = 0\quad \Rightarrow \Re (z_1+z_2=-1)\quad \Rightarrow \Im (z_1+z_2) = 0$
• $z_2=\bar{z}_1\quad \Rightarrow \Re (z_1) = \Re (z_2) = -\frac{1}{2}$ $\quad \quad \quad \quad \space \Rightarrow \Im (z_1) = -\Im (z_2) = \sin{\frac{2n\pi}{7}} + \sin{\frac{4n\pi}{7}} - \sin{\frac{6n\pi}{7}} = \frac{\sqrt{7}}{2}$

If $z_1$ and $z_2$ are roots of $x^2+bx+c = 0$ , then:

• $b = -(z_1+z_2) = -(-1) = 1$
• $c = z_1z_2 = (-\frac{1}{2}+\frac{\sqrt{7}}{2}i)(-\frac{1}{2}-\frac{\sqrt{7}}{2}i) = \frac {1}{4}+\frac {7}{4} = 2$

Therefore, $b^2-4c = 1^2 - 4(2) = \boxed{-7}$

Given the quadratic equation: $(x-(\alpha+\alpha^2+\alpha^4))(x-(\alpha^3+\alpha^5+\alpha^6))=ax^2+bx+c=0$ Making computation taking into account $\alpha^7=1$ we have: $a=1 \quad b=-(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6)\quad c=(\alpha+\alpha^2+\alpha^4)(\alpha^3+\alpha^5+\alpha^6)=3+(\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6)$

we know that athe seventh-roots of the unity are the solutions so the equation: $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)=0$

Since $\alpha$ is not real it is a root of the second factor of the expression above, so we have: $1+\alpha+\alpha^2+\alpha^3+\alpha^4+\alpha^5+\alpha^6=0$ Then we have: $a=1 \qquad b=1 \qquad c=2 \quad \Rightarrow \quad b^2-4ac=-7$

If $\alpha$ is a seventh root of unity, we have $\alpha^7 - 1 =0$ which can be factorized as $(\alpha - 1)(1+\alpha + \alpha^2 +\alpha^3 + \alpha^4 + \alpha^5 + \alpha^6) = 0$ .

Since $\alpha$ is a non-real root, then $(\alpha - 1 )$ is non-zero, and $1+\alpha + \alpha^2 +\alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 = 0$ .

Let the roots of the monic quadratic equation be $x_1$ and $x_2$ . Then $1 + x_1 + x_2 = 0$ , which gives us $b = -( x_1 + x_2) = 1$ .

Further $c = x_1 \cdot x_2 = (\alpha + \alpha^2 + \alpha^4)(\alpha^3 + \alpha^5 + \alpha^6) = \alpha^4 + \alpha^5 + \alpha^6 + 3\alpha^7 + \alpha^8 + \alpha^9 + \alpha^9$ $= \alpha^4 \times (1+\alpha + \alpha^2 +\alpha^3 + \alpha^4 + \alpha^5 + \alpha^6) + 2 \alpha^7 = 0 + 2 = 2$

Lastly, a monic quadratic equation has $a = 1$ , and the discriminant becomes $b^2 - 4 a c = 1 - 4\cdot 1 \cdot 2 = -7$