Key Concepts and Formulas

• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Complex Numbers and Polar Form: A complex number $z = x + iy$ can be represented in polar form as $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$, where $r = |z| = \sqrt{x^2 + y^2}$ is the modulus and $\theta = \arg(z)$ is the argument.
• De Moivre's Theorem: For a complex number in polar form $z = r(\cos\theta + i\sin\theta)$, its $n$-th power is $z^n = r^n(\cos(n\theta) + i\sin(n\theta))$. Also, if $\alpha = re^{i\theta}$, then $\alpha^n + \overline{\alpha}^n = 2r^n\cos(n\theta)$.

Step 1: Finding the Roots of the Quadratic Equation

We are given the quadratic equation $x^{2}+\sqrt{6} x+3=0$. We will use the quadratic formula to find its roots.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=\sqrt{6}$, and $c=3$, we get: $x = \frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4(1)(3)}}{2(1)} = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2} = \frac{-\sqrt{6} \pm \sqrt{-6}}{2} = \frac{-\sqrt{6} \pm i\sqrt{6}}{2}$ Thus, the roots are: $\alpha = \frac{-\sqrt{6} + i\sqrt{6}}{2} = -\frac{\sqrt{6}}{2} + i\frac{\sqrt{6}}{2}$ $\beta = \frac{-\sqrt{6} - i\sqrt{6}}{2} = -\frac{\sqrt{6}}{2} - i\frac{\sqrt{6}}{2}$

Step 2: Converting Roots to Polar Form

Converting $\alpha = -\frac{\sqrt{6}}{2} + i\frac{\sqrt{6}}{2}$ to polar form $re^{i\theta}$:

1. Calculate the Modulus ($r$): $r = |\alpha| = \sqrt{\left(-\frac{\sqrt{6}}{2}\right)^2 + \left(\frac{\sqrt{6}}{2}\right)^2} = \sqrt{\frac{6}{4} + \frac{6}{4}} = \sqrt{\frac{12}{4}} = \sqrt{3}$

2. Calculate the Argument ($\theta$): Since the real part is negative and the imaginary part is positive, $\alpha$ lies in the second quadrant. The reference angle is $\arctan\left(\left|\frac{\text{Im}(\alpha)}{\text{Re}(\alpha)}\right|\right) = \arctan\left(\left|\frac{\sqrt{6}/2}{-\sqrt{6}/2}\right|\right) = \arctan(1) = \frac{\pi}{4}$. So, $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

Thus, $\alpha = \sqrt{3}e^{i\frac{3\pi}{4}}$. Since $\beta$ is the complex conjugate of $\alpha$, $\beta = \sqrt{3}e^{-i\frac{3\pi}{4}}$.

Step 3: Applying De Moivre's Theorem and Simplifying

We want to evaluate $\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$ Using $\alpha^n + \beta^n = 2r^n\cos(n\theta)$ with $r = \sqrt{3}$ and $\theta = \frac{3\pi}{4}$: $E = \frac{2(\sqrt{3})^{23}\cos\left(23\cdot\frac{3\pi}{4}\right) + 2(\sqrt{3})^{14}\cos\left(14\cdot\frac{3\pi}{4}\right)}{2(\sqrt{3})^{15}\cos\left(15\cdot\frac{3\pi}{4}\right) + 2(\sqrt{3})^{10}\cos\left(10\cdot\frac{3\pi}{4}\right)}$ $E = \frac{(\sqrt{3})^{23}\cos\left(\frac{69\pi}{4}\right) + (\sqrt{3})^{14}\cos\left(\frac{42\pi}{4}\right)}{(\sqrt{3})^{15}\cos\left(\frac{45\pi}{4}\right) + (\sqrt{3})^{10}\cos\left(\frac{30\pi}{4}\right)}$

Step 4: Further Simplification and Angle Reduction

We can factor out $(\sqrt{3})^{10}$ from both the numerator and the denominator: $E = \frac{(\sqrt{3})^{10}\left((\sqrt{3})^{13}\cos\left(\frac{69\pi}{4}\right) + (\sqrt{3})^{4}\cos\left(\frac{42\pi}{4}\right)\right)}{(\sqrt{3})^{10}\left((\sqrt{3})^{5}\cos\left(\frac{45\pi}{4}\right) + \cos\left(\frac{30\pi}{4}\right)\right)}$ $E = \frac{(\sqrt{3})^{13}\cos\left(\frac{69\pi}{4}\right) + (\sqrt{3})^{4}\cos\left(\frac{21\pi}{2}\right)}{(\sqrt{3})^{5}\cos\left(\frac{45\pi}{4}\right) + \cos\left(\frac{15\pi}{2}\right)}$

Now we simplify the angles:

• $\frac{69\pi}{4} = \frac{(64+5)\pi}{4} = 16\pi + \frac{5\pi}{4} \implies \cos\left(\frac{69\pi}{4}\right) = \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
• $\frac{21\pi}{2} = \frac{(20+1)\pi}{2} = 10\pi + \frac{\pi}{2} \implies \cos\left(\frac{21\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$
• $\frac{45\pi}{4} = \frac{(40+5)\pi}{4} = 10\pi + \frac{5\pi}{4} \implies \cos\left(\frac{45\pi}{4}\right) = \cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
• $\frac{15\pi}{2} = \frac{(14+1)\pi}{2} = 7\pi + \frac{\pi}{2} \implies \cos\left(\frac{15\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0$

Substitute these values back into the expression: $E = \frac{(\sqrt{3})^{13}\left(-\frac{\sqrt{2}}{2}\right) + (\sqrt{3})^{4}(0)}{(\sqrt{3})^{5}\left(-\frac{\sqrt{2}}{2}\right) + (0)} = \frac{(\sqrt{3})^{13}\left(-\frac{\sqrt{2}}{2}\right)}{(\sqrt{3})^{5}\left(-\frac{\sqrt{2}}{2}\right)} = \frac{(\sqrt{3})^{13}}{(\sqrt{3})^{5}} = (\sqrt{3})^{13-5} = (\sqrt{3})^8$

Step 5: Final Calculation

$E = (\sqrt{3})^8 = (3^{1/2})^8 = 3^4 = 81$

Common Mistakes & Tips

• Always double-check the quadrant of the complex number when finding the argument.
• Remember to correctly apply De Moivre's Theorem.
• Be careful when reducing angles using periodicity. Ensure you are accounting for the $\pi$ shifts.

Summary

We found the roots of the quadratic equation, converted them to polar form, and then used De Moivre's theorem to simplify the expression. By carefully simplifying the trigonometric terms and using exponent rules, we arrived at the final answer.

Final Answer

The final answer is $\boxed{81}$, which corresponds to option (D).