QJEEM23Ma1886: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1886 Solution: Let be the roots of the quadratic equation . Then is equal to :

Given quadratic equation: $x^{2}+\sqrt{6} x+3=0$

We can find the roots using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$ , $b = \sqrt{6}$ , and $c = 3$ .

Substituting these values into the quadratic formula, we have:

$x = \frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4(1)(3)}}{2(1)}$

Simplifying further :

$x = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2}$

$x = \frac{-\sqrt{6} \pm \sqrt{-6}}{2}$

Since the discriminant is negative, the roots are complex numbers. We can express them using the imaginary unit $i$ :

$x = \frac{-\sqrt{6} \pm \sqrt{6}i}{2}$

$\Rightarrow$ $x = \frac{-1}{2}\sqrt{6} \pm \frac{1}{2}\sqrt{6}i$

$\therefore$ $\alpha, \beta=\sqrt{3} \mathrm{e}^{ \pm \frac{3 \pi \mathrm{i}}{4}}$ .

The required expression can be rewritten in terms of the argument of the exponential form of the roots, which simplifies the calculation:

$\begin{aligned}& =\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\\end{aligned}$

Here, the exponential power of $\sqrt{3}$ in the numerator is larger by 8 compared to the denominator, so we can divide the numerator and denominator by $(\sqrt{3})^{8} = 81$ to simplify:

$\begin{aligned}& =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\\end{aligned}$

Since the cosine function has a period of $2\pi$ , we can reduce the arguments of the cosine function in the numerator and denominator.

We have $69\pi/4 = \pi/4 + 17\pi = \pi/4$ ,

$42\pi/4 = 2\pi/4 + 10\pi = \pi/2$ ,

$45\pi/4 = \pi/4 + 11\pi = \pi/4$ , and

$30\pi/4 = 2\pi/4 + 7\pi = \pi/2$ .

Therefore, the required expression simplifies to :

$\begin{aligned}& =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{\pi}{2}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{\pi}{2}\right)} \\\\& =\frac{(\sqrt{3})^{15}\sqrt{2}+(\sqrt{3})^{6}\cdot 0}{(\sqrt{3})^{7}\sqrt{2}+(\sqrt{3})^{2}\cdot 0} \\\\& =\frac{(\sqrt{3})^{15}\sqrt{2}}{(\sqrt{3})^{7}\sqrt{2}} \\\\& = (\sqrt{3})^{15-7} \\\\& = (\sqrt{3})^{8} \\\\& = 81.\end{aligned}$