Let , be the roots of the quadratic equation x^2+√6x+3=0.

Question

Let $\alpha ,\beta $ be the roots of the quadratic equation ${x}^{2}+\sqrt{6}x+3=0$. Then $\frac{{\alpha }^{23}+{\beta }^{23}+{\alpha }^{14}+{\beta }^{14}}{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}}$ is equal to :

Accepted Answer

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 :

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

Solutions · 1