My Attempt:
Use
$$\alpha+\beta=-\frac{b}{a}$$
We get
$$\sin{\theta}+\cos{\theta}=\frac{1}{\sqrt{3}}$$
Square
$$\begin{align}(\sin{\theta}+\cos{\theta})^2&=\sin^2{\theta}+\cos^2{\theta}+2\sin{\theta}\cos{\theta}\\&=1+2\sin{\theta}\cos{\theta}
\\&=\frac{1}{3}
\end{align}$$
We get
$$\sin{\theta}\cos{\theta}=-\frac{1}{3}$$
Now, use
$$\alpha\times\beta=\frac{c}{a}$$
We get
$$\sin{\theta}\cos{\theta}=-\frac{1}{3}=\frac{k}{\sqrt{3}}$$
Therefore
$$3k=\sqrt{3}\\k=\frac{\sqrt{3}}{3}$$
The answer is $k=-\frac{\sqrt{3}}{3}$, where did I go wrong?
Best Answer
Your solution is correct (up to the minus sign which was clarified in the comments) unless $\alpha := \sin\theta = \cos\theta$. In this case $\alpha$ is only one solution and you don't know the other. Hence, you cannot use Vieta's formulas. In this case, $\alpha = \pm\tfrac 1{\sqrt 2}$ and $k$ has a different value than $-\sqrt 3/3$, namely $k=\tfrac{-\sqrt 3\pm\sqrt 2}{2}$.