2
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ Notice $sin(\theta)cos(\theta) = -\frac{1}{3} = \frac{k}{\sqrt{3}}$, then it comes $-3k = \sqrt{3}$, not $3k = \sqrt{3}$ $\endgroup$ – Larry Sep 15 '18 at 14:40
  • 3
    $\begingroup$ You forget a minus sign in the last 3rd line. It should be $3k = \color{red}{-}\sqrt 3$. $\endgroup$ – xbh Sep 15 '18 at 14:41
  • $\begingroup$ In the last part $$\sin{\theta}\cos{\theta}=-\frac{1}{3}=\frac{k}{\sqrt{3}}\Rightarrow 3k=-\sqrt{3}\Rightarrow k=\frac{-\sqrt{3}}{3}$$ $\endgroup$ – Sepideh Abadpour Sep 15 '18 at 15:00
  • $\begingroup$ @OP Would you please check my answer to close this question? $\endgroup$ – amsmath Sep 18 '18 at 18:33
0
$\begingroup$

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 $k=\tfrac{-\sqrt 3\pm\sqrt 2}{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.