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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?

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    $\begingroup$ 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, 2018 at 14:40
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    $\begingroup$ You forget a minus sign in the last 3rd line. It should be $3k = \color{red}{-}\sqrt 3$. $\endgroup$
    – xbh
    Sep 15, 2018 at 14:41
  • $\begingroup$ In the last part $$\sin{\theta}\cos{\theta}=-\frac{1}{3}=\frac{k}{\sqrt{3}}\Rightarrow 3k=-\sqrt{3}\Rightarrow k=\frac{-\sqrt{3}}{3}$$ $\endgroup$ Sep 15, 2018 at 15:00
  • $\begingroup$ @OP Would you please check my answer to close this question? $\endgroup$
    – amsmath
    Sep 18, 2018 at 18:33

1 Answer 1

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

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