# $\sqrt{2} x^2 - \sqrt{3} x +k=0$ with solutions $\sin\theta$ and $\cos\theta$, find k

If the equation $$\sqrt{2} x^2 - \sqrt{3} x +k=0$$ with $$k$$ a constant has two solutions $$\sin\theta$$ and $$\cos\theta$$ $$(0\leq\theta\leq\frac{\pi}{2})$$, then $$k=$$……

My approach is suggested below but I am not sure how to continue.

Since $$\sin\theta$$ and $$\cos\theta$$ are two solutions of the equation,

Then we have,

$$\sqrt{2} \sin^2\theta - \sqrt{3} \sin\theta +k=0$$ .....Equation (1)

$$\sqrt{2} \cos^2\theta - \sqrt{3} \cos\theta +k=0$$ .....Equation (2)

$$\sqrt{2} (\sin^2\theta + \cos^2\theta) - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$$

$$\sqrt{2} - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$$

The answer key provided is $$\frac{\sqrt{2}}{4}$$. I think I am probably on the right track here but not sure how I should proceed with $$\sin\theta$$ and $$\cos\theta$$ next. Please help.

$$\sin\theta+\cos\theta=\dfrac{\sqrt3}{\sqrt2}$$ and $$\sin\theta\cos\theta=\dfrac k{\sqrt2}$$.

$$(\sin\theta+\cos\theta)^2-2\sin\theta\cos\theta=1$$

$$\dfrac32-\sqrt2 k=1$$

$$k=\dfrac 1{2\sqrt2}$$

• How did you know $\sin\theta+\cos\theta=\dfrac{\sqrt3}{\sqrt2}$ and $\sin\theta\cos\theta=\dfrac k{\sqrt2}$ above? Commented Jun 16, 2019 at 13:33
• Vieta's formulas: If $\alpha$ and $\beta$ are the two roots of the quadratics equation $ax^2+bx+c=0$, then $\alpha\beta=\frac{-b}{a}$ and $\alpha\beta=\frac ca$. Commented Jun 16, 2019 at 13:35
• Using $\sin\theta+\cos\theta=\frac{\sqrt3}{\sqrt 2}$ and $\sqrt2-\sqrt3(\sin\theta+\cos\theta)+2k=0$, we can conclude that $\sqrt2-\sqrt3(\frac{\sqrt3}{\sqrt 2})+2k=0$ and find the value of $k$. Commented Jun 16, 2019 at 13:37
• Vieta's formulas are pretty new to me but thank you, you help a lot! Commented Jun 16, 2019 at 13:46
• If $\alpha$ and $\beta$ are the roots, then $ax^2+bx+c=a(x-\alpha)(x-\beta)=ax^2-a(\alpha+\beta)x+a\alpha\beta$. By comapring the coefficients, $b=-a(\alpha+\beta)$ and $c=a\alpha\beta$. Commented Jun 16, 2019 at 13:48

By the Viete we obtain: $$1=\left(\sqrt{\frac{3}{2}}\right)^2-\sqrt2k.$$ Can you end it now?

I'll use $$y$$ in place of $$\theta$$ . We know that sum of roots of $$ax^2+bx+c$$ is $$\frac{-b}{a}$$ $$siny+cosy=\frac{\sqrt{3}}{\sqrt{2}}$$ also $$siny+cosy=\sqrt{2}sin(y+\frac{\pi}{4})$$ thus $$sin(y+\frac{\pi}{4})=\frac{\sqrt{3}}{2}=sin(\frac{\pi}{3})=sin(\frac{2\pi}{3})$$ thus $$y=\frac{\pi}{12}$$ or $$\frac{5\pi}{12}$$ thus $$sin(\frac{\pi}{12})cos(\frac{\pi}{12})=\frac{1}{2}sin(\frac{\pi}{6})=\frac{k}{\sqrt{2}}$$ thus $$k=\frac{sqrt{2}}{4}$$