$\sqrt{2} x^2 - \sqrt{3} x +k=0$ 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)
Add (2) to (1),
$\sqrt{2} (\sin^2\theta + \cos^2\theta) - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$ 
$\sqrt{2} - \sqrt{3} (\sin\theta + \cos\theta) +2k=0$ 
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 $\sin\theta\cos\theta=\dfrac k{\sqrt2}$.
$(\sin\theta+\cos\theta)^2-2\sin\theta\cos\theta=1$
$\dfrac32-\sqrt2 k=1$
$k=\dfrac 1{2\sqrt2}$
