How to solve $\sqrt{6}\cdot x^4 - (\sqrt{3}+\frac{3}{2}\sqrt{2})x^2 +\frac{3}{2} = 0 $

Question

Of course I could put this is mathematica/wolframalpha or use a formula, but I think in here is a trick how to solve it very simply, but I can't figure it out. Help/Hints very appreciated

$\sqrt{6}\cdot x^4 - (\sqrt{3}+\frac{3}{2}\sqrt{2})x^2 +\frac{3}{2} = 0 \Leftrightarrow $

$x^2(\sqrt{6}x^2-\sqrt{3}+ \frac{3}{2}\sqrt{2})= -\frac{3}{2}\Leftrightarrow $

Answer 1

1. Set $y=x^2$.

2. Find $y$, using the quadratic formula.

3. Solve for $x$, from $y=x^2$.

Answer 2

As there is no $x^3$ or $x$ term you can treat it as a quadratic in $x^2$.

Let $t=x^2$ as dvix suggested.

$$\sqrt{6}\cdot t^2-\sqrt{3}t-\frac{3}{2}\sqrt{2}t+\frac{3}{2}=0$$

Then factorize as suggested by G.Sassetelli.

$$\sqrt{3}t(\sqrt{2}t-1)-\frac{3}{2}(\sqrt{2}t-1)=0$$

$$\left(\sqrt{3}t-\frac{3}{2}\right)\left(\sqrt{2}t-1\right)=0$$

$$t=\frac{\sqrt{3}}{2}\text{ or }t=\frac{1}{\sqrt{2}}$$

Note: This makes me think the original question was trig related so it might have been useful to include your steps which let up to the equation you were trying to solve.

$$x^2=\frac{\sqrt{3}}{2}\text{ or }x^2=\frac{1}{\sqrt{2}}$$

$$x=\frac{\sqrt[4]{3}}{\sqrt{2}}\text{ or }x=\frac{1}{\sqrt[4]{2}}$$

Or with rational denominators:

$$x=\frac{\sqrt[4]{12}}{2}\text{ or }x=\frac{\sqrt[4]{8}}{2}$$