f(x) = [2x-x³ if 0 <= x <= sqrt(2), 0 otherwise. Calculate the median.

The equation derived from your work is:

$$4m^2 - m^4 = 2$$

Rearrange this into a standard quartic form, which can be treated as a quadratic in $m^2$:

$$m^4 - 4m^2 + 2 = 0$$

Let $y = m^2$. Substituting this into the equation gives a quadratic equation in $y$:

$$y^2 - 4y + 2 = 0$$

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-4$, and $c=2$:

$$y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}$$ $$y = \frac{4 \pm \sqrt{16 - 8}}{2}$$ $$y = \frac{4 \pm \sqrt{8}}{2}$$ $$y = \frac{4 \pm 2\sqrt{2}}{2}$$ $$y = 2 \pm \sqrt{2}$$

Since $y = m^2$, we have two possible values for $m^2$:

$$m^2 = 2 + \sqrt{2} \quad \text{or} \quad m^2 = 2 - \sqrt{2}$$

Taking the square root for $m$:

$$m = \sqrt{2 + \sqrt{2}} \quad \text{or} \quad m = \sqrt{2 - \sqrt{2}}$$

The probability density function is defined for $0 \le x \le \sqrt{2}$. The median $m$ must lie within this interval.
We know that $\sqrt{2} \approx 1.414$.
For $m = \sqrt{2 + \sqrt{2}}$:
$m \approx \sqrt{2 + 1.414} = \sqrt{3.414} \approx 1.848$. This value is greater than $\sqrt{2}$, so it is not a valid median.
For $m = \sqrt{2 - \sqrt{2}}$:
$m \approx \sqrt{2 - 1.414} = \sqrt{0.586} \approx 0.765$. This value is between 0 and $\sqrt{2}$, so it is the valid median.

The median is:

$$\sqrt{2 - \sqrt{2}}$$