# Finding a pdf using transformations

Let $X$ be a random variable that is $X \sim \mathrm{Unif}(0,1) = 1$. Use a transformation method to find the pdf of $U = X(1 - X)$.

I tried solving for $X$ and I got $X = \dfrac{1 \pm \sqrt{1-4U}}{2}$

But the actual pdf is $\dfrac{2}{\sqrt{1-4U}}$, so I can't just take $X'$

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• $U$ can take values in $(0,\frac14)$ and two different values of $X$ give the same $U$
• $\Pr(X \le x)=x$ and $\Pr(X \ge x)=1-x$ for $x$ in $(0,1)$
• $\Pr(U \le u) = \Pr\left(X \le \frac{1 - \sqrt{1-4u}}{2} \right) +\Pr\left(X \ge \frac{1 + \sqrt{1-4u}}{2} \right) = 1 - \sqrt{1-4u}$
• $p(u) = \frac{d }{du} \Pr(U \le u)= \dfrac{2}{\sqrt{1-4u}}$.