Let $a_1,a_0$ be i.i.d. real random variables with uniform distribution in $[-1,1]$. I'm interested in the random zeros of the polynomial $$p(x) = x^2 + a_1x + a_0. $$
One thing (between many) thing I'm particularly interested is in computing the expected value of it's zeros. It's possible to write exactly what are the zeros $z_1,z_2$ of $p(x)$, they are $$ z_1 = \frac{-a_1+\sqrt{a_1^2-4a_0}}{2} $$ and $$ z_2 = \frac{-a_1-\sqrt{a_1^2-4a_0}}{2}. $$
Therefore what I want is to calculate $$E[z_1] = E\Bigg[\frac{-a_1+\sqrt{a_1^2-4a_0}}{2}\Bigg] = \frac{-E[a_1] + E\Big[{\sqrt{a_1^2-4a_0}}\Big]}{2} = \frac{E\Big[{\sqrt{a_1^2-4a_0}}\Big]}{2}.$$
Note that I only need to calculate $E[z_1]$, for $E[z_2] = \overline{E[z_1]}$. Also, doing some experimentation with Matlab, it looks like $E[z_1] = i$.
I want to confirm that result with a mathematical proof, if possible. Thank you.
Edit: From my own computation, we can see that the correct relation between the expected values should be $E[z_2] = -E[z_1]$, not $E[z_2] = \overline{E[z_1]}$.
$$E[z_1]=\frac{E\Big[{\sqrt{a_1^2-4a_0}}\Big]}2\\=\frac18\left(\int_{-1}^1\int_{-1}^1\sqrt{x^2-4y}dydx\right)\\=\frac1{48}\left(\int_{-1}^1(x^2+4)^{\frac32}-(x^2-4)^{\frac32}dx\right)\\=\frac1{24}\left(\int_0^1(4+x^2)^\frac32dx+i\int_0^1(4-x^2)^\frac32dx\right)\\=\frac{1}{24}\left(\left(\frac{11\sqrt5}4+6\sinh^{-1}\frac12\right)+i\left(\frac{9\sqrt3}4+\pi\right)\right)\\\approx0.3765+0.2933i$$