Probability that a quadratic polynomial with random coefficients has real roots

Hint: You are looking for the volume of the $(a,b,c) \in [0,1]^3$ such that $b^2 \geq 4ac$.

Hints: First consider $B^2 \geq 4AC$. Now, if $U$ is uniform$(0,1)$, then $-\log(U)$ is exponential$(1)$; further, the sum of two independent exponential$(1)$ random variables has pdf $x e^{-x}$, $x > 0$. Thus, using the law of total probability, the answer can be found by solving an elementary one dimensional integral. I've confirmed the result by simulations.

EDIT: Specifically, $$ {\rm P}(B^2 - 4AC \ge 0) = \int_{\log 4}^\infty {[1 - e^{ - (x - \log 4)/2} ]xe^{ - x} \,{\rm d}x} = \frac{{5 + 3\log 4}}{{36}} \approx 0.2544134. $$ The answer was confirmed using Monte Carlo simulations: $N=10^8$ runs gave average of $0.25444043$.

EDIT: Note that it is quite easy to determine, moreover, the distribution of the product of $n$ independent uniform$(0,1)$ random variables. Indeed, let $U_1,\ldots,U_n$ be independent uniform$(0,1)$ variables. Write $$ U_1 \cdots U_n = \exp \Big[ - \sum\nolimits_{i = 1}^n { - \log U_i } \Big]. $$ Since the $-\log U_i$ are independent exponential$(1)$ variables, $U_1 \cdots U_n$ is merely distributed as $e^{-X}$, where $X$ has gamma pdf $x^{n-1}e^{-x}/(n-1)!$, $x > 0$.

EDIT: Elaborating in response to the OP's request (see the first comment below).

Actually, the hint was supposed to send you in a slightly different direction, that is to consider the probability $$ {\rm P}\bigg( - \log B \le \frac{{( - \log A) + ( - \log C) - \log 4}}{2}\bigg), $$ or $$ {\rm P}\bigg(X \le \frac{{Y - \log 4}}{2}\bigg), $$ where $X$ is exponential$(1)$ and $Y$, independent of $X$, has gamma pdf $f_Y (x) = xe^{-x}$, $x > 0$. Then, by the law of total probability (and using that $X$ and $Y$ are independent), the above probability is given by $$ \int_0^\infty {{\rm P}\bigg(X \le \frac{{Y - \log 4}}{2}\bigg|Y = x\bigg)f_Y (x)\,{\rm d}x} = \int_0^\infty {{\rm P}\bigg(X \le \frac{{x - \log 4}}{2}\bigg)xe^{ - x} \,{\rm d}x}, $$ and so substituting the exponential$(1)$ cdf yields the desired integral.

I liked Shai Covo's approach of taking the product distribution to a sum. This will help to extend it to larger number of variables as well.

In this particular case, we can also use the following method (if one is averse to exponents :-)).

Given $A,B$ and $C$ are i.i.d and Uniform(0,1), \begin{eqnarray*} f_{A}(a) &=& \begin{cases} 1, & a \in [0,1] \\ 0, & otherwise \end{cases} \\ f_{B}(b) &=& \begin{cases} 1, & b \in [0,1] \\ 0, & otherwise \end{cases} \\ f_{C}(c) &=& \begin{cases} 1, & c \in [0,1] \\ 0, & otherwise \end{cases} \end{eqnarray*}

The product $U=AC$ follows, \begin{eqnarray*} \mathbb{P}\left(U \le u \right) &=& \mathbb{P}\left( AC \le u \right) \\ &=& \int_{c=0}^{1}{\mathbb{P}\left( A \le \frac{u}{c} \right) f_{C}(c) dc} \\ &=& \int_{c=0}^{u}{\mathbb{P}\left( A \le \frac{u}{c} \right) f_{C}(c) dc} + \int_{c=u}^{1}{\mathbb{P}\left( A \le \frac{u}{c} \right) f_{C}(c) dc} \\ &=& \int_{c=0}^{u}{ dc} + \int_{c=u}^{1}{ \frac{u}{c} dc} \\ &=& u +(1 \log 1- u \log u) \\ &=& u(1-\log u ) \end{eqnarray*}

Note: We can get the product distribution as well, upon differentiation: viz., $f_{U}(u)=-\log(u), 0\le u\le 1$.

Now the requisite condition for real roots is, \begin{eqnarray*} \mathbb{P}\left(B^{2}-4 A C \ge 0 \right) &=& \mathbb{P}\left( A C \le \frac{B^{2}}{4} \right) \\ &=& \int_{0}^{1}{\mathbb{P}\left( A C \le \frac{b^{2}}{4} \right) f_{B}(b) db} \\ &=&\int_{0}^{1}{ \frac{b^{2}}{4} \left(1-\log \frac{b^{2}}{4} \right) db} \\ &=& \left[ \frac{b^{3}}{36} \left(5 + \log 64 - 3 \log b^{2} \right) \right]_{0}^{1} \\ &=& \frac{1}{36} \left(5+ \log 64\right) \\ &\approx& 0.2544134189822131. \end{eqnarray*}