Probability of getting imaginary roots Let $X$ and $Y$ be independent and identically distributed uniform random variable over $(0,1)$. Let $S=X+Y$. Find the probability that the quadratic equation $9x^2+9Sx+1=0$ has no real roots.  
What I attempted 
For no real roots we must have $$(9S)^2-4\cdot 1\cdot 9<0$$
So, we need $P[(9S)^2-4.1.9<0]=P(9S^2<4)=P(S^2<\frac{4}{9})=P(-\frac{2}{3}<S<\frac{2}{3})=P(0<S<\frac{2}{3})$
[As $S$ can never be less than $0$]  
Now, $\displaystyle P\left(0<S<\frac{2}{3}\right)=P\left(0<X+Y<\frac{2}{3}\right)=\int_{0}^{ \frac{2}{3}}\int_{0}^{ \frac{2}{3}-x}f_X(x)\cdot f_Y(y)\,dy\,dx= \int_{0}^{ \frac{2}{3}}\int_{0}^{ \frac{2}{3}-x}\,dy\,dx. $  
Now, $\displaystyle \int_{0}^{ \frac{2}{3}}\int_{0}^{ \frac{2}{3}-x}\,dy\,dx=\int_{0}^{ \frac{2}{3}}\left(\frac{2}{3}-x\right)\,dx=\frac{1}{2}\left(\frac{4}{9}\right)=\frac{2}{9} $    
Am I correct ?

Yes.
And nowadays, it's easy to check for gross errors by simulation. Here is a MATLAB simulation:
>> n = 1e8; sum((9*sum(rand(n,2),2)).^2-36 < 0)/n
ans =
    0.2223

In the real world, it's always good to check, or at least partially check, your work by different methods.
