The roots are both real when the discriminant of the quadratic is positive:
[tex]a^2-4(a+15)=a^2-4a-60=(a-10)(a+6)>0[/tex]
When [tex]a<-6[/tex], say [tex]a=-7[/tex], you have [tex](-7-10)(-7+6)=17>0[/tex].
When [tex]-6<a<10[/tex], say [tex]a=0[/tex], you have [tex](0-10)(0+6)=-60<0[/tex].
When [tex]a>10[/tex], say [tex]a=11[/tex], you have [tex](11-10)(11+6)=17>0[/tex].
So the quadratic will have two real roots whenever [tex]a<-6[/tex] or [tex]a>10[/tex]. The probability of this occurring is
[tex]\mathbb P((a<-6)\lor(a>10))=\mathbb P(a<-6)+\mathbb P(a>10)[/tex]
The density function for this random variable is
[tex]f_X(x)=\begin{cases}\dfrac1{11-(-15)}=\dfrac1{26}&\text{for }-15\le x\le11\\\\0&\text{otherwise}\end{cases}[/tex]
so
[tex]\mathbb P(a<-6)+\mathbb P(a>10)=\displaystyle\int_{-15}^{-6}\dfrac{\mathrm dx}{26}+\int_{10}^{11}\dfrac{\mathrm dx}{26}=\dfrac{(-6-(-15))+(11-10)}{26}=\dfrac5{13}[/tex]
![General Continuous Random Variable problem. Given a is uniformly distributed over [-15,11], what is the probability that the roots of the equation are both real class=](https://us-static.z-dn.net/files/d01/ea001fa549929188a169355f4e0b9cbf.png)
