Finding expected value of $|X_1+X_2|$, given that the two roots $X_1, X_2$ of $X^2 + 2BX + 1 =0 $ are real 
$X^2 + 2BX + 1 =0 $
  The random variable B is normally distrubuted with mean zero and unit variance. 
Given that the two roots $X_1, X_2$ are real, find, giving your answers to three s.f. the expected value of $|X_1+X_2|$.

So that I found the prob. of the quadratic equation having real roots when $|B|>1$, giving a value of 0.3174
Now the sum of the root is $-2B$, so the expectation of $2|B|$ 
$$
\frac {1} {\sqrt {2\pi}} \int_{-\infty}^{-1} (\color{red}{-2x})e^{-\frac 1 2 x^2} + \frac {1} {\sqrt {2\pi}} \int_{1}^{\infty} (\color{red} {2x})e^{-\frac 1 2 x^2}
$$
Why multiply by $-x$ for the negative limit? 
the definition of expected value is $\int_{-\infty}^{\infty} xf(x)$  .
Also the question is asking expected value, not a conditional probability.
Should I also divide it by the probability of having a real roots even if it is asking for the expected value?
Initially I thought expected value of $|B|$ would simply equal to $B$, since normal distribution graph with B as its x-axis is symmetrical. But it is clearly wrong. How should I interpret this?
Many thanks in advance  
 A: I go through the solution step by step:
The quantity to be calculated is $$E[|X_1+X_2||X_1,X_2\text{ are reals}]=2E[|B|||B|>1]$$
where $B$ is a standard Gaussian r.v.
In order to calculate the $E[|B|||B|>1]$ we need the corresponding conditional distribution:
$$P(|B|<x||B|>1)=\frac{P(|B|<x,|B|>1)}{P(|B|>1)}.$$
(1) 
$$P(|B|>1)=P(B<-1)+P(B>1)=2\Phi(-1),$$
where $\Phi$ is the cdf of the standard Gaussian distribution.
(2) $$P(|B|<x,|B|>1)=\begin{cases}
P(B<-1)+P(1< B <x)& \text{ if }& x>1\\
P(B<-1)&\text{ if }& -1< x \le 1\\
P(B<x)&\text{ if }& x\le -1\\
\end{cases}=$$
$$=\begin{cases}
2\Phi(-1)+\Phi(x)& \text{ if }& x>1\\
\Phi(-1)&\text{ if }& -1< x \le 1\\
\Phi (x)&\text{ if }& x\le -1\\
\end{cases}$$
The derivative of this function exists (except at $-1$ and $1$):
$$g(x)=\begin{cases}
\frac{1}{\sqrt {2\pi} } e^{-\frac 1 2 x^2}& \text{ if } x>1 \text{ or } x<-1\\
0& \text{ othewise}
\end{cases}.$$
With this the conditional expectation we are looking for is 
$$E[|X_1+X_2||X_1,X_2\text{ are reals}]=2E[|B|||B|>1]=          $$
$$=\int_{-\infty}^{\infty}|x|g(x)dx=\color {red}{2}\frac{\int_{-\infty}^1\frac{1}{\sqrt {2\pi} } \color{red}{|x|}e^{-\frac 1 2 x^2}dx+\int_{1}^{\infty}x\frac{1}{\sqrt {2\pi} } \color{red}{|x|}e^{-\frac 1 2 x^2}dx}{\color{red}{2}\Phi(-1)}=$$
$$=\color {red}{2}\frac{\int_{-\infty}^1\frac{1}{\sqrt {2\pi} } \color{red}{(-x)}e^{-\frac 1 2 x^2}dx+\int_{1}^{\infty}x\frac{1}{\sqrt {2\pi} } \color{red}{x}e^{-\frac 1 2 x^2}dx}{\color{red}{2}\Phi(-1)}=$$
$$=\frac{\color{red}{-}\int_{-\infty}^1\frac{1}{\sqrt {2\pi} } xe^{-\frac 1 2 x^2}dx+\int_{1}^{\infty}x\frac{1}{\sqrt {2\pi} } xe^{-\frac 1 2 x^2}dx}{\Phi(-1)}=$$
$$=\frac{\color{red}{2}\int_{1}^{\infty}\frac{1}{\sqrt {2\pi} } xe^{-\frac 1 2 x^2}dx}{\Phi(-1)}.$$
Further edited upon request
Finally, let's evaluate the last integral:
$$\int_{1}^{\infty} xe^{-\frac 1 2 x^2}dx.$$
Introducing the new variable $u=x^2$ we get $x=\sqrt u$, $\frac{dx}{du}=\frac{1}{2}\frac{1}{\sqrt u}$, also, when $x=1$ then $u=1$, and when $x=\infty$ then $u=\infty$. So, 
$$\int_{1}^{\infty} xe^{-\frac 1 2 x^2}dx=\frac{1}{2}\int_1^{\infty}\sqrt u\frac{1}{\sqrt u}e^{-\frac{1}{2}u}=\frac{1}{\sqrt e}.$$
That is,
$$E[|X_1+X_2||X_1,X_2\text{ are reals}]=\frac{1}{\Phi(-1)}\color{red}{2}\frac{1}{\sqrt {2\pi}}\frac{1}{\sqrt e}=\frac{1}{\Phi(-1)}\sqrt{\frac{2}{\pi e}}.$$
