Problem source

Find the probability that the quadratic equation 
\[
X^{2}+2BX+1=0
\]
has real roots when $B$ is normally distributed with zero mean and
unit variance. 

Given that the two roots $X_{1}$ and $X_{2}$ are real, find: 
\begin{questionparts}
\item the probability that both $X_{1}$ and $X_{2}$ are greater than $\frac{1}{5}$;
\item the expected value of $\left|X_{1}+X_{2}\right|$; 
\end{questionparts}
giving your answers to three significant figures.

Solution source

The roots are $X_1, X_2 = -B \pm \sqrt{B^2-1}$

\begin{questionparts}
\item The smallest root will be $-B - \sqrt{B^2-1}$. For this to be larger than $\frac15$ we must have,

\begin{align*}
&& -B -\sqrt{B^2-1} &\geq \frac15 \\
\Rightarrow && -B - \frac15 &\geq \sqrt{B^2 - 1} \\
\Rightarrow && B^2 + \frac25 B + \frac1{25} &\geq B^2 - 1 \\
\Rightarrow && \frac25 B \geq -\frac{26}{25} \\
\Rightarrow && B \geq -\frac{13}{5}
\end{align*}

Therefore $-\frac{13}5 \leq B \leq -1$.

Therefore we want:

\begin{align*}
\frac{\P(-\frac{13}5 \leq B \leq -1)}{\P(B < -1) + \P(B > 1)} &= \frac{\Phi(-1) - \Phi(-\frac{13}{5})}{\Phi(-1)+1-\Phi(1)} \\
&= \frac{0.1586\ldots - 0.0046\ldots}{0.1586\ldots + 1- 0.8413\ldots} \\
&= 0.4853\ldots \\
&= 0.485 \,\,(3 \text{ s.f.})
\end{align*}

\item $X_1 + X_2 = -2B$.

Therefore we want:

\begin{align*}
\mathbb{E}(|X_1 + X_2| &= \mathbb{E}(|2B|)  \\
&= 2 \l\frac{1}{2\Phi(-1)} \int_1^{\infty} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B+\frac{1}{2\Phi(-1)} \int_{-\infty}^{-1} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B \r \\
&= \frac{4}{2\Phi(-1)} \int_1^{\infty} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B \\
&=\frac{4}{2\sqrt{2 \pi} \Phi(-1)} \left [ -e^{-\frac12 B^2}\right]_1^{\infty} \\
&= \frac{4}{2\sqrt{2 \pi} \Phi(-1) \sqrt{e}} \\
&= 3.0502\ldots \\
&= 3.05\,\, (3\text{ s.f.})
\end{align*}
\end{questionparts}