Proof $\rho(X,Y)=\pm1$, only for special cases I'm having some difficulties with the following proof:

Only onder special circumstances it can be the case that $\rho(X,Y)^1=\pm1$, and these circumstances are explored by considering the proof of the Cauchy-Schwartz inequality more carefully. Let $a=\operatorname{var}(X)$, $b=2\operatorname{cov}(X,Y)$, $c=\operatorname{var}(Y)$ and suppose that $\rho(X,Y)=\pm1$. Then $\operatorname{var}(X)\operatorname{var}(Y)\neq0$ and
  $$b^2-4ac=4\operatorname{var}(X)\operatorname{var}(Y)\big[\rho(X,Y)^2-1\big]=0,$$
  and so the quadratic equation
  $$as^2+bs+c=0$$
  has two equal real roots, at $s=\alpha$, say. Therefore, ...

I find the proof overall a bit weird, because I don't know how they came up with it. But I can follow the algebraic steps and such, so in essence I'm ok with the proof.
It continues as follows:

Therefore, $W=\alpha[X-\mathbb E(X)]+[Y-\mathbb E(Y)]$ satisfies
  $$
\mathbb E(W^2)=a\alpha^2+b\alpha+c=0,
$$
  giving that $\mathbb P(W=0)=1$, and showing that (essentially) $Y=-\alpha X+\beta$, where $\beta=\alpha\mathbb E(X)+\mathbb E(Y)$. A slightly more careful treatment discriminates between the values $+1$ and $-1$ for $\rho(X,Y)$:
  $$
\begin{aligned}
\rho(X,Y)&=1\quad\text{if and only if }\mathbb P(Y=\alpha X+\beta)=1\text{ for some real }\alpha\text{ and }\beta\text{ with }\alpha>0\\
\rho(X,Y)&=-1\quad\text{if and only if }\mathbb P(Y=\alpha X+\beta)=1\text{ for some real }\alpha\text{ and }\beta\text{ with }\alpha<0.\\
\end{aligned}
$$
\begin{align*}
\rho(X,Y)&=1&&
\begin{aligned}[t]
\text{if and only if }\mathbb P(Y=\alpha X+\beta)=1\\
\text{for some real $\alpha$ and $\beta$ with }\alpha>0,
\end{aligned}\\
\rho(X,Y)&=-1&&
\begin{aligned}[t]
\text{if and only if }\mathbb P(Y=\alpha X+\beta)=1\\
\text{for some real $\alpha$ and $\beta$ with }\alpha<0.
\end{aligned}[t]
\end{align*}

So, I can follow everything, except for the last bit where they are giving the conditions for $\rho(X,Y)=1$ and $\rho(X,Y)=-1$.
Say we look at $\rho(X,Y)=1$. It follows that $\operatorname{cov}(X,Y)=\sqrt{\operatorname{var}(X)\operatorname{var}(Y)}$. What else can I use to show that these are indeed the conditions?
 A: This proof is a paraphrased version of what is in Casella and Berger's Statistical Inference, p. 172.
I use $\mu_{T}$ and $\sigma^2_{T}$ to denote the mean and variance respectively of a random variable $T$, and given another random variable $U$, $\sigma_{TU} = \text{Cov}(T, U)$.
p. 172 approaches this proof using a more brute-force method and does not use Cauchy-Schwarz: consider
$$h(t) = \mathbb{E}\left\{[(X - \mu_X)t + (Y - \mu_Y)]^2\right\}\text{.}$$
After some work, it can be shown that 
$$h(t) = t^2\sigma^2_{X} + 2t\sigma_{XY}+\sigma^2_{Y} = at^2+bt+c\text{.}$$
This is a quadratic with respect to $t$. Notice that $h$ is an expected value of a random variable which is always nonnegative: hence $h(t) \geq 0$ for all $t$. Because of this, $h$ has at most one root $r$ such that $h(r) = 0$. Therefore, it either has no roots or has one root; in both cases, the discriminant is non-positive; i.e.,
$$b^2 - 4ac = (2\sigma_{XY})^2-4(\sigma^2_{X})(\sigma^{2}_{Y}) \leq 0\text{.}$$
Dividing by $\sigma_{X}\sigma_{Y} = \sqrt{\sigma^2_{X}\sigma^{2}_{Y}}$ gives
$$\dfrac{(2\sigma_{XY})^2}{\sigma_{X}\sigma_{Y}}-4\dfrac{(\sigma^2_{X})(\sigma^{2}_{Y})}{\sigma_{X}\sigma_{Y}} = 4\sigma_{XY}\cdot\rho(X, Y)-4\sigma_{X}\sigma_{Y}=4\sigma_X\sigma_Y\cdot [\rho(X, Y)^2-1]\leq 0\text{.}$$
Now multiply both sides by $\sigma_X\sigma_Y$ to get 
$$4\sigma^2_X\sigma^2_Y\cdot[\rho(X, Y)^2-1] \leq 0\text{.}$$
Solving for $\rho(X, Y)$ gives
$$\rho(X, Y)^2 \leq 1\text{.}$$
In the case of equality, we have $\rho(X, Y)^2 = 1$, or $\rho(X, Y) = \pm 1$. Using the link above, we have to only have one root for this to happen, which is $\alpha$, as explained in your question.
