Question about zeroes of simple 2nd order FIR filter
The reason for this is that the maths centres on a false assumption:


  We know that the zeroes need to be complex conjugates


Do the zeros need to be complex conjugates? Can they not be 2 real zeros? 

I sympathise here because I also read this book and the maths portion ended up utterly confusing me because of very odd methods.

A more general method would be at this point:

$0=1+\alpha_1z^{-1}+\alpha_2z^{-2}$

Rearrange thus:

$0=\frac{z^{2}+\alpha_1z^{1}+\alpha_2}{z^{2}}$

Here on the top we have a quadratic equation, which you can use any quadratic solver for. 

For your first case, I arrive at 0.635 $\pm$ 0.637i, which is a pair of complex numbers denoting the conjugate zeros - correct.

For your second, I arrive at two real zeros, one at 3.36 and one at 0.24 - so it is most definitely able to be computed! 

And finally, you will have noted the $z^{2}$ term lying on the bottom, which indicates that there is a square pole at 0+0i. If these zeros were poles then there would similarly be a square zero at 0+0i.

All of these results can be verified with the zplane function in MATLAB.

It took me some time with this book and checking results with the zplane function in matlab to realise that his methods of working really don't work on a general basis, and frankly overcomplicate things. It has however been awhile since I've looked at it so maybe I similarly misinterpreted his maths