Shamir Secret Sharing: Why cannot we recover polynomial's root if we have $t-1$ shares? Imagine we have $t-1$ shares in $(t,n)$ shamir secret sharing. So at least $t$ shares are needed. 

Question: Why cannot we use $t-1$ shares to find a root of the polynomial and then recover the secret?
Shamir Secret Sharing (SSS) is based on constructing a polynomial of degree $t-1$, whose independent term is the secret $S$. Each share is actually a point of the polynomial.

The security of SSS is based on the fact that, when one wants to interpolate a polynomial of degree $t-1$, one needs at least $t$ points of the polynomial. It can be seen graphically that trying to interpolate with less points (e.g., with $t-1$ points) is impossible since infinite polynomials can be drawn through those points. The Wikipedia article for SSS illustrates this very well.

In cryptographic applications you will not have "infinite" polynomials, but all the possible polynomials of degree $t$ for the given field. That means that knowing $t-1$ points is useless as all possible polynomials are equally plausible. This is called "perfect secrecy".