Proving a function has real roots I am not interested in finding roots but interested in proving that the function has real roots.
Suppose a function $f(x) = x^2 - 1$
This function obviously has real roots.
$x = {-1, 1}$
How could I prove this without actually finding the roots? 
Trial and error could work, number theory even? (modulus etc?) Calculus, any methods?
Thanks!

For a continuous function $f$, the Intermediate Value Theorem applies, so if $f(a) > 0$ and $f(b) < 0$ (or the other way around), and $a < b$, then $f$ has a zero somewhere in the interval $(a, b)$.
In your example, computing $f(\pm 2) = 3$ and $f(0) = -1$ gives that $f$ has one root each in $(-2, 0)$ and $(0, +2)$.
In the special case that $f$ is a quadratic polynomial, say, $f(x) = ax^2 + bx + c$, then $f$ has real roots iff $b^2 - 4 a c \geq 0$ (it has a single double root if equality holds). We call this quantity the discriminant of the polynomial.
One can devise additional criteria for special cases of polynomials: All odd polynomials have at least one real root. If $f(x) = a_n x^n + \cdots a_1 x + a_0$ is a polynomial of even degree $n$ and $a_0 a_n < 0$, then $f$ has at least one negative and one positive root. Similar considerations give rise to Descartes' Rule of Signs, which gives upper bounds on the number of positive or negative roots.
