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!
 A: Since $f(0)=-1<0$ and $f(2)=3>0$, it is clear that $f(x)=0$ for some $x\in[0,2]$ by the intermediate value theorem.
Similarly, you can see that there is a root of $f$ for some $x\in[-2,0]$.
In general, it is hard to find zeroes for an arbitrary function, even if it is continuous. The intermediate value theorem doesn't work for functions that only touch the $y=0$ axis, for example for $f(x)=x^2$.
A: One way is using the discriminant of the quadratic equation:
$$\sqrt{b^2-4ac}$$
If the value inside the square root is greater than 0, then there are two real roots
If it is equal to 0, there is one real root
If it is less than 0, it has imaginary roots
A: 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.
