It is known that:
A positive discriminant indicates that the quadratic has two distinct real number solutions. A discriminant of zero indicates that the quadratic has a repeated real number solution. A negative discriminant indicates that neither of the solutions are real numbers.
but why is that? it looks obvious in the formula because of the +- sign, and by definition of the sqrt.
I'm looking for a formal proof for that. Can someone refer me to where I can read about that?
Thanks
Suppose the leading coefficient of the quadratic is $1$ (if not, it must be a number that isn't zero or we haven't got a quadratic, so we can divide by it). Then we can write the quadratic equation as $$ x^2+2bx+c = 0. $$ Now, we want to end up with only one $x$ term on the left; the way to do this is to use the identity $$ (x+b)^2 - b^2 = x^2+2bx $$ to rewrite the equation as $$ (x+b)^2-b^2+c = 0, $$ or $$ (x+b)^2 = b^2-c, $$ and you will recognise the right-hand side as ($1/4$ of) the discriminant of the original equation. $y^2$ is positive unless $y=0$, so there are two real roots if $b^2-c>0$, and one repeated root ($x=-b$) if $b^2-c=0$.