# Determining number of solutions by discriminant

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

• A formal proof ? Please specify what this should be , and why you are not satisfied with the argument you have given. – Peter Mar 2 '17 at 12:37

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 $b^2-c=0$.
• There isn't an equivalent of completing the square for cubics. One can define the discriminant as the product of the squares of the differences of the roots, which, with some calculations involving symmetric polynomials in the roots, gives $-4b^3-27c^3$ as the discriminant. (There doesn't seem to be an easier way of deriving this.) Quadratics are easy, cubics are a pain, quartics are horrible, and quintics generally impossible. It's normally easier to use that the derivative also vanishes at a repeated root. See also Cardano's method for cubics. – Chappers Mar 2 '17 at 21:21