Solution to Quadratic Equation

Theorem
An algebraic equation of the form $ax^2 + bx + c = 0$ is called a quadratic equation.

It has solutions $\displaystyle x = \frac {-b \pm \sqrt {b^2 - 4 a c}} {2a}$.

Discriminant
The expression $b^2 - 4 a c$ is called the discriminant of the equation.

Let $a, b, c \in \R$.

Then the quadratic equation $a x^2 + b x + c = 0$ has:
 * Two real solutions if $b^2 - 4 a c > 0$;
 * One real solution if $b^2 - 4 a c = 0$;
 * Two complex solutions in $\C$ if $b^2 - 4 a c < 0$, and those two solutions are complex conjugates.

Note that this is a special case of the general discriminant, although it is important to note that the general formula is given for monic polynomials.

Direct Proof
Let $ax^2 + bx + c = 0$. Then:


 * If the discriminant $b^2 - 4 a c > 0$ then $\sqrt {b^2 - 4 a c}$ has two values and the result follows.


 * If the discriminant $b^2 - 4 a c = 0$ then $\sqrt {b^2 - 4 a c} = 0$ and $\displaystyle x = \frac {-b} {2 a}$.


 * If the discriminant $b^2 - 4 a c < 0$, then we can write it as:

$b^2 - 4 a c = \left({-1}\right) \left|{b^2 - 4 a c}\right|$

Thus $\sqrt {b^2 - 4 a c} = \pm i \sqrt {\left|{b^2 - 4 a c}\right|}$, and the two solutions are:

$\displaystyle x = \frac {-b} {2 a} + i \frac {\sqrt {\left|{b^2 - 4 a c}\right|}} {2 a}, x = \frac {-b} {2 a} - i \frac {\sqrt {\left|{b^2 - 4 a c}\right|}} {2 a}$

and once again the result follows.