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
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.

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.

Also defined as
Some older treatments of this subject report this as:


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


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

but this has fallen out of fashion.