Solution to Quadratic Equation

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

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

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 $$x = \frac {-b} {2 a}$$.


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

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

and once again the result follows.