Quadratic equation in one variable is the equation with standard form $\color{purple}{{{a}{{x}}^{{2}}+{b}{x}+{c}={0}}}$.

$a$, $b$ and $c$ are some numbers and $x$ is variable. Note, that $a$ can't be zero.

Quadratic equation ${a}{{x}}^{{2}}+{b}{x}+{c}={0}$ is called incomplete, if either ${b}$ or ${c}$ (or both) equals 0.

Such equations can be easily solved without advanced methods.

Example 1. Solve ${{x}}^{{2}}-{81}={0}$.

## Solving Quadratic Equations by Completing the Square

Completing the square is a method for solving quadratic equation.

Before we into the method itself, let's start from a simple example.

Example 1. Solve equation ${{\left({x}-{3}\right)}}^{{2}}={16}$.

## Quadratic Equation Formula and the Discriminant

Quadratic Equation Formula can be derived from the steps for completing the square (actually, this formula is a general case).

Let's see how to do it.

1. Start from the equation ${a}{{x}}^{{2}}+{b}{x}+{c}={0}$.
2. Divide both sides by ${a}$: ${{x}}^{{2}}+\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}={0}$.
3. Move constant term to the right: ${{x}}^{{2}}+\frac{{b}}{{a}}{x}=-\frac{{c}}{{a}}$.
4. Add ${{\left(\frac{{b}}{{2}}{a}\right)}}^{{2}}=\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}$ to both sides of the equation: ${{x}}^{{2}}+\frac{{b}}{{a}}{x}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=-\frac{{c}}{{a}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}$.
5. Rewrite left hand side: ${{\left({x}+\frac{{b}}{{{2}{a}}}\right)}}^{{2}}=-\frac{{c}}{{a}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}$.
6. Simplify right hand side: $-\frac{{c}}{{a}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=\frac{{-{c}\cdot{\color{red}{{{4}{a}}}}}}{{{a}\cdot{\color{red}{{{4}{a}}}}}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=\frac{{-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}+\frac{{{b}}^{{2}}}{{{4}{{a}}^{{2}}}}=\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}$.
7. Write the final equation: ${{\left({x}+\frac{{b}}{{{2}{a}}}\right)}}^{{2}}=\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}$.
8. Solve the equation: ${x}+\frac{{b}}{{{2}{a}}}=\sqrt{{\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}}}$ or ${x}+\frac{{b}}{{{2}{a}}}=-\sqrt{{\frac{{{{b}}^{{2}}-{4}{a}{c}}}{{{4}{{a}}^{{2}}}}}}$.
9. Above equations have roots ${x}_{{1}}=\frac{{-{b}+\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}$ and ${x}_{{2}}=\frac{{-{b}-\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}$.
10. We can write it even more compactly: ${x}_{{{1},{2}}}=\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}$.

Expression ${{b}}^{{2}}-{4}{a}{c}$ is called the discriminant of the quadratic equation.

## Viet Theorem

Viet Theorem. If quadratic equation ${a}{{x}}^{{2}}+{b}{x}+{c}={0}$ (reduced form is ${{x}}^{{2}}+\frac{{b}}{{a}}+\frac{{c}}{{a}}={0}$) has roots ${p}$ and ${q}$, then ${\color{green}{{{p}+{q}=-\frac{{b}}{{a}}}}}$, ${\color{ma\genta}{{{p}{q}=\frac{{c}}{{a}}}}}$, i.e. sum of the roots equals second coefficient, taken with opposite sign, and product of roots equals constant.