Category: Quadratic Equations

What is Quadratic Equation

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.

Incomplete Quadratic Equations

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.