Quadratic Equations in One Variable

Definition

A quadratic equation in $\mathrm{x}$ is any equation that may be written in the form
$\mathrm{a}{\mathrm{x}}^{\mathrm{2}}\mathrm{+}\mathrm{b}\mathrm{x}\mathrm{+}\mathrm{c}\mathrm{=}\mathrm{0}$, where $\mathrm{a}\mathrm{,}$ $\mathrm{b}$, and $\mathrm{c}$ are coefficients and $\mathrm{a}\mathrm{\ne }\mathrm{0}$.

Note that if $\mathrm{a}\mathrm{=}\mathrm{0}$, then the equation would simply be a linear equation, not quadratic.

Examples

${\mathrm{x}}^{\mathrm{2}}\mathrm{+}\mathrm{2}\mathrm{x}\mathrm{=}\mathrm{4}$ is a quadratic since it may be rewritten in the form $\mathrm{a}{\mathrm{x}}^{\mathrm{2}}\mathrm{+}\mathrm{b}\mathrm{x}\mathrm{+}\mathrm{c}\mathrm{=}\mathrm{0}$ by
applying the Addition Property of Equality and subtracting 4 from both sides of $\mathrm{=}$.

$\mathrm{(}\mathrm{2}\mathrm{+}\mathrm{x}\mathrm{)}\mathrm{(}\mathrm{3}\mathrm{-}\mathrm{x}\mathrm{)}\mathrm{=}\mathrm{0}$ is a quadratic since it may be rewritten in the form $\mathrm{a}{\mathrm{x}}^{\mathrm{2}}\mathrm{+}\mathrm{b}\mathrm{x}\mathrm{+}\mathrm{c}\mathrm{=}\mathrm{0}$
by applying the Distributive Property to multiply out all terms and then combining
like terms.

${\mathrm{x}}^{\mathrm{2}}\mathrm{\text{-}}\mathrm{3}\mathrm{=}\mathrm{0}$ is a quadratic since it has the form $\mathrm{a}{\mathrm{x}}^{\mathrm{2}}\mathrm{+}\mathrm{b}\mathrm{x}\mathrm{+}\mathrm{c}\mathrm{=}\mathrm{0}$ with $\mathrm{b}\mathrm{=}\mathrm{0}$ in this case.

$\mathrm{3}{\mathrm{x}}^{\mathrm{2}}\mathrm{-}\frac{\mathrm{2}}{\mathit{x}}\mathrm{+}\mathrm{4}\mathrm{=}\mathrm{0}$ is not a quadratic since it has the term $\frac{\mathrm{2}}{\mathit{x}}$. The term $\frac{\mathrm{2}}{\mathit{x}}$ is the
same as $\mathrm{2}{\mathrm{x}}^{\mathrm{-}\mathrm{1}}$, and quadratics do not have $\mathrm{x}$ raised to any power other than 1 or 2.

Just remember: Quadratics always have an ${\mathit{x}}^{\mathrm{2}}$ term, possibly an x-term, and

possibly a constant term! If your equation has an ${\mathit{x}}^{\mathrm{2}}$ term or will have an ${\mathit{x}}^{\mathrm{2}}$ term
after multiplying out, it may be a quadratic, provided the other terms fit the form.

Solving Quadratic Equations-Method 1 -Factoring

The easiest way to solve a quadratic equation is to solve by factoring, if possible.

Here are the steps to solve a quadratic by factoring:

1. Write your equation in the form $\mathrm{a}{\mathrm{x}}^{\mathrm{2}}\mathrm{+}\mathrm{b}\mathrm{x}\mathrm{+}\mathrm{c}\mathrm{=}\mathrm{0}$ by applying the Distributive

Properly, Combine Like Terms, and apply the Addition Property of Equality to

move terms to one side of $\mathrm{=}$.

2. Factor your equation by using the Distributive Property and the appropriate

factoring technique. Note: Any type of factoring relies on the Distributive$\mathrm{}$Property.

3. Let each factor $\mathrm{=}\mathrm{0}$ and solve. This is possible because of the Zero Product$\mathrm{}$Law.

Example: Solve $\mathrm{(}\mathrm{3}\mathrm{x}\mathrm{+}\mathrm{4}\mathrm{)}\mathrm{x}\mathrm{=}\mathrm{7}$

$\mathrm{(}\mathrm{3}\mathrm{x}\mathrm{+}\mathrm{4}\mathrm{)}\mathrm{x}\mathrm{=}\mathrm{7}$ Given

$\mathrm{3}{\mathrm{x}}^{\mathrm{2}}\mathrm{+}\mathrm{4}\mathrm{x}\mathrm{=}\mathrm{7}$ by the Distributive Property

$\mathrm{3}{\mathrm{x}}^{\mathrm{2}}\mathrm{+}\mathrm{4}\mathrm{x}\mathrm{-}\mathrm{7}\mathrm{=}\mathrm{0}$ by the Addition Property of Equality

Now, factor $\mathrm{3}{\mathrm{x}}^{\mathrm{2}}\mathrm{+}\mathrm{4}\mathrm{x}\mathrm{-}\mathrm{7}\mathrm{=}\mathrm{0}$

This factors as $\mathrm{(}\mathrm{3}\mathrm{x}\mathrm{+}\mathrm{?}\mathrm{)}\mathrm{(}\mathrm{x}\mathrm{\text{-}}\mathrm{?}\mathrm{)}\mathrm{=}\mathrm{0}$ or $\mathrm{(}\mathrm{3}\mathrm{x}\mathrm{-}\mathrm{?}\mathrm{)}\mathrm{(}\mathrm{x}\mathrm{+}\mathrm{?}\mathrm{)}\mathrm{=}\mathrm{0}$ where the two unknown
numbers multiply to-7 when we use the Distributive Property to multiply out.

Also the first two terms must multiply out to $\mathrm{3}{\mathrm{x}}^{\mathrm{2}}$. The middle products must add
up to $\mathrm{4}\mathrm{x}$.

$\mathrm{(}\mathrm{3}\mathrm{x}\mathrm{+}\mathrm{7}\mathrm{)}\mathrm{\left(}\mathrm{x}\mathrm{\text{-}}\mathrm{1}\mathrm{\right)}\mathrm{=}\mathrm{0}$ gives us middle products $\mathrm{7}\mathrm{x}\mathrm{}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}\mathrm{-}\mathrm{3}\mathrm{x}$ adding up to $\mathrm{4}\mathrm{x}$.

By the Zero Product Law, we can state
$\mathrm{3}\mathrm{x}\mathrm{+}\mathrm{7}\mathrm{=}\mathrm{0}$ and $\mathrm{x}\mathrm{\text{-}}\mathrm{1}\mathrm{=}\mathrm{0}$.

Solve these two equations by using the Addition Property of Equality and the
Division Property of Equality.

$\mathrm{3}\mathrm{x}\mathrm{+}\mathrm{7}\mathrm{=}\mathrm{0}$

$\mathrm{3}\mathrm{x}\mathrm{}\mathrm{=}\mathrm{}\mathrm{-}\mathrm{7}$

$\mathit{x}\mathrm{}\mathrm{=}\mathrm{}\frac{\mathrm{-}\mathrm{7}}{\mathrm{3}}$

$\mathrm{A}\mathrm{l}\mathrm{s}\mathrm{o}\mathrm{}$

$\mathrm{x}\mathrm{-}\mathrm{1}\mathrm{}\mathrm{=}\mathrm{}\mathrm{0}\mathrm{}$

$\mathrm{i}\mathrm{.}\mathrm{e}\mathrm{}\mathrm{}\mathrm{x}\mathrm{=}\mathrm{1}$