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\section{Quadratic Equations\label{Quadratic Equations}}

Applied problems often lead to equations which are not linear. For example,
the motion of a projectile thrown or fired upward (such as a football or
bullet) is determined by a second-degree equation. Also, in many cases,
distance-rate-time problems actually lead to second-degree equations rather
than linear ones.\bigskip

\subsection{Definition of Quadratic Equations}

A \textbf{quadratic equation\label{Definition of Quadratic Equation}} is an
equation equivalent to one of the form 
\[
ax^{2}+bx+c=0 
\]
where $a$, $b$, and $c$ are real numbers and $a\neq 0$.\bigskip

\subsection{Solving Quadratic Equations by Factoring\label%
{Solving Quadratic Equations by Factoring}}

\subsubsection{Zero-product property\protect\vspace{1pt}\label%
{Zero-product Property of Real Numbers}}

Some quadratic equations can be solved by factoring and using the \textbf{%
zero-product property} of real numbers: 
\[
AB=0\text{ if and only if }A=0\text{ or }B=0 
\]
\ 

\subsubsection{Solving a simple quadratic equation}

Consider an equation of the form 
\[
x^{2}-c=0\text{, }c\geq 0
\]
Since  $c=\left( \sqrt{c}\right) ^{2}$, you can solve this one as follows: 
\begin{eqnarray*}
x^{2}-\left( \sqrt{c}\right) ^{2} &=&0 \\
\left( x+\sqrt{c}\right) \left( x-\sqrt{c}\right)  &=&0\text{ Factor left
hand side using the difference of squares} \\
x+\sqrt{c} &=&0\text{ or }x-\sqrt{c}=0\text{ Using the zero-product property}
\\
x &=&-\sqrt{c}\text{ or }x=\sqrt{c}
\end{eqnarray*}
This is abbreviated 
\[
x=\pm \sqrt{c}
\]
Thus the solutions of the equation $x^{2}=c$, $c\geq 0$, are 
\[
x=\sqrt{c}\text{\quad and\quad }x=-\sqrt{c}
\]
.\bigskip 

\subsection{Solving Quadratic Equations by Completing the Square\label%
{Solving Quadratic Equations by Completing the Square}}

If a quadratic equation does not factor readily, you can solve it using the
technique of \textbf{completing the square}. This means you add a constant
to an expression to make it a perfect square, and then use the technique of
taking square roots of each side as shown above.

\subsubsection{Completing the Square}

To make $x^{2}+bx$ a perfect square, add $\left( \frac{b}{2}\right) ^{2}$: 
\[
x^{2}+bx+\left( \frac{b}{2}\right) ^{2}=\left( x+\frac{b}{2}\right) ^{2} 
\]
\bigskip

\subsection{The Quadratic Formula\label{Quadratic Formula}}

You can use the technique of completing the square to solve the general
quadratic equation 
\[
ax^{2}+bx+c=0
\]
First, divide each side by $a$%
\[
x^{2}+\frac{b}{a}x+\frac{c}{a}=0
\]
Now subtract $\frac{c}{a}$ from each side so the constant term will appear
only on the right: 
\[
x^{2}+\frac{b}{a}x=-\frac{c}{a}
\]
Now complete the square; the coefficient of $x$ is $\dfrac{b}{a}$, so you
want to add $\left( \dfrac{\frac{b}{a}}{2}\right) ^{2}$ or $\left( \dfrac{b}{%
2a}\right) ^{2}$ to each side: 
\[
x^{2}+\frac{b}{a}x+\left( \frac{b}{2a}\right) ^{2}=-\frac{c}{a}+\left( \frac{%
b}{2a}\right) ^{2}
\]
Then simplify and solve: 
\begin{eqnarray*}
\left( x+\frac{b}{2a}\right) ^{2} &=&\frac{-c}{a}+\frac{b^{2}}{4a^{2}} \\
\left( x+\frac{b}{2a}\right) ^{2} &=&\frac{-4ac+b^{2}}{4a^{2}} \\
\left( x+\frac{b}{2a}\right) ^{2} &=&\left( \frac{\sqrt{b^{2}-4ac}}{2a}%
\right) ^{2} \\
x+\frac{b}{2a} &=&\frac{\sqrt{b^{2}-4ac}}{2a}\text{ or }x+\frac{b}{2a}=-%
\frac{\sqrt{b^{2}-4ac}}{2a} \\
x &=&-\frac{b}{2a}+\frac{\sqrt{b^{2}-4ac}}{2a}\text{ or }x=-\frac{b}{2a}-%
\frac{\sqrt{b^{2}-4ac}}{2a} \\
x &=&\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}
\end{eqnarray*}

\subsubsection{The Discriminant\label{Discriminant of a quadratic equation}}

Because of the form of the quadratic formula, you can determine whether or
not the quadratic equation $ax^{2}+bx+c=0$ has solutions, and whether it has
only one solution, by examining the quantity $D=b^{2}-4ac$. This is called
the \textbf{discriminant }of the quadratic equation.The discriminant has the
following property:

\begin{itemize}
\item  If $D>0$, then the equation has two distinct real solutions.

\begin{itemize}
\item[Example]  Solve the following equation using the quadratic formula 
\[
x^{2}+x-1=0
\]

\item[Solution]  Here, $a=1$, $b=1$, and $c=-1$, so: 
\begin{eqnarray*}
x &=&\frac{-1+\sqrt{1^{2}-4\left( 1\right) \left( -1\right) }}{2\left(
1\right) }\text{ or }x=\frac{-1-\sqrt{1^{2}-4\left( 1\right) \left(
-1\right) }}{2\left( 1\right) } \\
x &=&\frac{-1+\sqrt{5}}{2}\text{ or }x=\frac{-1-\sqrt{5}}{2}
\end{eqnarray*}
\end{itemize}

\item  If $D=0$, then the equation has exactly one real solution.

\begin{itemize}
\item[Example]  Solve the following equation using the quadratic formula 
\[
x^{2}+2z+1=0
\]

\item[Solution]  Here, $a=1$, $b=2$, and $c=1$, so: 
\begin{eqnarray*}
x &=&\frac{-2+\sqrt{2^{2}-4\left( 1\right) \left( 1\right) }}{2\left(
1\right) }\text{ or }x=\frac{-2-\sqrt{2^{2}-4\left( 1\right) \left( 1\right) 
}}{2\left( 1\right) } \\
x &=&\frac{-2+0}{2}\text{ or }x=\frac{-2-0}{2} \\
x &=&-1\text{ or }x=-1
\end{eqnarray*}
\end{itemize}

\item  If $D<0$, then the equation has no real solution.

\begin{itemize}
\item[Example]  Solve the following equation using the quadratic formula 
\[
x^{2}+x+1=0
\]

\item[Solution]  Here, $a=1$, $b=1$, and $c=1$, so: 
\begin{eqnarray*}
x &=&\frac{-1+\sqrt{1^{2}-4\left( 1\right) \left( 1\right) }}{2\left(
1\right) }\text{ or }x=\frac{-1-\sqrt{1^{2}-4\left( 1\right) \left( 1\right) 
}}{2\left( 1\right) } \\
x &=&\frac{-1+\sqrt{-3}}{2}\text{ or }x=\frac{-1-\sqrt{-3}}{2}
\end{eqnarray*}
But since there is no real number whose square is $-3$, there is no real
solution to this equation.\bigskip 
\end{itemize}
\end{itemize}

\subsection{Problem Solving with Quadratic Equations\label%
{Problem Solving with Quadratic Equations}}

\subsubsection{The Path of a Projectile}

An object thrown or fired straight upward at an initial speed of $v_{0}%
\dfrac{\unit{ft}}{\unit{s}}$ will reach a height of $h\unit{ft}$ after $t%
\unit{s}$, where $h$ and $t$ are related by the formula 
\[
h=-16t^{2}+v_{0}t
\]

Suppose a bullet is shot straight upward with an initial speed of $800\dfrac{%
\unit{ft}}{\unit{s}}$. When does the bullet fall back to ground level?
Ground level corresponds to $h=0$, so you must solve the equation 
\begin{eqnarray*}
0 &=&-16t^{2}+800t \\
0 &=&-16t(t-50) \\
-16t &=&0\text{ or }t-50=0 \\
t &=&0\text{ or }t=50
\end{eqnarray*}
Thus the height is $0$ both when $t=0$, when it is initially fired, and when 
$t=50$; so it falls back to ground level after $50$ seconds.\bigskip 

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