Quadratic equations and the discriminant — adapted from OpenStax College Algebra

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\title{Quadratic Equations --- Worked Examples}
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\noindent\textit{Examples adapted from \textbf{OpenStax College Algebra} (CC BY 4.0). Source: \url{https://openstax.org/details/books/college-algebra}.}

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\section*{1.\ The quadratic formula}

The solutions of $ax^2 + bx + c = 0$ (with $a \neq 0$) are given by
\[
  x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\]

The expression $\Delta = b^2 - 4ac$ is called the \emph{discriminant} and determines the number of real solutions:
\begin{itemize}[nosep]
  \item $\Delta > 0$ : two distinct real solutions,
  \item $\Delta = 0$ : one repeated real solution,
  \item $\Delta < 0$ : no real solutions (two complex solutions).
\end{itemize}

\section*{2.\ Worked example}

\textbf{Solve} $2x^2 - 5x - 3 = 0$.

We have $a = 2$, $b = -5$, $c = -3$.
\begin{align*}
  \Delta &= (-5)^2 - 4 \cdot 2 \cdot (-3) = 25 + 24 = 49 \\
  x &= \frac{-(-5) \pm \sqrt{49}}{2 \cdot 2} = \frac{5 \pm 7}{4}
\end{align*}

Therefore $x_1 = 3$ and $x_2 = -\dfrac{1}{2}$.

\section*{3.\ Practice problems}

\begin{enumerate}[label=\textbf{\arabic*.}]
  \item Solve $x^2 - 4x + 4 = 0$.
  \item Solve $3x^2 + 2x - 1 = 0$.
  \item Solve $x^2 + x + 1 = 0$. (Hint: check the discriminant.)
\end{enumerate}

\section*{4.\ Application: projectile motion}

A ball is thrown upward from a height of $1.5$\,m with an initial velocity of $12$\,m/s. Its height (in meters) after $t$ seconds is given by
\[
  h(t) = -4.9 t^2 + 12 t + 1.5.
\]

\begin{enumerate}[label=\textbf{\arabic*.}]
  \item At what time does the ball reach its maximum height? (Use $t = -\dfrac{b}{2a}$.)
  \item What is the maximum height reached?
  \item When does the ball hit the ground? Solve $h(t) = 0$ and keep only the positive root.
\end{enumerate}

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\noindent\hrulefill\\
\textit{Source: this material is adapted from OpenStax College Algebra, Chapter 2, available under a Creative Commons Attribution 4.0 license. Original at \url{https://openstax.org/books/college-algebra-2e}.}

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