HBA #1

HBA #1

Do this assignment to get an easy hour by arrangement credit. First read the "LaTeX Tutorial" post to learn some basic LaTeX commands (LaTeX is a mathematical typesetting language that creates beautifully typeset mathematical documents).

Next, reply to this post using some basic LaTeX commands to create the quadratic formula in "display style" as shown here:

$$x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$$

Note: You can get the $\pm$ symbol with the LaTeX command \pm.

Don't forget to put your name in the post so I can give you credit.

That's all there is to it.

Short LaTeX Tutorial

A Short \(\LaTeX\) Primer

I will expand this list as the need arises. A more complete primer can be found here.

Use the following \(\LaTeX\) commands to enter the following basic mathematical elements:

• All latex commands start with a \ symbol and must be between single or double dollar signs
• Mathematics inline within text must be between $ signs. For example, this equation $2x\sqrt{x^2+3} = 5 - x$ is inline and was typeset as:
  
  ...this equation $2x\sqrt{x^2+3} = 5 - x$ is inline and...
  
  
• Mathematics displayed, centered on its own line must be between $$ signs. For example, this equation $$2x\sqrt{x^2+3} = 5 - x$$ is displayed, and was typeset as:
  
  ...this equation $$2x\sqrt{x^2+3} = 5 - x$$ is displayed...
  
  
• \frac{a}{b} for \(\frac{a}{b}\)
• \sqrt{a} for \(\sqrt{a}\)
• \sqrt[n]{a} for $\sqrt[n]{a}$
• _ for subscripts. For example a_1 for \(a_1\) but if the subscript is more than a single character, then you must put it in braces as in a_{34} for \(a_{34}\).
• ^ for superscripts (exponents and such). For example x^2 for \(x^2\). As with subscripts if the superscript is more than a single character, enclose it in braces as in x^{23} for \(x^{23}\).
• \int x^2\;dx for $\int x^2\;dx$; \int_2^3 x^2\;dx for $\int_2^3 x^2\;dx$. The \; gives a bit of extra space.
• \sum a_i for $\sum a_i$; \sum_{i=1}^{\infty} a_i for $\sum_{i=1}^{\infty} a_i$.
• \vec{v} for $\vec{v}$ (vectors).
• \bvec{v} for $\bvec{v}$ (bold vector notation).
• \lang{3,4,5} for $\lang{3,4,5}$ (vectors in component form)
• You can use \cdot (center dot) for dot products as in $\vec{v}\cdot\vec{w}$.
• You can use \times for cross products as in $\vec{v}\times\vec{w}$.
• For absolute value (or norm of a vector), use |\vec{v}| as in $|\vec{v}|$.

Here is an example of a more complicated expression entered using $\LaTeX$. The code $\sqrt{\vec{v}\cdot\vec{v}} = |\vec{v}|$ creates $\sqrt{\vec{v}\cdot\vec{v}} = |\vec{v}|$.

The Quadratic Formula

In order to solve a quadratic equation (a second degree polynomial equation) using the quadratic formula, we must first put the equation into the "standard form" $ax^2 + bx + c=0$. Once this is done, we simply substitute the coefficients $a, b, \text{and }, c$ into the formula

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The quantity $b^2 - 4ac$ under the radical sign is called the "discriminant" of the equation. Its sign can tell us about the number and type of solutions of the equation:

• If $b^2 - 4ac > 0$, then there are two distinct real solution of the equation.
• If $b^2 - 4ac < 0$, then there are two complex conjugate solutions of the equation.
• If $b^2 - 4ac = 0$, then the equation has a single real number solution, $\frac{-b}{2a}$.

 As an example, suppose the equation is $(2x - 3)(x + 1) = x^2 - 9$. First, we must put the equation into standard form ($ax^2 + bx + c = 0$):

\begin{align}
(2x - 3)(x + 1) = x^2 - 9 &\implies 2x^2 - x - 3 = x^2 - 9\\
&\implies x^2 - x + 6 = 0
\end{align}

Now we can identify the coefficients $a = 1$, $b = -1$ and $c = 6$. The discriminant $b^2 - 4ac$ has value $(-1)^2 - 4(1)(6) = 1 - 24 = -23$. Therefore, the solutions are complex conjugates given by:

\begin{equation}
x = \frac{-(-1) \pm \sqrt{-23}}{2(1)} = \frac{1 \pm \sqrt{-23}}{2} =
\frac{1}{2} \pm \frac{\sqrt{23}}{2}\,i
\end{equation}