Mathematics Formulas


Below are a few of the basic mathematical formulas of secondary school mathematics. To display these formulas in this WordPress site I used Latex. I have a previous post on this called Quick LaTeX Plugin. Currently I am using the plugin MathJax-LaTeX.

Here is a fun YouTube video looking at mathematics from above. It’s called The Map of Mathematics and it’s by Domain of Science. Here is a video that covers a lot of matematics that this post is just illuding to. It’s called Learn Precalculus and it’s by Derek Banas.

Basics

 a = a

 

 1a = (1)(a) = a

 

 (a)(0) = 0

 

 \dfrac{a}{1} = a

 

 2a = a + a

 

 ab = ba

 

 a + b = b + a

 

 a^2 = aa

 

 a^1 = a

 

 a^0 = 1

 

 \dfrac{a}{a} = 1

 

 \dfrac{ab}{a} = b

 

 (\dfrac{a}{b}) (\dfrac{c}{d}) = \dfrac{ac}{bd}

 

 \dfrac{a}{b} = (\dfrac{a}{1})(\dfrac{1}{b})

 

 (\dfrac{a}{b}) + (\dfrac{c}{b}) = \dfrac{a + c}{b}     Notice that the denominators are the same (both b)

 

 \dfrac{a}{0}     is undefined

Proof

Before going any further I want to make a point. To be successful at solving equations you should attempt to master the basics and understand substitution. Also, if you have an equation, you know that whatever you do to one side you must do to the other. So if you do something that changes the value of one side you need to do the same to the other. However, if you are not changing the value of one side you do not need to do the same to the other side. Suppose all you are doing is changing a multiplied by a to a squared, then go ahead and do that to the side that has a multiplied by a and leave the other side as is. This is what I mean by substitution. You are substituting one expression for an expression that is equivalent. You can do this because of the following rule shown above.

 a^2 = aa

Let’s prove the formula at the bottom of the list under the section Exponents and Radicals using the basic rules above and some rules below.

 b^{\frac{m}{n}} = \sqrt[n]{b^m}

The objective here is to change the left side of the equation to be in the format shown on the right side using the basic rules listed above, fully showing our work.

Since     1a = (1)(a) = a   and    \dfrac{a}{b} = (\dfrac{a}{1})(\dfrac{1}{b})<br />   then

 b^{\frac{m}{n}} = b^{\frac{m}{1} \frac{1}{n}}

Since  {a^{mn} = (a^m)}^n   then

 b^{\frac{m}{1} \frac{1}{n}} = (b^m)^ \frac{1}{n}

Since  \sqrt[n]{a} = a^{\frac{1}{n}}

 (b^m)^ \frac{1}{n} =  \sqrt[n]{b^m}

Here below is the summary without showing as many steps.

 b^{\frac{m}{n}} = (b^m)^ \frac{1}{n} =  \sqrt[n]{b^m}

Let b = 2, m = 9, n = 3 and substitute those numbers into each side of the equation and both side should equal 8.

Expansions

 {(a + b)}^2 = a^2 + 2ab + b^2     FOIL is first, outside, inside and last.

 {(a + b)}^3 = a^3 + 3a^2b + 3ab^2 + b^3  (a + b)(b - a) = a^2 - b^2  {(a + b + c)}^2 = a^2 + b^2 + c^2 +2ab + 2ac + 2bc

Exponents and Radicals

 a^ma^n = a^{m+n}  {(a^m)}^n = a^{mn}  {(ab)}^n = a^nb^n

 \dfrac{a^m}{a^n} = a^{m - n}     where a does not equal 0

 (\dfrac{a}{b})^n = \dfrac{a^n}{b^n}      where b does not equal 0

 a^{-n} = \dfrac{1}{a^n}  \sqrt{a} = a^{\frac{1}{2}}  \sqrt[3]{a} = a^{\frac{1}{3}}  \sqrt[n]{a} = a^{\frac{1}{n}}  (\sqrt[n]{a})^n = a  \sqrt[n]{ab} = (\sqrt[n]{a})(\sqrt[n]{b})

 \sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}     where b does not equal 0

 b^{\frac{m}{n}} = \sqrt[n]{b^m}

Logarithms

If  y = b^x then x = log_b y   for b > 0

 

 log_b b = 1

 

 log_b 1 = 0

 

 log_b (uv) = log_b u + log_b v

 

 log_b u^n = (n)log_b u

 

 b^{log_b x} = x  log_b {b^x} = x  log_a {(\frac{1}{x})} = -log_a x

Exponents and Logarithms

 e = \lim_{x\to\infty} {\left( 1 + \dfrac{1}{x} \right)}^x  e = 2.71828 ...  ln(e) = 1  e^{ln(a)} = a  log_e e = 1  log_e a = ln a  log_b a = \dfrac{\ln a}{\ln b}

Quadratic Equation

 ax^2 + bx + c     where a does not equal 0

The discriminant is    b^2 - 4ac

A parabola has the form    y = ax^2 + bx + c

and the vertex is    x = \dfrac{-b}{2a}

and the focus    \dfrac{a}{4}

Summation

 \sum_{i=1}^{\infty}\frac{1}{i} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + ...  \sum_{x=1}^3 x = 1 + 2 + 3 = 6

Calculus Derivative

\dfrac{d}{dx}\left(x^2\right) = 2x

Calculus Integral

 \int 2x\,dx = x^2 + C

Simplifying Equations

2x^2 + 3(x-1)(x-2) = 2x^2 + 3(x^2-3x+2) = 2x^2 + 3x^2 - 9x + 6 = 5x^2 - 9x + 6

Factoring

Difference of Squares
 m^2 - n^2 = (m + n)(m - n)
Simple Trinomials
 x^2 - 8x + 15 = (x - 5)(x - 3)

 

Complex Trinomials
2x^2 + 5x + 2 = 2x^2 + 4x + x + 2 = 2x(x + 2) + (x +2) = (x + 2)(2x + 1)