This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.

- Welcome
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- Place Value in Decimal Number Systems
- Arithmetic Operations
- Basic Laws
- Operations on Signed numbers
- Order of Operations
- Some Useful Basic Numeracy
- Decimals
- Fractions
- Percents
- Ratios and Proportions
- Exponents
- Statistics
- Factoring
- Rearranging Formulas
- Solving Linear Equations
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- Simple Interest
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- Equivalent Values in Compound Interest
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- Place Value in Decimal Number Systems
- Arithmetic Operations
- Order of Operations
- Basic Laws
- Prime Factorisation and Least Common Multiple
- Fractions
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- Percents
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- Units of Measures
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- Basic Laws
- Operations with Numbers
- Prime Factorisation and Least Common Multiple
- Fractions
- Exponents
- Reducing Radicals
- Factoring
- Rearranging Formulas
- Solving Linear Equations
- Areas and Volumes of Figures
- Congruence and Similarity
- Functions
- Domain and Range of Functions
- Basics of Graphing
- Transformations
- Graphing Linear Functions
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- Solving Systems of Linear Equations
- Solving Quadratic Equations
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- Trigonometry
- Graphing Trigonometric Functions
- Graphing Circles and Ellipses
- Exponential and Logarithmic Functions
- Complex Numbers
- Number Bases in Computer Arithmetic
- Linear Algebra
- Calculus
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- Modular Numbers and Cryptography
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- Upgrading / Pre-Health
- Basic Laws
- Place Value in Decimal Number Systems
- Decimals
- Significant Digits
- Prime Factorisation and Least Common Multiple
- Fractions
- Percents
- Ratios and Proportions
- Exponents
- Reducing Radicals
- Metric Conversions
- Factoring
- Solving Linear Equations1
- Solving Quadratic Equations
- Polynomial Long Division
- Exponential and Logarithmic Functions
- Statistics

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- Arithmetic Operations
- Order of Operations
- Place Value in Decimal Number Systems
- Decimals
- Fractions
- Percents
- Ratios and Proportions
- Interpreting Drug Orders
- Oral Dosages
- Dosage Based on Size of the Patient
- Parenteral Dosages
- Intravenous (IV) Administration
- Infusion Rates for Intravenous Piggyback (IVPB) Bag
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There are three rules a radical expression must obey in order to be considered fully simplified.

- No square numbers as factors under the root sign.
- No fractions under the root sign.
- No square roots in the denominator.

Note: A square number is one that can be obtained by taking the square of a smaller number. For example, 4, 9, 16, 25, 36, and 49 are square numbers because \( \qquad 2^2 = 4 ,\qquad 3^2 = 9 ,\qquad 4^2 = 16 ,\qquad \) and so on...

`Example`

The following radicals are NOT fully simplified.

\[ \sqrt{18}, \quad \sqrt{\frac{1}{2}}, \quad \frac{3}{\sqrt{5}} \]

- \(\sqrt{18}\) because \(9\) is a factor of \(18\) and \(9\) is a square number (\(3^2\)).
- \( \sqrt{\frac{1}{2}} \) because there is a fraction underneath the square root.
- \( \frac{3}{\sqrt{5}} \) because the \(\sqrt5\) is in the denominator.

We will discuss how to simplify these types of radicals later on in the lesson.

To remove square factors from below square root signs, try to find the largest square factor of the number and factor the number accordingly.

`Example`

Let's simplify \(\sqrt{200}\).

`Solution`

Notice that \(200 = 100 \times 2\), so

\[ \begin{align*} \sqrt{200} &= \sqrt{100 \times 2} \\ &= \sqrt{100}\times\sqrt{2} \\ &= 10\sqrt{2} \end{align*} \]

But what if you don't notice that \(100\) is the largest square factor? If you have trouble identifying the largest square factor, you can always use the prime factorisation of the number to reduce. See the section on Prime Factorisation and Least Common Multiple for more details on this.

First, find the prime factorisation of 200.

\[ \begin{align*} 200 &= 100 \times 2 \\ &= 50 \times 2 \times 2 \\ &= 25 \times 2 \times 2 \times 2 \\ &= 5 \times 5 \times 2 \times 2 \times 2 \end{align*} \]

Therefore,

\[ \begin{align*} \sqrt{200} &= \sqrt{5 \times 5 \times 2 \times 2 \times 2} \\ &= \sqrt{5} \times \sqrt{5} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2} \end{align*} \]

Now, notice that

\[ \sqrt{5} \times \sqrt{5} = (\sqrt{5})^2 = 5 \qquad \text{ and likewise,} \qquad \sqrt{2} \times \sqrt{2} = 2 \]

so

\[ \begin{align*} \sqrt{200} &= \sqrt{5} \times \sqrt{5} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2} \\ &= 5 \times 2 \times \sqrt2 \\ &= 10\sqrt2 \end{align*} \]

Once the number is written out as a product of square roots of prime numbers, the numbers can be __paired up and reduced__ as shown in the example above.

`Example`

Suppose you find \(\sqrt{3}\times\sqrt{3}\times\sqrt{3}\times\sqrt{3}\times\sqrt{3}\) in the reduction. There are five \(\sqrt3\)'s in this expression, so group them as \((\sqrt{3}\times\sqrt{3})\times(\sqrt{3}\times\sqrt{3})\times\sqrt{3}\) and reduce as above to get \(9\sqrt3\)

Fractions under roots can be be dealt with using the following property:

\[ \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \]

`Examples`

- \( \sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{\sqrt{4}} = \dfrac{\sqrt{3}}{2} \)
- \( \sqrt{\dfrac{7}{3}} = \dfrac{\sqrt{7}}{\sqrt{3}} \)

Notice that removing the square root from below the root sometimes leads to a square root below the denominator as in the second example above. This can be fixed by multiplying above and below by the root term in the denominator.

`Example`

Using the above example, we ended up at:

\[ \sqrt{\frac{7}{3}} = \frac{\sqrt{7}}{\sqrt{3}} \]

with a \(\sqrt3 \) in the denominator. To remove it, we multiply above and below by \(\sqrt3\).

\begin{align} \frac{\sqrt{7}}{\sqrt{3}} &= \frac{\sqrt{7}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ &= \frac{\sqrt{7}\sqrt3}{3} \\ &= \frac{\sqrt{21}}{3} \end{align}

`Example`

Simplify the radical expression \(\sqrt{\dfrac{3}{160}}\)

`Solution`

- Apply the radical to both numerator and denominator. \[ \sqrt{\dfrac{3}{160}} = \dfrac{\sqrt3}{\sqrt{160}} \]
- Reduce \(\sqrt{160}\). \[ \sqrt{160} = \sqrt{16}\sqrt{10} = 4\sqrt{10} \]
- Remove the \(\sqrt{10}\) from the denominator.

\begin{align} \sqrt{\dfrac{3}{160}} &= \dfrac{3}{4\sqrt{10}} \\ &= \dfrac{\sqrt3}{4\sqrt{10}} \times \dfrac{\sqrt{10}}{\sqrt{10}} \\ &= \frac{\sqrt{30}}{40} \end{align}

Now the expression satisfies all the rules to be considered fully simplified.

- Last Updated: Mar 27, 2023 5:08 PM
- URL: https://libraryguides.centennialcollege.ca/mathhelp
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