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.

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- Basic Laws
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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 18, 2024 3:47 PM
- URL: https://libraryguides.centennialcollege.ca/mathhelp
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