### Numbers: Powers and roots

### Standard notation of roots

We have seen some calculation rules for *roots* and *roots of fractions*. Using these calculation rules we can simplify roots.

When a root occurs in an expression, the expression is in **standard notation** if it meets the following requirements.

*The number in front of the radical sign is as large as possible.*

This means that we eliminate the squares from the number inside the radical sign. Keep in mind that we only want one square root. Therefore, you should not 'simplify' #\sqrt{14}# to #\sqrt{2}\cdot\sqrt{7}#. We do the following.

\[\sqrt{12}=\sqrt{2^2 \cdot 3}=\sqrt{2^2} \cdot \sqrt{3}=2\sqrt{3}\]*There are no fractions inside the radical sign.*We use that the root of a fraction equals the fraction of the roots. \[\sqrt{\frac{2}{3}}=\frac{\sqrt{2}}{\sqrt{3}}\]*There are no roots in the denominator of the fraction.*

We achieve this by multiplying the numerator and the denominator of the fraction by the root from the denominator.

\[\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}=\frac{\sqrt{3}}{3}\]*The rational numbers in the expression are simplified fractions.*\[\frac{3}{\sqrt{3}}=\frac{3\sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}\]

**Examples**

\[\begin{array}{rcl}\sqrt{56}&=&\sqrt{2^2 \cdot 14} \\ &=& \sqrt{2^2} \cdot \sqrt{14} \\ &=& 2 \sqrt{14} \\ \\\sqrt{108}&=&\sqrt{2^2 \cdot 3^2 \cdot 3} \\&=& \sqrt{2^2} \cdot \sqrt{3^2} \cdot \sqrt{3} \\ &=& 2 \cdot 3 \cdot \sqrt{3} \\ &=& 6 \sqrt{3} \\ \\ \dfrac{2}{\sqrt{5}}&=& \dfrac{2 \sqrt{5}}{\sqrt{5} \cdot \sqrt{5}}\\ &=& \dfrac{2 \sqrt{5}}{5} \\ \dfrac{3}{\sqrt{63}}&=&\dfrac{3}{\sqrt{3^2\cdot7}}\\&=& \dfrac{3}{3\cdot\sqrt{7}}\\&=&\dfrac{1}{\sqrt{7}}\\&=& \dfrac{\sqrt{7}}{\sqrt{7}\cdot\sqrt{7}}\\&=&\dfrac{\sqrt{7}}{7}\end{array}\]

#\begin{array}{rcl}

\sqrt{125}&=&\sqrt{25\times 5} \\ &&\phantom{xxx}\blue{125 \text{ written as a product inside the radical symbol}} \\

&=& \sqrt{{5}^2\times 5} \\ &&\phantom{xxx}\blue{25 \text{ written as a square}} \\

&=& \sqrt{{5}^2} \times \sqrt{5} \\ &&\phantom{xxx}\blue{\text{calculation rule: the root of a product is the product of roots}} \\

&=&5\sqrt{5} \\ &&\phantom{xxx}\blue{\text{root of a square is the absolute value of the number}} \end{array}#

**Pass Your Math**independent of your university. See pricing and more.

Or visit omptest.org if jou are taking an OMPT exam.