Calculating with square roots

Algebra: Calculating with exponents and roots

Calculating with square roots

The following rules are very convenient when simplifying expressions with square roots.

\[\sqrt{\blue a \cdot \green b}=\sqrt{\blue a} \cdot \sqrt{\green b}\]

Example

\[\begin{array}{rcl}\sqrt{\blue 4 \cdot \green x}&=&\sqrt{\blue 4} \cdot \sqrt{\green x}\\ &=& 2 \cdot \sqrt{\green x} \end{array}\]

Not for adding

There is no similar rule for adding square roots. Hence:

\[ \sqrt{a+b} \; \red{\ne} \; \sqrt{a} + \sqrt{b} \]

Example

\[\sqrt{9+16} \;\red{\not=}\; \sqrt{9} + \sqrt{16} \]

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

Example

\[\begin{array}{rcl}\sqrt{\dfrac{\blue 4}{\green x}}&=&\dfrac{\sqrt{\blue 4}}{\sqrt{\green x}}\\ &=& \dfrac{2}{\sqrt{\green x}} \end{array}\]

For #\blue a \geq 0# we have:

\[\sqrt{\blue a^2}=\blue a\]

and

\[\left(\sqrt{\blue a}\right)^2=\blue a\]

Examples

\[\begin{array}{rcl}\sqrt{\blue x^4}&=&\sqrt{\blue x^2 \cdot \blue x^2}\\ &=& \sqrt{\blue x^2} \cdot \sqrt{\blue x^2} \\ &=& \blue x \cdot \blue x \\&=&\blue x^2\\ \\ \left(\sqrt{\blue x}\right)^4 &=& \left(\left(\sqrt{\blue x}\right)^2\right)^2 \\ &=& \blue x^2\end{array}\]

In the first rule, when #\blue a# is negative, we will also have a positive number under the root sign. Since a square root is always positive, #\blue a# cannot be negative.

Because #\sqrt{(-5)^2} =5 \ne -5#.

We can formulate this rule for negative #\blue a#, we then find:

\[\sqrt{\blue a^2}=\abs{\blue a}\]

Here, #\abs{\blue a}# is the absolute value of #\blue a#, which means the value is always positive. Hence, #\abs{-5}=5#.

Assume that #z \ge 0# and #d \ge 0#. Write without a square root: #\sqrt{9 \cdot z^{6} \cdot d^{6}}#

#3 \cdot z^3 \cdot d^3#

#\begin{array}{rcl}
\sqrt{9 \cdot z^{6} \cdot d^{6}}&=& \sqrt{9} \cdot \sqrt{z^{6}} \cdot \sqrt{d^{6}} \\
&& \phantom{xxx}\blue{\text{rule }\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}}\\
&=& 3 \cdot z^3 \cdot d^3\\
&& \phantom{xxx}\blue{\text{rule }\sqrt{a^2}=a \text{ and square root calculated }}
\end{array}#

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