Numbers: Powers and roots
Square roots
Because #\blue3# squared equals #\orange9#, we call #\blue3# the square root of #\orange9#. We write this as:
\[\sqrt{\orange9}=\blue3\]
The following generally applies:
The square root of a #\orange{\textit{number}}# is a #\blue{\textit{non-negative number}}# that equals the #\orange{\textit{number under the}}# #\orange{\textit{radical sign } (\sqrt{\phantom{1}})}# #\text{when squared}#.
Note that we mentioned that the square root must be a non-negative number. Therefore, #\sqrt{\orange9}# #\red{\ne}# #\blue{-3}#, even though #\left(\blue{-3}\right)^2=\orange9#.
Examples
\[\begin{array}{rclc}\sqrt{\orange1}&=&\blue1 &\text{because }\blue1^2=\orange1 \text{ and } \blue1 \geq 0\\ \\\sqrt{\orange4}&=&\blue2 &\text{because }\blue2^2=\orange4 \text{ and } \blue2 \geq 0 \\ \\\sqrt{\orange9}&=&\blue3 &\text{because }\blue3^2=\orange9 \text{ and } \blue3 \geq 0\\ \\\sqrt{\orange{16}}&=&\blue4 &\text{because }\blue4^2=\orange{16} \text{ and } \blue4 \geq 0 \end{array}\]
The square roots in the examples all happen to be integers, but this does not apply to all square roots.
According to the definition of the square root, it holds that \[\left(\sqrt{\orange2}\right)^2=\orange2\]
Because #1^2=1# and #2^2=4#, we note that #1 \lt \sqrt{\orange2} \lt 2#.
Using a calculator, we find the approximation #\sqrt{\orange2} \approx 1.41423562....#
Later we will see that #\sqrt{\orange2}# cannot be written as a fraction, and that it has an infinite number of decimal places. When exercises ask you not to round off your answer, #\sqrt{\orange2}# is therefore considered a final answer. Just like #1#, #\tfrac{1}{2}# and #0.6#, #\sqrt{\orange2}# is a number.
So far, we have only seen square roots of non-negative numbers. That is because the square root of negative numbers does not exist.
According to the definition of the square root, #\sqrt{\orange{-1}}# should be a number that is equal to #\orange{-1}# when squared.
However, if we square a positive number, the result is always a positive number. Therefore, a positive number squared can never be #\orange{-1}#.
A negative number squared is also always a positive number. Therefore, a negative number squared can also never be #\orange{-1}#.
This means that #\sqrt{\orange{-1}}# does not exist. We can only extract roots of non-negative numbers.
\[\begin{array}{rcl}(\text{positive})^2&=&\text{positive} \;\times \text {positive} \\ &=&\text{positive} \\ \\ (\text{negative})^2&=&\text{negative} \;\times \text {negative} \\ &=&\text{positive} \end{array}\]
When we calculate #\sqrt{16}#, we are looking for a non-negative number which equals #16# when squared. In this case: \[4^2=16\] Therefore, #\sqrt{16}=4#.
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