Higher degree roots

Algebra: Calculating with exponents and roots

Higher degree roots

Roots are also called square roots, to distinguish them from higher degree roots, which we will define here. Unlike with square roots, it's possible to take the root of a negative number.

Please check that #(\green{-2})^\purple{3}= \blue{-8}#.
Therefore, the cube root is #\sqrt[\purple{3}]{\blue{-8}}=\green{-2}#.
This is the only number #\green x# for which we have #\green{x}^\purple{3} = \blue{-8}#.
Also #\green{2}^\purple 3=8# does not equal #\blue{-8}#.

\[\begin{array}{c}
\sqrt[\purple 3]{\blue{-8}}=\green{-2} \\
\text{since} \\
(\green{-2})^\purple 3 = \blue {-8}
\end{array}\]

Fourth degree roots have more in common with square roots. This is because #\orange 2# and #\orange 4# are both even numbers.

Please check that #(\green 2)^\orange 4=\blue{16}#.
Therefore, the fouth degree root is #\sqrt[\orange 4]{\blue {16}}=\green 2#.
But also #({-2})^\orange 4=\blue{16}#.
Hence, there are two numbers #\green{x}# for which we have #\green x^\orange 4 = \blue {16}#.
The solution to this fourth degree root is #\green{2}# because #\green{2} \geq 0#.

\[\begin{array}{c}
\sqrt[\orange 4]{\blue{16}}=\green{2} \\
\text{since} \\
\green{2}^\orange 4 = \blue{16} \quad \text{and} \quad \green 2 \geq 0
\end{array}\]

If #\orange n# is even, then the #\orange n#-th degree root #\sqrt[\orange n]{\blue a}# of #\blue a# is the number #\green x# for which we have

\[\green x^\orange n=\blue a \quad \text{and} \quad \green x\geq 0 \]

Examples

\[\begin{array}{c}\sqrt[\orange 4]{\blue{81}} = \green 3 \\
\text{since} \\
\green 3^\orange 4 = \blue{81} \quad \text{and} \quad \green 3 \geq 0\end{array}\]

If #\purple n# is odd, then the restriction #\green x\geq 0# does not apply anymore.

The #\purple n#‍-‍th degree root #\sqrt[\purple n]{\blue a}# of #\blue a# is the number #\green x# for which we have

\[\green x^\purple n=\blue a \]

\[\begin{array}{c}\sqrt[\purple 3]{\blue{27}} = \green 3 \\
\text{since} \\
\green 3^\purple 3 = \blue{27}\end{array}\]

Above we saw that if you take the root of#\sqrt[\orange n]{\blue a}# with #\orange n# being even, we will never find a negative number #\green x#. In this case, we also have that #\blue a# itself cannot be negatieve, hence, #\blue a \geq 0#.

If #\orange n# is even, then the higher degree root of a number #\sqrt[\orange n]{\blue a}# is only meaningful if #\blue a\geq 0#. Hence, you can never enter a number like #\blue{-1}# for #\blue a#.

Expressions and algebraic rules with these kind of roots are only valid if the expression below the roots sign is not negative. For example, for the expression #\sqrt[\orange 6]{\blue{ab}}# we must have #\blue{ab}\geq 0#.

Example

\[\begin{array}{c}
(\sqrt[\orange{4}]{\blue{x}})^4 = x \\
\text{this rule implicitely assumes that }\\ \blue{x} \geq 0 \\ \text{since \(\orange{4}\) is even}\end{array}\]

In the examples below it will become clear which values for #x# are allowed to be subtituted under a higher degree root.

Substitute #x=8# in #\sqrt[5]{x-1}#. Is this possible?

Yes

The root is of an odd degree, being #5#. This means that any value of #x# may be substituted, including when #x=8#.

New example