# Roots and Rational Exponents: a summary

Although addition and multiplication are commutative, exponentiation is not: swapping the value in the base with the value in the exponent will produce a different result (unless, of course, they are the same value): $2^3 \ne 3^2$

Therefore, two different inverse functions are needed to solve equations that involve exponential expressions:
– roots, to undo exponents
logarithms, to undo bases

Just as there are many versions of the addition function (one for each number you might wish to add), and many versions of the “logarithm” function (each with a different base), there are many versions of the “root” function: one for each exponent value to be undone.

### Notation

The symbol for a root is $\sqrt{~~~~}$, and is referred to as a “radical“.  It consists of a sort of check mark on the left, followed by a horizontal line, called a “vinculum”, that serves as a grouping symbol (like parentheses) to let the reader know exactly what the root applies to. The expression under the radical is called the “radicand“. Roots may (or may not) be labelled with an “index“, which tells us the exponent that they undo.  The index appears above the notch at the left of the radical. $\sqrt{x~}$

In the expression above, “x” is the radicand. It is under a radical with an index of 3, and is read as “the cube root of x”.

Square roots (roots with an index of 2) are used more often than most other roots, so the notation for them has been simplified a bit by omitting the index: $\sqrt{25~}~=~\sqrt{25~}$

If no index is shown, the index is assumed to be 2 (a square root).

### Roots as Inverse Functions

A “square root” undoes an exponent of 2. In geometric terms, the square root of the area of a square tells you the length of each of the four sides of the square: $\sqrt{16~}~=4~$

In the equation above, “16” is the radicand. It is under a radical with an index of 2. A square with an area of 16 will have sides of length 4.

The square root of “four squared” is four. In other words, if you see inverse functions applied to one another, there is no need to evaluate the innermost expression before simplifying: $\sqrt{4^2~}~=4~$

A “cube root” undoes an exponent of 3: $\sqrt{2^3~}~=~2$

When a root’s index matches the value of an exponent, they are inverse functions. One “undoes” the other: $\sqrt{(x-7)^5~}~=~x-7\\*~\\*(\sqrt{x-7}~)^5~=~x-7$

The order in which the two inverse functions appear does not matter. They will always undo one another as long as  nothing else is “in the way”, as illustrated in the two examples above. Some examples of situations where this is not the case follow. Each has something that is “in the way” (the 3), that prevents them from being examples of a “function applied to its inverse”: $\sqrt{3+(x-7)^2~}\\*~\\*\sqrt{3\cdot (x-7)^2~}\\*~\\*(3+\sqrt{x-7~}~)^2\\*~\\*(3\sqrt{x-7~}~)^2$

While some of the above can be simplified further using the Laws of Roots (to be discussed below), they cannot be simplified as they appear above using the “inverse functions cancel each other out” concept.

### Roots as Rational Exponents

In the same way that subtraction can be rewritten as the addition of a negative, or division can be rewritten as multiplication by the reciprocal, a root can be rewritten as a rational exponent (an exponent that is a ratio of two integers). Consider the following true statement: $\sqrt{5^2~}=5^1$

If we set up a similar situation with the square of five, but ask ourselves how we could use an exponent to undo the square, we get: $(5^2)^x=5^1$

The laws of exponents tell us that when we raise an exponential expression to a power, we should multiply the powers, so we can rewrite the previous expression as: $5^{2\cdot x}=5^1$

If the above expression is to be true, the two exponents must equal each another: $2\cdot x=1\\*~\\*x=\frac{1}{2}$

In other words, a squared quantity that is raised to the one half power will end up with an exponent of one, which is the same as no exponent at all. Taking a square root has the same effect as raising to the one half power. Taking a cube root (index of 3) has the same effect as raising to the one third power.

It is often easier to manipulate expressions when roots are converted into fractional exponents. The following shows a variety of ways in which one expression can be rewritten: $\sqrt{(x+7)^5~}\\*~\\* =((x+7)^5)^\frac{1}{2}\\*~\\* =(x+7)^\frac{5}{2}\\*~\\* =(x+7)^{2.5}\\*~\\* =(x+7)^2 \cdot (x+7)^\frac{1}{2}\\*~\\* =(x+7)^2\sqrt{x+7~}$

When you encounter a rational exponent, the numerator is the power and the denominator is the root. The order in which you evaluate the power and the root does not matter; either order will produce the same result: $x^\frac{2}{3}~=~(\sqrt{x~})^2~=~\sqrt{x^2~}$

### Laws of Roots

Since roots can be expressed as fractional exponents, the laws of exponents apply to roots as well. Since an exponent can be distributed over a product or quotient, a root can be too. Note that each of these rules can be used either left-to-right or right-to-left.

The root of a product is the product of the roots $(ab)^\frac{1}{2}=a^\frac{1}{2}b^\frac{1}{2}$ $\sqrt{a \cdot b~}=\sqrt{a}\sqrt{b}$

The root of a quotient is the quotient of the roots $\left(\dfrac{a}{b}\right)^\frac{1}{2}=\dfrac{a^\frac{1}{2}~}{b^\frac{1}{2}~}$ $\sqrt{\dfrac{a}{b}~}=\dfrac{\sqrt{a}}{\sqrt{b}}$

Recall that exponents do not distribute over sums or differences, and so neither do fractional exponents or roots. Watch out for the following situations in which many people are tempted to rewrite expressions in ways that are NOT equivalent: $\sqrt{a+b~}\ne\sqrt{a}+\sqrt{b}$ $\sqrt{a-b~}\ne\sqrt{a}-\sqrt{b}$

### Simplifying Roots

Many students who confidently recite the square root of numbers like 4 or 25 may initially find simplifying roots of less friendly numbers to feel challenging. Two approaches help simplify the process:

1. Break the problem down into small pieces, and work on one piece at a time
2. Inverse functions undo one another, “freeing” what was inside the innermost function

Learning to simplify roots is often the first occasion that students are forced to rely on a function’s nature as an inverse. A purely rote approach does not work on all problems. While previous operations or functions may have been explained as inverses, students have often managed to continue solving such problems in rote ways. So, an extra helping of patience and perseverance are useful when grappling with how two functions can “undo” one another for the first time.

To simplify a root, it is very helpful to rewrite the expression under the root as a product of prime factors. All that practice rewriting a number using its prime factorization when in middle school will be useful now: $\sqrt{24~}\\*~\\* =\sqrt{2 \cdot 2 \cdot 2 \cdot 3~}\\*~\\* =\sqrt{(2 \cdot 2) \cdot 2 \cdot 3~}\\*~\\* =\sqrt{(2 \cdot 2)}\sqrt{2 \cdot 3~}\\*~\\* =\sqrt{2^2} \sqrt{2 \cdot 3~}\\*~\\* =2\sqrt{6~}$

A square root undoes squaring, so to simplify a square root your need to find factors that appear twice. By rewriting each quantity in its prime factorization, you make it easy to find repeated factors – as illustrated above. Once pairs of factors have been identified, they can be broken out into a separate radical using the Laws of Roots, then simplified by recognizing that a square root undoes a square.

The exact same process applies to ANY root, except that instead of looking for pairs of identical factors, the index of the root tells you how many instances of a factor are necessary for it to be able to be simplified: $\sqrt{24~}\\*~\\* =\sqrt{2 \cdot 2 \cdot 2 \cdot 3~}\\*~\\* =\sqrt{(2 \cdot 2 \cdot 2) \cdot 3~}\\*~\\* =\sqrt{(2 \cdot 2 \cdot 2)~}\sqrt{ 3~}\\*~\\* =\sqrt{2^3~}\sqrt{ 3~}\\*~\\* =2\sqrt{3~}$

And as expressions become even more complex, the same process continues to apply: $\sqrt{360 x^3 y^5 ~}\\*~\\* =\sqrt{2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot (x^2) \cdot x \cdot (y^2) \cdot (y^2) \cdot y ~}\\*~\\* =\sqrt{2 \cdot 2~} \sqrt{3 \cdot 3~} \sqrt{x^2 ~} \sqrt{y^2 ~} \sqrt{y^2 ~} \sqrt{2 \cdot 5 \cdot x \cdot y ~}\\*~\\* =2 \cdot 3 \cdot x \cdot y \cdot y \cdot \sqrt{2 \cdot 5 \cdot x \cdot y ~}\\*~\\* =6xy^2 \sqrt{10xy ~}$

As with most math problems, there is more than one valid way to arrive at the answer… and some ways will be more efficient than others.  But don’t worry about efficiency until after you have mastered the concepts, and are confident that you can arrive at a correct answer using whatever approach you are most comfortable with. So, for example, the above problem could also be simplified as follows: $\sqrt{360 x^3 y^5 ~}\\*~\\* =\sqrt{(36) \cdot 10 \cdot (x^2) \cdot x \cdot (y^4) \cdot y ~}\\*~\\* =\sqrt{36~} \sqrt{x^2 ~} \sqrt{y^4 ~} \sqrt{10 \cdot x \cdot y ~}\\*~\\* =6 \cdot x \cdot y^2 \cdot \sqrt{10 \cdot x \cdot y ~}\\*~\\* =6xy^2 \sqrt{10xy ~}$

or using rational exponents: $\sqrt{360 x^3 y^5 ~}\\*~\\* =(360 x^3 y^5)^\frac{1}{2}\\*~\\* =(36 \cdot 10 \cdot x^3 \cdot y^5)^\frac{1}{2}\\*~\\* =36^\frac{1}{2} \cdot 10^\frac{1}{2} \cdot (x^3)^\frac{1}{2} \cdot (y^5)^\frac{1}{2}\\*~\\* =6 \cdot 10^\frac{1}{2} \cdot x^\frac{3}{2} \cdot y^\frac{5}{2}\\*~\\* =6 \cdot \sqrt{10~} \cdot x \sqrt{x~} \cdot y^2 \sqrt{y}\\*~\\* =6xy^2 \sqrt{10xy ~}$ ## By Whit Ford

Math tutor since 1992. Former math teacher, product manager, software developer, research analyst, etc.

1. Shaun says: