The material on this page is incomplete. Please help contribute and expand this article. This article needs serious refurnishing!

This article discusses the topic of radicals and fractional exponents.

The Radicals

A radical expression is an expression under a division house like structure that indicates the inverse of exponentiation. These are commonly identified with "the nth root of…". The most common of these roots is the square root which is the inverse of squaring. This means that the square root of 5 squared is simply 5. Radicals are indicated like this:

\begin{align} \sqrt[n]{b}=a \end{align}

The n is the index, the b is the radicand, and the a is the base of exponentiation. This actually is taking the inverse of an=b. The radical symbol, $\sqrt[n]{}$, usually describes the principal root of a number. That is, the radical only refers to the odd root unless the index is odd. The reason for principal roots is that roots with an even index have two answers: one positive and one negative. A principal root only takes the positive of those. The way to indicate the negative root is with the negative sign in front of the radical. Sometimes, the symbol $\pm$ is used. This shows both roots of the radical.

There is a difference, however, when trying to solve an equation like x2=4. The answer, after taking the inverse of squaring (the square root), you get $x = \pm 2$. However, this cannot happen with the principal root. What is actually happening is that when you take the square root of x2, you actually get |x|. Observe the process:

\begin{align} x^2 = 4 \\ \caption{Principal Root} \qquad \sqrt{x^2} = \sqrt{4} \\ |x| = 2 \\ \caption{Thus} \qquad x = 2 \quad or \quad x = -2 \end{align}

This shows where the plus or minus 2 actually comes from. The reason we need the |x| is because x could be any number. Suppose we had the square root of two squared. The result is obviously two. Now, suppose we had the square root of negative two squared. The answer is not negative two but two, rather. So, to make sure x is not negative, absolute values are used. The rule for this is if the radical has an even index and the simplified result has an odd power, then absolute values must be applied.


There are few special rules when dealing with radicals. Fortunately, these rules can help simplify radicals.
First, when you multiply the n root of a number by the n root of another number, you end up with the n root of the product of the two numbers.

\begin{align} \sqrt[n]{a}\sqrt[n]{b} = \sqrt[n]{ab} \end{align}

For example, say we have $\sqrt{4}\sqrt{16}$. The answer is obvious: 2*4=8. However, if we use our rule, we can see a similarity. Say we combined the radicals to make $\sqrt{4 \times 16}$. The product of 4 and 16 is 64. So, we can see that $\sqrt{64} = 8$ which is what we got before. That is how the little rule works.

The second rule is when you are taking the n root of a fraction, you can say that you are taking the n root of the numerator and the n root of the denominator.

\begin{align} \sqrt[n]{a \over b} = {\sqrt[n]{a} \over \sqrt[n]{b}} \end{align}

So, let us say that you want to find $\sqrt[3]{64 \over 8}$. You could simplify 64/8 to 8, and then find the cube root of 8 which is 2. Or, you could take the cube root of the numerator and the cube root of the denominator and come up with the same answer. ${\sqrt[3]{64} \over \sqrt[3]{8}} = {4 \over 2} = 2$

The next rule states that radicals can also be written as a rational exponent. When you take the n root of a number, you are actually taking the radicand to the n-1 power. Here is the rule.

\begin{align} \sqrt[n]{b} = b^{1/n} \end{align}

So, $\sqrt{25} = 25^{1/2}$. This can become incredibly useful when simplifying radicals. Knowing this rule allows you to combine radicals of different indexes. All you have to do is convert each radical to its exponential form and then apply the laws of exponents.

\begin{align} \sqrt[n]{a}\sqrt[m]{b} = a^{1/n}b^{1/m} = a^{m/nm}b^{n/nm} = (a^mb^n)^{1/nm} = \sqrt[nm]{a^mb^n} \\ \sqrt{3}\sqrt[3]{5} = 3^{1/2} \times 5^{1/3} = 3^{3/6} \times 5^{2/6} = (3^3 \times 5^2)^{1/6} =\sqrt[6]{675} \end{align}

The next operation is almost like combining like terms. Sometimes, you can pretend that the radical is a variable. When you do this, you can add and subtract radicals when the radicands are equivalent.

\begin{align} a\sqrt[n]{x}+b\sqrt[n]{x} = (a+b)\sqrt[n]{x} \end{align}

So, using an obvious example, let's take $2\sqrt{4}+4\sqrt{4}$. The answer is obviously 2(2)+4(2) which is 12. Well, using the rule, we can get $6\sqrt{4}$. This is equivalent to 6(2) which is also 12.

The other rule deals with radicals under radicals. We can use this rule to help get rid of radical signs to simplify the expression. However, this rule only works if the radicals are directly stacked. If there is something outside a radical within the parent radical, then the formula does not work.

\begin{align} \sqrt[n]{\sqrt[m]{a}} = \sqrt[mn]{a} \end{align}

This one does not work:

\begin{align} \sqrt[n]{a\sqrt[m]{b}} \ne \sqrt[mn]{ab} \end{align}

To simplify that, you need to put it into factional exponent form.

\begin{align} \sqrt[n]{a\sqrt[m]{b}} = (a(b)^{1/m})^{1/n} = a^{1/n}b^{1/mn} = \sqrt[n]{a}\sqrt[mn]{b} \end{align}

This represents a lot of the operations radicals are capable of. These operations is used for simplification and solving some polynomials.

Negative Indexes


Simplifying radicals is the process of bringing them into a more readable format. The goal of simplification is to get rid of as much of the radical as possible. This means reducing the radicand to a smaller size and using single radical houses. A radical is completely simplified when all of the follow has been met:

  • All perfect n numbers have been extracted out of a radical
  • The index is the smallest it can be
  • The radicand has no fractions
  • There are no radicals in the denominator of a fraction

There are many types of radical simplification techniques that can be used to simplify.

Perfect n Extraction

Perfect n extraction is the process of taking out a perfect n number from the radical. This is the general premise of simplifying radicals. This is what it might look like.

\begin{align} \sqrt[3]{a^5} = \sqrt[3]{a^3 \times a^2} = a\sqrt[3]{a^2} \end{align}

What you are doing is looking for the most a3's in the above example as possible. Since $\sqrt[3]{a^3}$ is a, you can take it out and multiply it by the rest of the radical. Consider the number example below.

\begin{align} \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \end{align}

If you plug in the original square root of 20 into the calculator, you get 4.472136. If you press two times the square root of five, you also get 4.472136.

Factor Tree

Perfect n extraction requires some training to get perfect every time. The other method of simplifying radicals is using factor trees. This is essentially the same thing as perfect n extraction, but this shows what happens and gives a set process.

First, you want to find the prime factorization of the radicand. This will give you all the primes for the number. Next, you group all the equal numbers together.

\begin{align} \sqrt[3]{23625} \qquad\qquad\qquad\; \\ \sqrt[3]{15} \qquad\qquad \sqrt[3]{1575}\qquad\quad\; \\ \sqrt[3]{3}\quad \sqrt[3]{5} \quad\;\;\; \sqrt[3]{25} \quad\;\;\; \sqrt[3]{63}\qquad \\ \sqrt[3]{3}\quad \sqrt[3]{5} \;\;\; \sqrt[3]{5} \;\; \sqrt[3]{5} \quad \sqrt[3]{9} \;\;\; \sqrt[3]{7}\;\;\; \\ \sqrt[3]{3}\quad \sqrt[3]{5} \;\;\; \sqrt[3]{5} \;\; \sqrt[3]{5} \; \sqrt[3]{3}\sqrt[3]{3} \; \sqrt[3]{7}\;\;\; \end{align}

Now group together

\begin{align} \sqrt[3]{3}\sqrt[3]{3}\sqrt[3]{3}\sqrt[3]{5}\sqrt[3]{5}\sqrt[3]{5}\sqrt[3]{7} \end{align}

Next, you would count how many of each number you have. You have three threes, three fives, and one seven. Notice your index: 3. That means that for every group of three common numbers, you can dispose of their radicals to form that number. So, you yield a final result of:

\begin{align} 3 \times 5 \times \sqrt[3]{7} = 15\sqrt[3]{7} \end{align}

The reason this works is because you can combine radicals of the same index with multiplication. Remember our first rule? So, $\sqrt[3]{3}\sqrt[3]{3}\sqrt[3]{3} = \sqrt[3]{3 \times 3 \times 3} = \sqrt[3]{27} = 3$. So, for other indexes, you can do that same thing for every common group of n.

Index Reduction

Rule Implementation

Fancy Form of One

This method is used to rationalize fractions with radicals in the denominator. For this to be utilized, the denominator must be a single term expression. One of the rules for simplification is to leave no radicals in the denominator, so if we have something like $6 \over \sqrt{6}$, we need to simplify it.

If this were an equation, we could simply multiply by $\sqrt{6}$. However, this is an expression, and the only thing we can multiply by is one. The reason we can multiply by one is because multiplying by one never changes the value of the expression. So we know two things: we must multiply by $\sqrt{6}$, and we can only multiply by one.

A compromise must be made. This is where we multiply by a fancy form of one. A fancy form of one is any expression equivalent to one; fancy forms of one are usually unsimplified. Here are some examples.

\begin{align} \frac{2}{2}, \frac{10+3}{10+3}, \frac{\sqrt{5}}{\sqrt{5}}, \frac{3^2}{3^2}, \ln e^{\ln e \end{align}

So, if we multiply by a fancy form of one, we can cancel the denominator and therefore fix the radical. To fix the problem, we need to multiply by $\sqrt{6} \over \sqrt{6}$. When we do that, the numerator becomes $6 \times \sqrt{6} = 6\sqrt{6}$, and the denominator becomes $\sqrt{6} \times \sqrt{6} = 6$. The entire fraction then simplifies to the nice $\sqrt{6}$.


Sometimes, a radical's denominator is not nice looking like a simple $\sqrt{6}$. The radical's denominator has the potential of being $\sqrt{5}+3$. In this case, rationalizing the denominator is not as simple as multiplying by $\sqrt{5} \over \sqrt{5}$. You must conjugate the expression.

A conjugate is a pair that fits evenly with one another. In mathematics, this would involve two expressions when put together by some operation works out to yield a convenient answer. For conjugating a fraction like $5 \over \sqrt{3}+6$, it is best to observe the difference of squares property.

A difference of squares is any expression in the form a2-b2. When factored, this conveniently yields (a+b)(a-b). This happens because the middle 2ab term is canceled out because of a plus-minus add out. The goal in conjugation of radicals is to get rid of the radical all together.

Consider $(\sqrt{3}+6)(\sqrt{3}-6)$. This looks a lot like the difference of squares. When you simplify the expression, you get $3-6\sqrt{3}+6\sqrt{3}-36$. This yields the beautiful -33 with no radicals.

So, using the advanced fancy form of one, we can null the radical in the denominator and therefore simplify the expression.

\begin{align} {5 \over \sqrt{3}+6} \times {\sqrt{3}-6 \over \sqrt{3}-6} = {5\sqrt{3}-30 \over -33} \end{align}

Conjugation will only work with square roots. Other indexes cannot be simplified in this manner.

Radical Equations

Complex and Imaginary Units

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License