Definition of Rationalization of Radicals (mathematics)

The rationalization of radicals is a mathematical process that is carried out when there is a quotient with radicals or roots in the denominator. In this way, mathematical operations can be facilitated where quotients with radicals and other types of mathematical objects are involved.

Types of Quotients with Radicals

It is important to mention some types of quotients with radicals that can be rationalized. However, before getting fully into the streamlining process, a couple of important concepts need to be remembered. First, suppose we have the following expression: \(\sqrt[m]{n}\). This is the root \(m\) of the number \(n\), that is, the result of said operation is a number such that raising it to the power \(m\) gives us the number \(n\) as a result ). The power and the root are inverse operations, in such a way that: \(\sqrt[m]{{{n^m}}} = n\).
On the other hand, it is worth mentioning that the product of two equal roots is equal to the root of the product, that is to say that: \(\sqrt[m]{n}\sqrt[m]{p} = \sqrt[m ]{{np}}\). These two properties are going to be our best allies when rationalizing.

The most common and simple type of quotient with a radical that we can find is the following:

\(\frac{a}{{b\sqrt c }}\)

Where \(a\), \(b\) and \(c\) can be any real numbers. The rationalization process in this case consists of finding a way to obtain in the quotient the expression \(\sqrt {{c^2}} = c\) to get rid of the radical. In this case, it is enough to multiply by \(\sqrt c \) both the numerator and the denominator:

\(\frac{a}{{b\sqrt c }} = \frac{a}{{b\sqrt c }}\frac{{\sqrt c }}{{\sqrt c }} = \frac{{ a\sqrt c }}{{b\sqrt c \sqrt c }}\)

Remembering what was mentioned above, we know that \(\sqrt c \sqrt c = \sqrt {{c^2}} = c\). Therefore, we finally obtain that:
\(\frac{a}{{b\sqrt c }} = \frac{a}{{bc}}\sqrt c \)

In this way we have rationalized the previous expression. This expression is nothing more than a particular case of a general expression that is the following:

\(\frac{a}{{b\sqrt[n]{{{c^m}}}}}\)

Where \(a\), \(b\), \(c\) are any real numbers and \(n\), \(m\) are positive powers. The rationalization of this expression is based on the same principle as the previous one, that is, obtain the expression \(\sqrt[n]{{{c^n}}} = c\) in the denominator. We can achieve this by multiplying by \(\sqrt[n]{{{c^{n – m}}}}\) both the numerator and the denominator:

\(\frac{a}{{b\sqrt[n]{{{c^m}}}}} = \frac{a}{{b\sqrt[n]{{{c^m}}}} }\frac{{\sqrt[n]{{{c^{n – m}}}}}}{{\sqrt[n]{{{c^{n – m}}}}}} = \frac{{a\sqrt[n]{{{c^{n – m}}}}}}{{b\sqrt[n]{{{c^m}}}\sqrt[n]{{{c^{n – m}}}}}}\)

We can develop the product of radicals in the denominator as follows: \(\sqrt[n]{{{c^m}}}\sqrt[n]{{{c^{n – m}}}} = \sqrt[n]{{{c^m}{c^ {n – m}}}} = \sqrt[n]{{{c^{m + \left( {n – m} \right)}}}} = \sqrt[n]{{{c^n}}} = c\). Therefore, the rationalized quotient remains as:

\(\frac{a}{{b\sqrt[n]{{{c^m}}}}} = \frac{a}{{bc}}\sqrt[n]{{{c^{n – m}}}}\)

Another type of quotient with radicals that can be rationalized is the one in which we have a binomial with square roots in the denominator:

\(\frac{a}{{b\sqrt c \pm d\sqrt e }}\)

Where \(a\), \(b\), \(c\), \(d\) and \(e\;\)are any real numbers. The symbol \( ± \) indicates that the sign can be positive or negative. The denominator binomial can have both roots or only one, however, we use this case to obtain a more general result. The central idea to carry out the rationalization process in this case is the same as in the previous cases, only that in this case we will multiply both the numerator and the denominator by the conjugate of the binomial found in the denominator. The conjugate of a binomial is a binomial that has the same terms, but whose central symbol is opposite to the original binomial. For example, the conjugate of the binomial \(ux + vy\) is \(ux – vy\). That being said, we then have:

\(\frac{a}{{b\sqrt c \pm d\sqrt e }} = \frac{a}{{b\sqrt c \pm d\sqrt e }}\frac{{b\sqrt c \ mp d\sqrt e }}{{b\sqrt c \mp d\sqrt e }} = \frac{{a\left( {b\sqrt c \mp d\sqrt e } \right)}}{{\left( {b\sqrt c \pm d\sqrt e } \right)\left( {b \sqrt c \mp d\sqrt e } \right)}}\)

The symbol \( \mp \) indicates that the sign can be positive or negative, but it has to be opposite to the symbol of the denominator for the binomials to be conjugated. By developing the multiplication of binomials of the denominator we obtain that:

\(\frac{a}{{b\sqrt c \pm d\sqrt e }} = \frac{{a\left( {b\sqrt c \mp d\sqrt e } \right)}}{{{ b^2}\sqrt {{c^2}} + bd\sqrt {ce} – bd\sqrt {ce} – {d^2}\sqrt {{e^2}} }}\)

Finally we get that:

\(\frac{a}{{b\sqrt c \pm d\sqrt e }} = \frac{a}{{{b^2}c – {d^2}e}}\left( {b\ sqrt c \mp d\sqrt e } \right)\)

With this we have rationalized the quotient with radical. These quotients with radicals are the ones that can generally be rationalized. Next, we will see some examples of rationalization of radicals.

examples

Let's look at some examples of rationalization with quotients with radicals of the type mentioned above. First suppose we have the following quotient:

\(\frac{3}{{7\sqrt 2 }}\)

In this case it is enough to multiply the numerator and the denominator by \(\sqrt 2 \)

\(\frac{3}{{7\sqrt 2 }} = \frac{3}{{7\sqrt 2 }}\frac{{\sqrt 2 }}{{\sqrt 2 }} = \frac{3 }{{7\sqrt 2 \sqrt 2 }}\sqrt 2 = \frac{3}{{7\sqrt 4 }}\sqrt 2 = \frac{3}{{14}}\sqrt 2 \)

Now, suppose we have the following quotient with radical:

\(\frac{2}{{3\sqrt[6]{{{4^3}}}}}\)

In this case we have a sixth root of a cubic power. In the previous section we mentioned that if we have a radical of the form \(\sqrt[n]{{{c^m}}}\) in the denominator, we can rationalize the quotient by multiplying the numerator and denominator by \(\sqrt[n]{{{c^{n –m}}}}\). Comparing this with the case presented here we can realize that \(n = 6\), \(c = 4\) and \(m = 3\), therefore Therefore, we can rationalize the previous quotient by multiplying the numerator and the denominator by \(\sqrt[6]{{{4^3}}}\):

\(\frac{2}{{3\sqrt[6]{{{4^3}}}}} = \frac{2}{{3\sqrt[6]{{{4^3}}}} }\frac{{\sqrt[6]{{{4^3}}}}}{{\sqrt[6]{{{4^3}}}}} = \frac{2}{{3\sqrt[6]{{{4^3}}}\sqrt[6]{{{4^3}}}}}\sqrt[6]{{{4^3} }} = \frac{2}{{3\sqrt[6]{{{4^6}}}}}\sqrt[6]{{{4^3}}} = \frac{{\sqrt[6]{{{4^3}}}}}{6}\)

Finally, suppose we have the following function:

\(\frac{1}{{x + \sqrt x }}\)

As shown in the previous section, to rationalize this type of quotient with radicals, you have to multiply the numerator and denominator by the conjugate of the denominator. In this case the conjugate of the denominator would be \(x – \sqrt x \). Therefore, the expression would be as follows:

\(\frac{1}{{x + \sqrt x }}\frac{{x – \sqrt x }}{{x – \sqrt x }} = \frac{1}{{\left( {x + \sqrt x } \right)\left( {x – \sqrt x } \right)}}\left( {x – \sqrt x } \right)\)

Developing the multiplication of conjugate binomials of the denominator, we finally obtain that:

\(\frac{1}{{x + \sqrt x }} = \frac{{x – \sqrt x }}{{{x^2} – x}}\)

References

Aguilar Arturo, Bravo Fabián, Gallegos Herman, Cerón Miguel & Reyes Ricardo. (2009). Arithmetic. Mexico: Pearson Education.