Rationalizing a Denominator or Numerator

Quick Reference

Denominator form	Multiply by	Result
$\sqrt{A} \pm \sqrt{B}$	$\sqrt{A} \mp \sqrt{B}$	$A - B$
$C\sqrt{A} \pm D\sqrt{B}$	$C\sqrt{A} \mp D\sqrt{B}$	$C^{2}A - D^{2}B$
$\sqrt[3]{A} + \sqrt[3]{B}$	$\sqrt[3]{A^{2}} - \sqrt[3]{AB} + \sqrt[3]{B^{2}}$	$A + B$
$\sqrt[3]{A} - \sqrt[3]{B}$	$\sqrt[3]{A^{2}} + \sqrt[3]{AB} + \sqrt[3]{B^{2}}$	$A - B$

• Remark that $C\sqrt{A}-D\sqrt{B}$ and $C\sqrt{A}+D\sqrt{B}$ are conjugates of each other.
• If the denominator of a fraction is of the form $\sqrt{A}-\sqrt{B}$ (or $C\sqrt{A}-D\sqrt{B}$), we can rationalize the denominator by multiplying the numerator and denominator of the fraction by the conjugate $\sqrt{A}+\sqrt{B}$ (or $C\sqrt{A}+D\sqrt{B}$). For example,

$ \begin{aligned} \frac{13}{\sqrt{5}+3\sqrt{2}} & =\frac{13}{\sqrt{5}+3\sqrt{2}}\cdot\frac{\sqrt{5}-3\sqrt{2}}{\sqrt{5}-3\sqrt{2}} \\[6pt] & =\frac{13(\sqrt{5}-3\sqrt{2})}{5-3^{2}\times2} \\[6pt] & =-(\sqrt{5}-3\sqrt{2}) \\[6pt] & =3\sqrt{2}-\sqrt{5} \end{aligned} $

Similarly:

• If the denominator of a fraction is $\sqrt[3]{A}+\sqrt[3]{B}$, we multiply the numerator and denominator of the fraction by $\sqrt[3]{A^{2}}-\sqrt[3]{AB}+\sqrt[3]{B^{2}}$ and use the Sum of Cubes formula to get a denominator of $A+B$. (See Section: Special Product Formulas for the special product formulas).
• If the denominator of a fraction is $\sqrt[3]{A}-\sqrt[3]{B}$, we multiply the numerator and denominator of the fraction by $\sqrt[3]{A^{2}}+\sqrt[3]{AB}+\sqrt[3]{B^{2}}$ and use the Difference of Cubes formula to get a denominator of $A-B$. (See Section: Special Product Formulas for the special product formulas). For example:

$ \begin{aligned} \frac{5}{\sqrt[3]{3}-2} & =\frac{5}{\sqrt[3]{3}-\sqrt[3]{2^{3}}}\cdot\frac{\sqrt[3]{9}+\sqrt[3]{3\times2^{3}}+\sqrt[3]{2^{6}}}{\sqrt[3]{9}+\sqrt[3]{3\times2^{3}}+\sqrt[3]{2^{6}}} \\[6pt] & =\frac{5}{3-2^{3}}\cdot\left(\sqrt[3]{9}+\sqrt[3]{3\times2^{3}}+\sqrt[3]{2^{6}}\right) \\[6pt] & =-\sqrt[3]{9}-\sqrt[3]{2^{3}\times3}-2^{6/3} \\[6pt] & =-\sqrt[3]{9}-2\sqrt[3]{3}-4 \end{aligned} $

Sometimes we need to rationalize the numerator. This process is similar to rationalizing the denominator.

Frequently Asked Questions

What does it mean to rationalize a denominator?

Rationalizing a denominator means rewriting a fraction so that the denominator contains no radicals. This is done by multiplying the numerator and denominator by a suitable expression (usually the conjugate) that eliminates the radical from the denominator without changing the value of the fraction.

What are conjugates?

Two binomials are conjugates of each other if they have the form $\sqrt{A}+\sqrt{B}$ and $\sqrt{A}-\sqrt{B}$, that is, they are identical except for the sign between the two radical terms. Their product $(\sqrt{A}+\sqrt{B})(\sqrt{A}-\sqrt{B})=A-B$ contains no radicals. More generally, $C\sqrt{A}+D\sqrt{B}$ and $C\sqrt{A}-D\sqrt{B}$ are conjugates.

Why do we rationalize denominators?

Rationalized forms are easier to compare, simplify, and evaluate numerically. In calculus, rationalizing the numerator of a difference quotient (as in Example 2) is a standard technique for computing limits where direct substitution gives $\frac{0}{0}$.

How do you rationalize a cube root denominator?

For a denominator of the form $\sqrt[3]{A}-\sqrt[3]{B}$, multiply numerator and denominator by $\sqrt[3]{A^{2}}+\sqrt[3]{AB}+\sqrt[3]{B^{2}}$ and apply the Difference of Cubes formula $A^{3}-B^{3}=(A-B)(A^{2}+AB+B^{2})$ to get a rational denominator of $A-B$. For a denominator $\sqrt[3]{A}+\sqrt[3]{B}$, use $\sqrt[3]{A^{2}}-\sqrt[3]{AB}+\sqrt[3]{B^{2}}$ and the Sum of Cubes formula to get $A+B$.

When would you rationalize the numerator instead of the denominator?

Rationalizing the numerator is useful when you need the numerator to be free of radicals, most commonly in calculus when evaluating limits of the form $\dfrac{\sqrt{f(x)}-c}{x-a}$. The technique is identical: multiply top and bottom by the conjugate of the numerator.