• The binomial numbers $\sqrt{A}+\sqrt{B}$ and $\sqrt{A}-\sqrt{B}$ are called conjugates of each other.

For example, $\sqrt{2}+\sqrt{5}$ and $\sqrt{2}-\sqrt{5},$ or $\sqrt{3x+5}+\sqrt{2x+7}$ and $\sqrt{3x+5}-\sqrt{2x+7}$ are conjugates of each other.

• Because conjugates are the sum and difference of the same two terms, their product is the difference of the squares of these terms (see  Special Product Formulas); that is, $(\sqrt{A}-\sqrt{B})(\sqrt{A}+\sqrt{B})=(\sqrt{A})^{2}-(\sqrt{B})^{2}=A-B.$

• 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{align*} \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}}\\
& =\frac{13(\sqrt{5}-3\sqrt{2})}{5-3^{2}\times2}\\
& =-(\sqrt{5}-3\sqrt{2})\\
& =3\sqrt{2}-\sqrt{5}
\end{align*}

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 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 Special Product Formulas).

For example:
\begin{align*}
\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}}}\\
& =\frac{5}{3-2^{3}}\cdot\left(\sqrt[3]{9}+\sqrt[3]{3\times2^{3}}+\sqrt[3]{2^{6}}\right)\\
& =-\sqrt[3]{9}-\sqrt[3]{2^{3}\times3}-2^{6/3}\\
& =-\sqrt[3]{9}-2\sqrt[3]{3}-4.
\end{align*}

Example

Remove the square roots in the denominator

$\frac{1}{\sqrt{x+3}+\sqrt{x-2}}$

Solution

We multiply both top and bottom by $\sqrt{x+3}-\sqrt{x-2},$  giving

\begin{align*}
\frac{1}{\sqrt{x+3}+\sqrt{x-2}} & =\frac{1}{\sqrt{x+3}+\sqrt{x-2}}\cdot\frac{\sqrt{x+3}-\sqrt{x-2}}{\sqrt{x+3}-\sqrt{x-2}}\\
\\
{\color{blue} \small[ \text{ Let } A=\sqrt{x+3}, } &{\color{blue} \small  B=\sqrt{x-2},\text {and then use }(A-B)(A+B)=A^{2}-B^{2}]}\\
\\
& =\frac{\sqrt{x+3}-\sqrt{x-2}}{(\sqrt{x+3})^{2}-(\sqrt{x-2})^{2}}\\
& =\frac{\sqrt{x+3}-\sqrt{x-2}}{x+3-(x-2)}\\
& =\frac{1}{5}\left(\sqrt{x+3}-\sqrt{x-2}\right)
\end{align*}