Rationalizing a Denominator or Numerator

The binomials of the form $\sqrt{A}+\sqrt{B}$ and $\sqrt{A}-\sqrt{B}$
are called conjugates of each other. For example,

$\sqrt{2}+\sqrt{5}\qquad\text{and}\quad \sqrt{2}-\sqrt{5},$

or

$\sqrt{3x+5}+\sqrt{2x+7}\quad\text{and}\quad \sqrt{3x+5}-\sqrt{2x+7}$

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 Section: 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{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}}

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

& =-(\sqrt{5}-3\sqrt{2})

& =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}}}

& =\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{aligned}

$ > **Example** > > Remove the square roots in the denominator > $

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

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

\begin{aligned}

\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}}

& =\frac{\sqrt{x+3}-\sqrt{x-2}}{(\sqrt{x+3})^{2}-(\sqrt{x-2})^{2}} \quad (\footnotesize Let $A=\sqrt{x+3)$ and $B=\sqrt{x-2}$ and then use $(A-B)(A+B)=A^{2}-B^{2}$}

& =\frac{\sqrt{x+3}-\sqrt{x-2}}{x+3-(x-2)}

& =\frac{1}{5}\left(\sqrt{x+3}-\sqrt{x-2}\right).
\end{aligned}
$

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

Example

Rationalize the numerator of $\dfrac{\sqrt{9+h}-3}{h}$

Solution

To rationalize the numerator, we multiply both the numerator and the denominator by the conjugate of the numerator, $\sqrt{9+h} - 3$, which is $\sqrt{9+h} + 3$:

$ \begin{aligned} \dfrac{\sqrt{9+h}-3}{h}&=\dfrac{\sqrt{9+h}-3}{h}\frac{\sqrt{9+h}+3}{\sqrty)+h}+3} &=\frac{(\sqrt{9+h})^2-3^2}{h(\sqrt{9+h}+3)} &=\frac{9+h-9}{h(\sqrt{9+h}+3)} &=\frac{h}{h(\sqrt{9+h}+3)} &=\frac{1}{\sqrt{9+h}+3} \end{aligned} $