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, \[\begin{align} \frac{13}{\sqrt{5}+3\sqrt{2}} & =\frac{13}{\sqrt{5}+3\sqrt{2}}\cdot\frac{\sqrt{5}-3\sqrt{2}}{\sqrt{5}-3\sqrt{2}}\\ & =\frac{13(\sqrt{5}-3\sqrt{2})}{5-3^{2}\times2}\\ & =-(\sqrt{5}-3\sqrt{2})\\ & =3\sqrt{2}-\sqrt{5} \end{align}\]
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 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: \[\begin{align} \frac{5}{\sqrt[3]{3}-2} & =\frac{5}{\sqrt[3]{3}-\sqrt[3]{2^{3}}}\cdot\frac{\sqrt[3]{9}+\sqrt[3]{3\times2^{3}}+\sqrt[3]{2^{6}}}{\sqrt[3]{9}+\sqrt[3]{3\times2^{3}}+\sqrt[3]{2^{6}}}\\ & =\frac{5}{3-2^{3}}\cdot\left(\sqrt[3]{9}+\sqrt[3]{3\times2^{3}}+\sqrt[3]{2^{6}}\right)\\ & =-\sqrt[3]{9}-\sqrt[3]{2^{3}\times3}-2^{6/3}\\ & =-\sqrt[3]{9}-2\sqrt[3]{3}-4. \end{align}\]
Example 1.45 . 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 \[\begin{align} \frac{1}{\sqrt{x+3}+\sqrt{x-2}} & =\frac{1}{\sqrt{x+3}+\sqrt{x-2}}\cdot\frac{\sqrt{x+3}-\sqrt{x-2}}{\sqrt{x+3}-\sqrt{x-2}}\\ & =\frac{\sqrt{x+3}-\sqrt{x-2}}{(\sqrt{x+3})^{2}-(\sqrt{x-2})^{2}}\tag{\footnotesize Let $A=\sqrt{x+3}$ and $B=\sqrt{x-2}$ and then use $(A-B)(A+B)=A^{2}-B^{2}$}\\ & =\frac{\sqrt{x+3}-\sqrt{x-2}}{x+3-(x-2)}\\ & =\frac{1}{5}\left(\sqrt{x+3}-\sqrt{x-2}\right). \end{align}\]
- Sometimes we need to rationalize the numerator. This process is similar to rationalizing the denominator.
Example 1.46 . 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 \(\sqrt{9+h} + 3\) :
\[\begin{align} \dfrac{\sqrt{9+h}-3}{h}&=\dfrac{\sqrt{9+h}-3}{h}\frac{\sqrt{9+h}+3}{\sqrt{9+h}+3}\\ &=\frac{(\sqrt{9+h})^2-3^2}{h(\sqrt{9+h}+3)}\\ &=\frac{9+h-9}{h(\sqrt{9+h}+3)}\\ &=\frac{h}{h(\sqrt{9+h}+3)}\\ &=\frac{1}{\sqrt{9+h}+3} \end{align}\]