Answers: Rationalizing the denominator
These are the answers to Questions: Rationalizing the denominator.
Please attempt the questions before reading these answers!
Q1
1.1. \(\quad \dfrac{5}{\sqrt{3}} = \dfrac{5\sqrt{3}}{3}\)
1.2. \(\quad \dfrac{7}{2\sqrt{5}} = \dfrac{7\sqrt{5}}{10}\)
1.3. \(\quad \dfrac{11}{4\sqrt{7}} = \dfrac{11\sqrt{7}}{28}\)
1.4. \(\quad \dfrac{8}{5\sqrt{6}} = \dfrac{4\sqrt{6}}{15}\)
1.5. \(\quad \dfrac{3\sqrt{2}}{\sqrt{5}} = \dfrac{3\sqrt{10}}{5}\)
1.6. \(\quad \dfrac{9}{\sqrt{10}} = \dfrac{9\sqrt{10}}{10}\)
1.7. \(\quad \dfrac{\sqrt{7}}{\sqrt{3}} = \dfrac{\sqrt{21}}{3}\)
1.8. \(\quad \dfrac{\sqrt{2}}{\sqrt{6}} = \dfrac{\sqrt{3}}{3}\)
1.9. \(\quad \dfrac{12}{\sqrt{11}} = \dfrac{12\sqrt{11}}{11}\)
1.10. \(\quad \dfrac{\sqrt{8}}{\sqrt{2}} = 2\)
1.11. \(\quad \dfrac{15}{3\sqrt{7}} = \dfrac{5\sqrt{7}}{7}\)
1.12. \(\quad \dfrac{6\sqrt{3}}{\sqrt{10}} = \dfrac{3\sqrt{30}}{5}\)
1.13. \(\quad \dfrac{\sqrt{18}}{\sqrt{9}} = \sqrt{2}\)
1.14. \(\quad \dfrac{2\sqrt{5}}{\sqrt{12}} = \dfrac{\sqrt{30}}{3}\)
1.15. \(\quad \dfrac{4}{\sqrt{2}} = 2\sqrt{2}\)
1.16. \(\quad \dfrac{10}{5\sqrt{13}} = \dfrac{2\sqrt{13}}{13}\)
Q2
2.1. \(\quad \dfrac{5}{2 + \sqrt{3}} = 10 - 5\sqrt{3}\)
2.2. \(\quad \dfrac{7}{4 - \sqrt{2}} = \dfrac{4 + \sqrt{2}}{2}\)
2.3. \(\quad \dfrac{3}{\sqrt{5} + 1} = \dfrac{3\sqrt{5} - 3}{4}\)
2.4. \(\quad \dfrac{\sqrt{7}}{\sqrt{3} - 1} = \dfrac{\sqrt{21}+\sqrt{7}}{2}\)
2.5. \(\quad \dfrac{2 + \sqrt{5}}{1 - \sqrt{2}} = {-2 - 2\sqrt{2} - \sqrt{5} - \sqrt{10}}\)
2.6. \(\quad \dfrac{3\sqrt{2} + 5}{4 + \sqrt{6}} = \dfrac{12\sqrt{2} - 6\sqrt{3} + 20 - 5\sqrt{6}}{10}\)
2.7. \(\quad \dfrac{8}{3 - \sqrt{7}} = 12 + 4\sqrt{7}\)
2.8. \(\quad \dfrac{6}{2 + \sqrt{5}} = -12 + 6\sqrt{5}\)
2.9. \(\quad \dfrac{\sqrt{10}}{\sqrt{2} + 3} = \dfrac{3\sqrt{10} - 2\sqrt{5}}{7}\)
2.10. \(\quad \dfrac{2\sqrt{3} + 5}{\sqrt{7}-1} = \dfrac{2\sqrt{21} + 5\sqrt{7} + 2\sqrt{3} + 5}{6}\)
2.11. \(\quad \dfrac{\sqrt{6} - \sqrt{2}}{2 + \sqrt{5}} = -2\sqrt{6} + 2\sqrt{5} + 2\sqrt{2} - \sqrt{10}\)
2.12. \(\quad \dfrac{4 + \sqrt{3}}{5 - \sqrt{7}} = \dfrac{4\sqrt{7} + 5\sqrt{3} + \sqrt{21} + 20}{18}\)
2.13. \(\quad \dfrac{2}{4 - \sqrt{11}} = \dfrac{8 + 2\sqrt{11}}{5}\)
2.14. \(\quad \dfrac{\sqrt{8} + \sqrt{3}}{\sqrt{7} - 2} = \dfrac{2\sqrt{14} + 4\sqrt{2} + \sqrt{21} + 2\sqrt{3}}{3}\)
Q3
3.1. \(\quad\) To prove this equation, rationalize the denominator of the left hand side of the equation.
Since the denominator contains two square roots you can multiply the numerator and denominator by \(-2\sqrt{3} + \sqrt{5}\) or by \(2\sqrt{3} - \sqrt{5}\) to rationalize the denominator.
If you multiply the numerator and denominator by \(2\sqrt{3} - \sqrt{5}\) you get:
\[ \dfrac{\sqrt{11}}{2\sqrt{3} + \sqrt{5}} \cdot \dfrac{2\sqrt{3} - \sqrt{5}}{2\sqrt{3} - \sqrt{5}} = \dfrac{\sqrt{11}(2\sqrt{3} - \sqrt{5})}{(2\sqrt{3} + \sqrt{5})(2\sqrt{3} - \sqrt{5})} \]
Expanding the brackets in both the numerator and the denominator gives you:
\[ \dfrac{\sqrt{11}(2\sqrt{3} - \sqrt{5})}{(2\sqrt{3} + \sqrt{5})(2\sqrt{3} - \sqrt{5})} = \dfrac{2\sqrt{33} - \sqrt{55}}{(2\sqrt{3})^2 - 2\sqrt{15} + 2\sqrt{15} - (\sqrt{5})^2} \]
Simplifying the denominator then gives you:
\[ \dfrac{2\sqrt{33} - \sqrt{55}}{(2\sqrt{3})^2 - 2\sqrt{15} + 2\sqrt{15} - (\sqrt{5})^2} = \dfrac{2\sqrt{33} - \sqrt{55}}{4(3) - 5} \]
Simplifying further gives you the final answer and the right hand side of the equation you are proving:
\[ \dfrac{2\sqrt{33} - \sqrt{55}}{4(3) - 5} = \dfrac{2\sqrt{33} - \sqrt{55}}{7} \]
If you instead multiply the numerator and denominator by \(-2\sqrt{3} + \sqrt{5}\) you get:
\[ \dfrac{\sqrt{11}}{2\sqrt{3} + \sqrt{5}} \cdot \dfrac{-2\sqrt{3} + \sqrt{5}}{-2\sqrt{3} + \sqrt{5}} = \dfrac{\sqrt{11}(-2\sqrt{3} + \sqrt{5})}{(2\sqrt{3} + \sqrt{5})(-2\sqrt{3} + \sqrt{5})} \]
Expanding the brackets in both the numerator and the denominator gives you:
\[ \dfrac{\sqrt{11}(-2\sqrt{3} + \sqrt{5})}{(2\sqrt{3} + \sqrt{5})(-2\sqrt{3} + \sqrt{5})} = \dfrac{\sqrt{11}(-2\sqrt{3} + \sqrt{5})}{-(2\sqrt{3})^2 + 2\sqrt{15} - 2\sqrt{15} + (\sqrt{5})^2} \]
Simplifying the denominator gives you:
\[ \dfrac{\sqrt{11}(-2\sqrt{3} + \sqrt{5})}{-(2\sqrt{3})^2 + 2\sqrt{15} - 2\sqrt{15} + (\sqrt{5})^2} = \dfrac{- 2\sqrt{33} + \sqrt{55}}{5 - 4(3)} \]
Further simplifying the denominator then gives you:
\[ \dfrac{- 2\sqrt{33} + \sqrt{55}}{5 - 4(3)} = \dfrac{- 2\sqrt{33} + \sqrt{55}}{-7} \]
To get a positive denominator, multiplying both the numerator and the denominator by \(-1\) gives you the right hand side of the equation you are proving:
\[ \dfrac{- 2\sqrt{33} + \sqrt{55}}{-7} = \dfrac{2\sqrt{33} - \sqrt{55}}{7} \]
3.2. \(\quad \dfrac{5-\sqrt{2}}{\sqrt{10}-\sqrt{3}} = \dfrac{5\sqrt{10}+5\sqrt{3}-2\sqrt{5}-\sqrt{6}}{7}\)
Version history and licensing
v1.0: initial version created 12/24 by Maximilian Volmar.