Rationalizing the denominator is a fundamental algebraic technique used to remove radicals from the denominators of fractions. In this collection you will find fully worked exercises, arranged in increasing order of difficulty.
Exercise 1 — level ★☆☆☆☆
\[ \frac{1}{\sqrt{2}} \]
Answer
\[ \frac{\sqrt{2}}{2} \]
Solution
Multiply the numerator and the denominator by \(\sqrt{2}\):
\[ \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]
Exercise 2 — level ★☆☆☆☆
\[ \frac{3}{\sqrt{5}} \]
Answer
\[ \frac{3\sqrt{5}}{5} \]
Solution
Multiply the numerator and the denominator by \(\sqrt{5}\):
\[ \frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5} \]
Exercise 3 — level ★★☆☆☆
\[ \frac{2}{3\sqrt{7}} \]
Answer
\[ \frac{2\sqrt{7}}{21} \]
Solution
Multiply the numerator and the denominator by \(\sqrt{7}\):
\[ \frac{2}{3\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{2\sqrt{7}}{3 \cdot 7} = \frac{2\sqrt{7}}{21} \]
Exercise 4 — level ★★☆☆☆
\[ \frac{1}{\sqrt{2}+1} \]
Answer
\[ \sqrt{2}-1 \]
Solution
Multiply the numerator and the denominator by the conjugate \(\sqrt{2}-1\):
\[ \frac{1}{\sqrt{2}+1} \cdot \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1 \]
Exercise 5 — level ★★☆☆☆
\[ \frac{1}{\sqrt{5}-2} \]
Answer
\[ \sqrt{5}+2 \]
Solution
Multiply by the conjugate \(\sqrt{5}+2\):
\[ \frac{1}{\sqrt{5}-2} \cdot \frac{\sqrt{5}+2}{\sqrt{5}+2} = \frac{\sqrt{5}+2}{5-4} = \sqrt{5}+2 \]
Exercise 6 — level ★★★☆☆
\[ \frac{2}{\sqrt{3}-1} \]
Answer
\[ \sqrt{3}+1 \]
Solution
Multiply by the conjugate \(\sqrt{3}+1\):
\[ \frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{2(\sqrt{3}+1)}{3-1} = \sqrt{3}+1 \]
Exercise 7 — level ★★★☆☆
\[ \frac{4}{2+\sqrt{3}} \]
Answer
\[ 8-4\sqrt{3} \]
Solution
Multiply by the conjugate \(2-\sqrt{3}\):
\[ \frac{4(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{8-4\sqrt{3}}{4-3} = 8-4\sqrt{3} \]
Exercise 8 — level ★★★☆☆
\[ \frac{1}{\sqrt{2}+\sqrt{5}} \]
Answer
\[ \frac{\sqrt{5}-\sqrt{2}}{3} \]
Solution
Multiply by the conjugate \(\sqrt{5}-\sqrt{2}\):
\[ \frac{1}{\sqrt{2}+\sqrt{5}} \cdot \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} = \frac{\sqrt{5}-\sqrt{2}}{5-2} = \frac{\sqrt{5}-\sqrt{2}}{3} \]
Exercise 9 — level ★★★☆☆
\[ \frac{2}{\sqrt{7}+3} \]
Answer
\[ 3-\sqrt{7} \]
Solution
Multiply by the conjugate \(\sqrt{7}-3\):
\[ \frac{2(\sqrt{7}-3)}{(\sqrt{7}+3)(\sqrt{7}-3)} = \frac{2(\sqrt{7}-3)}{7-9} = 3-\sqrt{7} \]
Exercise 10 — level ★★★☆☆
\[ \frac{\sqrt{2}+1}{\sqrt{2}} \]
Answer
\[ \frac{2+\sqrt{2}}{2} \]
Solution
Multiply the numerator and the denominator by \(\sqrt{2}\):
\[ \frac{(\sqrt{2}+1)\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}} = \frac{2+\sqrt{2}}{2} \]
Exercise 11 — level ★★★★☆
\[ \frac{3}{\sqrt{2}-\sqrt{3}} \]
Answer
\[ -3(\sqrt{2}+\sqrt{3}) \]
Solution
Multiply by the conjugate \(\sqrt{2}+\sqrt{3}\):
\[ \frac{3(\sqrt{2}+\sqrt{3})}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})} = \frac{3(\sqrt{2}+\sqrt{3})}{2-3} = -3(\sqrt{2}+\sqrt{3}) \]
Exercise 12 — level ★★★★☆
\[ \frac{5}{2-\sqrt{5}} \]
Answer
\[ -10-5\sqrt{5} \]
Solution
Multiply by the conjugate \(2+\sqrt{5}\):
\[ \frac{5(2+\sqrt{5})}{(2-\sqrt{5})(2+\sqrt{5})} = \frac{10+5\sqrt{5}}{4-5} = -10-5\sqrt{5} \]
Exercise 13 — level ★★★★☆
\[ \frac{1}{\sqrt{3}+\sqrt{2}+1} \]
Answer
\[ \frac{2+\sqrt{2}-\sqrt{6}}{4} \]
Solution
First, rationalize with respect to the group \(\sqrt{3}+\sqrt{2}\), multiplying by \(\sqrt{3}+\sqrt{2}-1\):
\[ \frac{1}{\sqrt{3}+\sqrt{2}+1}\cdot\frac{\sqrt{3}+\sqrt{2}-1}{\sqrt{3}+\sqrt{2}-1}=\frac{\sqrt{3}+\sqrt{2}-1}{4+2\sqrt{6}} \]
Then rationalize the new denominator by multiplying by the conjugate \(4-2\sqrt{6}\). Expanding and simplifying gives:
\[ \frac{(\sqrt{3}+\sqrt{2}-1)(4-2\sqrt{6})}{(4+2\sqrt{6})(4-2\sqrt{6})}=\frac{2+\sqrt{2}-\sqrt{6}}{4} \]
Exercise 14 — level ★★★★★
\[ \frac{1}{\sqrt{5}+\sqrt{3}-\sqrt{2}} \]
Answer
\[ \frac{-3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{12} \]
Solution
Multiply by the partial conjugate \(\sqrt{5}+\sqrt{3}+\sqrt{2}\):
\[ \frac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{(\sqrt{5}+\sqrt{3}-\sqrt{2})(\sqrt{5}+\sqrt{3}+\sqrt{2})}=\frac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{6+2\sqrt{15}} \]
Now rationalize the new denominator by multiplying by \(6-2\sqrt{15}\). Expanding and simplifying gives:
\[ \frac{(\sqrt{5}+\sqrt{3}+\sqrt{2})(6-2\sqrt{15})}{(6+2\sqrt{15})(6-2\sqrt{15})}=\frac{-3\sqrt{2}+2\sqrt{3}+\sqrt{30}}{12} \]
Exercise 15 — level ★★★☆☆
\[ \frac{2}{\sqrt{x}+1} \]
Answer
\[ \frac{2(\sqrt{x}-1)}{x-1} \]
Solution
Conditions: the expression is defined for \(x\geq 0,\ x\neq1\).
Multiply by the conjugate \(\sqrt{x}-1\):
\[ \frac{2(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)} = \frac{2(\sqrt{x}-1)}{x-1} \]
Exercise 16 — level ★★★★☆
\[ \frac{\sqrt{x}+2}{\sqrt{x}-2} \]
Answer
\[ \frac{x+4\sqrt{x}+4}{x-4} \]
Solution
Conditions: the expression is defined for \(x\geq0,\ x\neq4\).
Multiply the numerator and the denominator by the conjugate \(\sqrt{x}+2\):
\[ \frac{(\sqrt{x}+2)^2}{(\sqrt{x}-2)(\sqrt{x}+2)} = \frac{x+4\sqrt{x}+4}{x-4} \]
Exercise 17 — level ★★★★☆
\[ \frac{1}{\sqrt{x+1}-\sqrt{x}} \]
Answer
\[ \sqrt{x+1}+\sqrt{x} \]
Solution
Conditions: the expression is defined for \(x\geq0\).
Multiply by the conjugate \(\sqrt{x+1}+\sqrt{x}\):
\[ \frac{\sqrt{x+1}+\sqrt{x}}{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})} = \frac{\sqrt{x+1}+\sqrt{x}}{(x+1)-x} = \sqrt{x+1}+\sqrt{x} \]
Exercise 18 — level ★★★★☆
\[ \frac{1}{\sqrt{x+2}+\sqrt{x+1}} \]
Answer
\[ \sqrt{x+2}-\sqrt{x+1} \]
Solution
Conditions: the expression is defined for \(x\geq-1\).
Multiply by the conjugate \(\sqrt{x+2}-\sqrt{x+1}\):
\[ \frac{\sqrt{x+2}-\sqrt{x+1}}{(\sqrt{x+2}+\sqrt{x+1})(\sqrt{x+2}-\sqrt{x+1})} = \frac{\sqrt{x+2}-\sqrt{x+1}}{(x+2)-(x+1)} = \sqrt{x+2}-\sqrt{x+1} \]
Exercise 19 — level ★★★★★
\[ \frac{\sqrt{x}-\sqrt{2}}{\sqrt{x}+\sqrt{2}} \]
Answer
\[ \frac{x+2-2\sqrt{2x}}{x-2} \]
Solution
Conditions: the expression is defined for \(x\geq0,\ x\neq2\).
Multiply the numerator and the denominator by the conjugate \(\sqrt{x}-\sqrt{2}\):
\[ \frac{(\sqrt{x}-\sqrt{2})^2}{(\sqrt{x}+\sqrt{2})(\sqrt{x}-\sqrt{2})} = \frac{x-2\sqrt{2x}+2}{x-2} \]
Exercise 20 — level ★★★★★
\[ \frac{1}{\sqrt{x}-\sqrt{x-1}} \]
Answer
\[ \sqrt{x}+\sqrt{x-1} \]
Solution
Conditions: the expression is defined for \(x\geq1\).
Multiply by the conjugate \(\sqrt{x}+\sqrt{x-1}\):
\[ \frac{\sqrt{x}+\sqrt{x-1}}{(\sqrt{x}-\sqrt{x-1})(\sqrt{x}+\sqrt{x-1})} = \frac{\sqrt{x}+\sqrt{x-1}}{x-(x-1)} = \sqrt{x}+\sqrt{x-1} \]