A progressive collection of 20 problems to learn how to eliminate radicals from denominators using conjugates and algebraic techniques.
Each exercise is solved step by step, with clear explanations and increasing difficulty, including expressions with parameters and domain restrictions.
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 original expression is defined for \(x\geq 0\). The rationalized form is equivalent for \(x\geq0,\ x\neq1\); for \(x=1\), the original expression equals \(1\).
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 original expression is defined for \(x\geq0\). The rationalized form is equivalent for \(x\geq0,\ x\neq2\); for \(x=2\), the original expression equals \(0\).
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 \(\sqrt{x}+\sqrt{x-1}\):
\[ \frac{\sqrt{x}+\sqrt{x-1}}{(\sqrt{x}-\sqrt{x-1})(\sqrt{x}+\sqrt{x-1})} = \frac{\sqrt{x}+\sqrt{x-1}}{x-(x-1)} = \sqrt{x}+\sqrt{x-1} \]