A collection of fully worked practice problems designed to help you simplify radicals, extract factors, and rationalize denominators.
You will also find step-by-step solutions for complex manipulations and radical equations, with explanations tailored to help you master the core concepts at your own pace.
Exercise 1 — level ★☆☆☆☆
\[ \sqrt{75} \]
Result
\[ 5\sqrt{3} \]
Solution
Key idea
Factor the radicand by extracting the greatest perfect square.
Factoring the radicand
\[ 75 = 25 \cdot 3 \]
Applying the property
\[ \sqrt{75} = \sqrt{25 \cdot 3} = \sqrt{25}\cdot\sqrt{3} = 5\sqrt{3} \]
Result
\[ \boxed{5\sqrt{3}} \]
Exercise 2 — level ★☆☆☆☆
\[ \sqrt{48} \]
Result
\[ 4\sqrt{3} \]
Solution
Factoring the radicand
\[ 48 = 16 \cdot 3 \]
Applying the property
\[ \sqrt{48} = \sqrt{16}\cdot\sqrt{3} = 4\sqrt{3} \]
Result
\[ \boxed{4\sqrt{3}} \]
Exercise 3 — level ★☆☆☆☆
\[ \sqrt{\dfrac{25}{4}} \]
Result
\[ \dfrac{5}{2} \]
Solution
Key idea
The square root of a fraction is the quotient of the square roots of the numerator and denominator.
Applying the property
\[ \sqrt{\frac{25}{4}} = \frac{\sqrt{25}}{\sqrt{4}} = \frac{5}{2} \]
Result
\[ \boxed{\dfrac{5}{2}} \]
Exercise 4 — level ★★☆☆☆
\[ 2\sqrt{3} + 5\sqrt{3} \]
Result
\[ 7\sqrt{3} \]
Solution
Key idea
Radicals with the same radicand are added like like terms.
Factoring out the common radical
\[ (2+5)\sqrt{3} = 7\sqrt{3} \]
Result
\[ \boxed{7\sqrt{3}} \]
Exercise 5 — level ★★☆☆☆
\[ \sqrt{12} + \sqrt{27} \]
Result
\[ 5\sqrt{3} \]
Solution
Key idea
Simplify each radical first, then add the like terms.
Simplifications
\[ \sqrt{12}=\sqrt{4\cdot3}=2\sqrt{3} \qquad \sqrt{27}=\sqrt{9\cdot3}=3\sqrt{3} \]
Sum
\[ 2\sqrt{3}+3\sqrt{3}=5\sqrt{3} \]
Result
\[ \boxed{5\sqrt{3}} \]
Exercise 6 — level ★★☆☆☆
\[ \sqrt{50} - \sqrt{18} \]
Result
\[ 2\sqrt{2} \]
Solution
Simplifications
\[ \sqrt{50}=\sqrt{25\cdot2}=5\sqrt{2} \qquad \sqrt{18}=\sqrt{9\cdot2}=3\sqrt{2} \]
Difference
\[ 5\sqrt{2}-3\sqrt{2}=2\sqrt{2} \]
Result
\[ \boxed{2\sqrt{2}} \]
Exercise 7 — level ★★☆☆☆
\[ \sqrt{2}\cdot\sqrt{8} \]
Result
\[ 4 \]
Solution
Applying the property
\[ \sqrt{2}\cdot\sqrt{8}=\sqrt{16}=4 \]
Result
\[ \boxed{4} \]
Exercise 8 — level ★★☆☆☆
\[ \sqrt{6}\cdot\sqrt{24} \]
Result
\[ 12 \]
Solution
Applying the property
\[ \sqrt{6}\cdot\sqrt{24}=\sqrt{144}=12 \]
Result
\[ \boxed{12} \]
Exercise 9 — level ★★★☆☆
\[ (\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3}) \]
Result
\[ 2 \]
Solution
Key idea
The product has the form \((a+b)(a-b)=a^2-b^2\), with \(a=\sqrt{5}\) and \(b=\sqrt{3}\).
Applying the difference of squares
\[ (\sqrt{5})^2-(\sqrt{3})^2=5-3=2 \]
Result
\[ \boxed{2} \]
Exercise 10 — level ★★★☆☆
\[ (2\sqrt{3})^2 \]
Result
\[ 12 \]
Solution
Calculation
\[ (2\sqrt{3})^2=4\cdot3=12 \]
Result
\[ \boxed{12} \]
Exercise 11 — level ★★★☆☆
\[ \sqrt{72}+\sqrt{32}-\sqrt{8} \]
Result
\[ 8\sqrt{2} \]
Solution
Simplifying each radical
\[ \sqrt{72}=6\sqrt{2} \qquad \sqrt{32}=4\sqrt{2} \qquad \sqrt{8}=2\sqrt{2} \]
Algebraic sum
\[ 6\sqrt{2}+4\sqrt{2}-2\sqrt{2}=8\sqrt{2} \]
Result
\[ \boxed{8\sqrt{2}} \]
Exercise 12 — level ★★★☆☆
\[ \dfrac{1}{\sqrt{2}} \]
Result
\[ \dfrac{\sqrt{2}}{2} \]
Solution
Rationalization
\[ \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}}=\frac{\sqrt{2}}{2} \]
Result
\[ \boxed{\dfrac{\sqrt{2}}{2}} \]
Exercise 13 — level ★★★☆☆
\[ \dfrac{3}{\sqrt{5}} \]
Result
\[ \dfrac{3\sqrt{5}}{5} \]
Solution
Rationalization
\[ \frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}=\frac{3\sqrt{5}}{5} \]
Result
\[ \boxed{\dfrac{3\sqrt{5}}{5}} \]
Exercise 14 — level ★★★☆☆
\[ (\sqrt{3}+\sqrt{2})^2 \]
Result
\[ 5+2\sqrt{6} \]
Solution
Applying \((a+b)^2=a^2+2ab+b^2\)
\[ (\sqrt{3})^2+2\sqrt{3}\cdot\sqrt{2}+(\sqrt{2})^2=3+2\sqrt{6}+2=5+2\sqrt{6} \]
Result
\[ \boxed{5+2\sqrt{6}} \]
Exercise 15 — level ★★★★☆
\[ \dfrac{5}{\sqrt{7}-\sqrt{2}} \]
Result
\[ \sqrt{7}+\sqrt{2} \]
Solution
Multiplying by the conjugate
\[ \frac{5}{\sqrt{7}-\sqrt{2}}\cdot\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}+\sqrt{2}}=\frac{5(\sqrt{7}+\sqrt{2})}{7-2}=\frac{5(\sqrt{7}+\sqrt{2})}{5}=\sqrt{7}+\sqrt{2} \]
Result
\[ \boxed{\sqrt{7}+\sqrt{2}} \]
Exercise 16 — level ★★★★☆
\[ \sqrt[3]{8}+\sqrt[3]{27} \]
Result
\[ 5 \]
Solution
Computing the cube roots
\[ \sqrt[3]{8}=\sqrt[3]{2^3}=2 \qquad \sqrt[3]{27}=\sqrt[3]{3^3}=3 \]
Sum
\[ 2+3=5 \]
Result
\[ \boxed{5} \]
Exercise 17 — level ★★★★☆
\[ \dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{2}} \]
Result
\[ \sqrt{3}+\sqrt{5} \]
Solution
Distributing the division
\[ \frac{\sqrt{6}}{\sqrt{2}}+\frac{\sqrt{10}}{\sqrt{2}}=\sqrt{\frac{6}{2}}+\sqrt{\frac{10}{2}}=\sqrt{3}+\sqrt{5} \]
Result
\[ \boxed{\sqrt{3}+\sqrt{5}} \]
Exercise 18 — level ★★★★☆
\[ \sqrt{5+2\sqrt{6}} \]
Result
\[ \sqrt{3}+\sqrt{2} \]
Solution
Finding \(a\) and \(b\)
We look for \(a,b\) such that \(a^2+b^2=5\) and \(ab=\sqrt{6}\): this gives \(a=\sqrt{3},\,b=\sqrt{2}\).
Rewriting the expression
\[ 5+2\sqrt{6}=(\sqrt{3})^2+2\sqrt{3}\sqrt{2}+(\sqrt{2})^2=(\sqrt{3}+\sqrt{2})^2 \]
Computation
\[ \sqrt{(\sqrt{3}+\sqrt{2})^2}=\sqrt{3}+\sqrt{2} \]
Result
\[ \boxed{\sqrt{3}+\sqrt{2}} \]
Exercise 19 — level ★★★★★
\[ \dfrac{1}{\sqrt{3}+\sqrt{5}} \]
Result
\[ \dfrac{\sqrt{5}-\sqrt{3}}{2} \]
Solution
Multiplying by the conjugate \((\sqrt{5}-\sqrt{3})\)
\[ \frac{1}{\sqrt{3}+\sqrt{5}}\cdot\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{(\sqrt{5})^2-(\sqrt{3})^2}=\frac{\sqrt{5}-\sqrt{3}}{2} \]
Result
\[ \boxed{\dfrac{\sqrt{5}-\sqrt{3}}{2}} \]
Exercise 20 — level ★★★★★
\[ \sqrt{8+2\sqrt{15}} \]
Result
\[ \sqrt{5}+\sqrt{3} \]
Solution
Finding \(a\) and \(b\)
We look for \(a,b\) such that \(a^2+b^2=8\) and \(ab=\sqrt{15}\): this gives \(a=\sqrt{5},\,b=\sqrt{3}\).
Rewriting the expression
\[ 8+2\sqrt{15}=(\sqrt{5}+\sqrt{3})^2 \]
Computation
\[ \sqrt{(\sqrt{5}+\sqrt{3})^2}=\sqrt{5}+\sqrt{3} \]
Result
\[ \boxed{\sqrt{5}+\sqrt{3}} \]
Exercise 21 — level ★★☆☆☆
\[ \dfrac{\sqrt{3}+1}{\sqrt{3}-1} \]
Result
\[ 2+\sqrt{3} \]
Solution
Multiplying by the conjugate
\[ \frac{\sqrt{3}+1}{\sqrt{3}-1}\cdot\frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{(\sqrt{3}+1)^2}{3-1}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3} \]
Result
\[ \boxed{2+\sqrt{3}} \]
Exercise 22 — level ★★☆☆☆
\[ \dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}} \]
Result
\[ \dfrac{7-2\sqrt{10}}{3} \]
Solution
Multiplying by the conjugate
\[ \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\cdot\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{(\sqrt{5}-\sqrt{2})^2}{5-2}=\frac{7-2\sqrt{10}}{3} \]
Expanding the numerator
\[ (\sqrt{5}-\sqrt{2})^2=5-2\sqrt{10}+2=7-2\sqrt{10} \]
Result
\[ \boxed{\dfrac{7-2\sqrt{10}}{3}} \]
Exercise 23 — level ★★★☆☆
\[ \dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1} \]
Result
\[ 4 \]
Solution
First fraction
\[ \frac{(\sqrt{3}+1)^2}{2}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3} \]
Second fraction
\[ \frac{(\sqrt{3}-1)^2}{2}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3} \]
Sum
\[ (2+\sqrt{3})+(2-\sqrt{3})=4 \]
Result
\[ \boxed{4} \]
Exercise 24 — level ★★★☆☆
\[ \sqrt{3+\sqrt{5}}\cdot\sqrt{3-\sqrt{5}} \]
Result
\[ 2 \]
Solution
Product under the radical
\[ \sqrt{(3+\sqrt{5})(3-\sqrt{5})}=\sqrt{9-5}=\sqrt{4}=2 \]
Result
\[ \boxed{2} \]
Exercise 25 — level ★★★☆☆
\[ \sqrt[3]{54}-\sqrt[3]{16}+\sqrt[3]{2} \]
Result
\[ 2\sqrt[3]{2} \]
Solution
Simplifications
\[ \sqrt[3]{54}=3\sqrt[3]{2} \qquad \sqrt[3]{16}=2\sqrt[3]{2} \]
Algebraic sum
\[ 3\sqrt[3]{2}-2\sqrt[3]{2}+\sqrt[3]{2}=(3-2+1)\sqrt[3]{2}=2\sqrt[3]{2} \]
Result
\[ \boxed{2\sqrt[3]{2}} \]
Exercise 26 — level ★★★☆☆
\[ \dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{2}-1} \]
Result
\[ 2\sqrt{2} \]
Solution
Rationalizing both terms
\[ \frac{1}{\sqrt{2}+1}=\sqrt{2}-1 \qquad \frac{1}{\sqrt{2}-1}=\sqrt{2}+1 \]
Sum
\[ (\sqrt{2}-1)+(\sqrt{2}+1)=2\sqrt{2} \]
Result
\[ \boxed{2\sqrt{2}} \]
Exercise 27 — level ★★★☆☆
\[ \sqrt{7+4\sqrt{3}} \]
Result
\[ 2+\sqrt{3} \]
Solution
Finding \(a\) and \(b\)
We seek \((a+b)^2=a^2+2ab+b^2=7+4\sqrt{3}\). With \(a=2,\,b=\sqrt{3}\), we get \(a^2+b^2=7\) and \(2ab=4\sqrt{3}\).
Rewriting the expression
\[ 7+4\sqrt{3}=(2+\sqrt{3})^2 \implies \sqrt{7+4\sqrt{3}}=2+\sqrt{3} \]
Result
\[ \boxed{2+\sqrt{3}} \]
Exercise 28 — level ★★★☆☆
\[ \sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}} \]
Result
\[ 2\sqrt{2} \]
Solution
Denesting
\[ \sqrt{5+2\sqrt{6}}=\sqrt{3}+\sqrt{2} \qquad \sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}\quad(\sqrt{3}>\sqrt{2}) \]
Difference
\[ (\sqrt{3}+\sqrt{2})-(\sqrt{3}-\sqrt{2})=2\sqrt{2} \]
Result
\[ \boxed{2\sqrt{2}} \]
Exercise 29 — level ★★★★☆
\[ \sqrt{2x+1}=x-1 \]
Result
\[ x=4 \]
Solution
Domain conditions
\(2x+1\geq0\) and \(x-1\geq0\), so \(x\geq1\).
Squaring both sides
\[ 2x+1=(x-1)^2=x^2-2x+1 \implies x^2-4x=0 \implies x(x-4)=0 \]
Checking and rejecting extraneous solutions
\(x=0\) is rejected (\(x\geq1\)). For \(x=4\): \(\sqrt{9}=3=4-1\) — valid.
Result
\[ \boxed{x=4} \]
Exercise 30 — level ★★★★☆
\[ \sqrt{x+5}-\sqrt{x-3}=2 \]
Result
\[ x=4 \]
Solution
Domain conditions
\(x\geq3\).
Strategy: sum-and-difference system
Let \(u=\sqrt{x+5}\), \(v=\sqrt{x-3}\), with \(u-v=2\) and \(u^2-v^2=8\).
\[ (u+v)\cdot2=8 \implies u+v=4 \]
Solving the system gives \(u=3,\,v=1\). From \(u^2=x+5\), we get \(x=4\).
Check
\[ \sqrt{9}-\sqrt{1}=3-1=2 \]
Result
\[ \boxed{x=4} \]
Exercise 31 — level ★★★★☆
\[ (\sqrt{3}+\sqrt{2})^4 \]
Result
\[ 49+20\sqrt{6} \]
Solution
First square
\[ (\sqrt{3}+\sqrt{2})^2=3+2\sqrt{6}+2=5+2\sqrt{6} \]
Second square
\[ (5+2\sqrt{6})^2=25+20\sqrt{6}+24=49+20\sqrt{6} \]
Result
\[ \boxed{49+20\sqrt{6}} \]
Exercise 32 — level ★★★★☆
\[ \sqrt[4]{48}\cdot\sqrt[4]{3} \]
Result
\[ 2\sqrt{3} \]
Solution
Product under the radical
\[ \sqrt[4]{48\cdot3}=\sqrt[4]{144}=\sqrt[4]{16\cdot9}=\sqrt[4]{16}\cdot\sqrt[4]{9}=2\cdot\sqrt[4]{9} \]
Simplifying \(\sqrt[4]{9}\)
\[ \sqrt[4]{9}=9^{1/4}=(3^2)^{1/4}=3^{1/2}=\sqrt{3} \]
Result
\[ \boxed{2\sqrt{3}} \]
Exercise 33 — level ★★★★☆
\[ \dfrac{\sqrt[3]{4}}{\sqrt[3]{2}}+\sqrt[3]{16}-\dfrac{2}{\sqrt[3]{4}} \]
Result
\[ 2\sqrt[3]{2} \]
Solution
Simplifying each term
\[ \frac{\sqrt[3]{4}}{\sqrt[3]{2}}=\sqrt[3]{2} \qquad \sqrt[3]{16}=2\sqrt[3]{2} \qquad \frac{2}{\sqrt[3]{4}}=\frac{2\sqrt[3]{2}}{2}=\sqrt[3]{2} \]
Algebraic sum
\[ \sqrt[3]{2}+2\sqrt[3]{2}-\sqrt[3]{2}=2\sqrt[3]{2} \]
Result
\[ \boxed{2\sqrt[3]{2}} \]
Exercise 34 — level ★★★★☆
\[ \sqrt[3]{2+\sqrt{5}}\cdot\sqrt[3]{2-\sqrt{5}} \]
Result
\[ -1 \]
Solution
Product under the cube root
\[ \sqrt[3]{(2+\sqrt{5})(2-\sqrt{5})}=\sqrt[3]{4-5}=\sqrt[3]{-1} \]
Final computation
Over the real numbers, \(\sqrt[3]{-1}=-1\) because \((-1)^3=-1\).
Result
\[ \boxed{-1} \]
Exercise 35 — level ★★★★★
\[ \sqrt{3x+1}-\sqrt{x+4}=1 \]
Result
\[ x=5 \]
Solution
Domain conditions
\(x\geq-\tfrac{1}{3}\).
Isolating one radical
\[ \sqrt{3x+1}=1+\sqrt{x+4} \]
First squaring
\[ 3x+1=1+2\sqrt{x+4}+(x+4)=x+5+2\sqrt{x+4} \]
\[ 2x-4=2\sqrt{x+4} \implies x-2=\sqrt{x+4}\quad(x\geq2) \]
Second squaring
\[ (x-2)^2=x+4 \implies x^2-5x=0 \implies x(x-5)=0 \]
Checking and rejecting extraneous solutions
\(x=0\) is rejected. For \(x=5\): \(\sqrt{16}-\sqrt{9}=4-3=1\) — valid.
Result
\[ \boxed{x=5} \]
Exercise 36 — level ★★★★★
\[ \sqrt{x+\sqrt{2x-1}}=\sqrt{2} \]
Result
\[ x=1 \]
Solution
Domain conditions
\(x\geq\tfrac{1}{2}\).
First squaring
\[ x+\sqrt{2x-1}=2 \implies \sqrt{2x-1}=2-x\quad(x\leq2) \]
Second squaring
\[ 2x-1=(2-x)^2=4-4x+x^2 \implies x^2-6x+5=0 \implies (x-1)(x-5)=0 \]
Checking and rejecting extraneous solutions
\(x=5\) is rejected (\(x\leq2\)). For \(x=1\): \(\sqrt{1+\sqrt{1}}=\sqrt{2}\) — valid.
Result
\[ \boxed{x=1} \]
Exercise 37 — level ★★★★★
\[ \dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}} \]
Result
\[ 1 \]
Solution
Key idea — telescoping sum
After rationalization, the general term becomes \(\sqrt{n+1}-\sqrt{n}\):
\[ \frac{1}{\sqrt{n}+\sqrt{n+1}}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n}=\sqrt{n+1}-\sqrt{n} \]
Applying it to the three terms
\[ (\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(2-\sqrt{3}) \]
Telescoping cancellation
\[ =2-1=1 \]
Result
\[ \boxed{1} \]
Exercise 38 — level ★★★★★
\[ \sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}} \]
Result
\[ \sqrt{2} \]
Solution
Key idea
These two nested radicals do not simplify neatly in closed form, so we compute the square of the difference.
Computing the square
\[ \left(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\right)^2=(6+\sqrt{11})+(6-\sqrt{11})-2\sqrt{(6+\sqrt{11})(6-\sqrt{11})} \]
\[ =12-2\sqrt{36-11}=12-2\sqrt{25}=12-10=2 \]
Taking the square root
The difference is positive (the first radical is larger than the second), so:
\[ \sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}=\sqrt{2} \]
Result
\[ \boxed{\sqrt{2}} \]
Exercise 39 — level ★★★★★
\[ \sqrt{9-4\sqrt{5}} \]
Result
\[ \sqrt{5}-2 \]
Solution
Finding \(a\) and \(b\)
We look for \(a,b\) such that \(a^2+b^2=9\) and \(ab=2\sqrt{5}\): this gives \(a=\sqrt{5},\,b=2\).
Rewriting the expression
\[ 9-4\sqrt{5}=5-4\sqrt{5}+4=(\sqrt{5}-2)^2 \]
Computing the square root
Since \(\sqrt{5}>2\):
\[ \sqrt{(\sqrt{5}-2)^2}=\sqrt{5}-2 \]
Result
\[ \boxed{\sqrt{5}-2} \]
Exercise 40 — level ★★★★★
\[ \sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}} \]
Result
\[ 2 \]
Solution
Simplifying \(\sqrt{3+2\sqrt{2}}\)
\[ 3+2\sqrt{2}=1+2\sqrt{2}+2=(1+\sqrt{2})^2 \implies \sqrt{3+2\sqrt{2}}=1+\sqrt{2} \]
Simplifying \(\sqrt{3-2\sqrt{2}}\)
\[ 3-2\sqrt{2}=(\sqrt{2}-1)^2 \implies \sqrt{3-2\sqrt{2}}=\sqrt{2}-1\quad(\sqrt{2}>1) \]
Difference
\[ (1+\sqrt{2})-(\sqrt{2}-1)=2 \]
Result
\[ \boxed{2} \]