Simplify Radicals
Questions with Solutions for Grade 10

Grade 10 questions on how to simplify radicals expressions with solutions are presented.

In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals

Examples

2) Product (Multiplication) formula of radicals with equal indices is given by \[ \Large{\color{red}{ \left( \sqrt[n]{x} \right) \cdot \left( \sqrt[n]{y} \right) = \sqrt[n]{x \cdot y}} } \] More examples on how to Multiply Radical Expressions. 3) Quotient (Division) formula of radicals with equal indices is given by \[ \Large{\color{red}{ \dfrac{\sqrt[n]{x}}{\sqrt[n]{y}} = \sqrt[n]{\dfrac{x}{y}} }} \]

More examples on how to Divide Radical Expressions.

4) You may add or subtract like radicals only

Example

5) You may rewrite expressions without radicals (to rationalize denominators) as follows

1. Example 1: \[ \color{red} { \sqrt{x} \cdot \sqrt{x} = x } \]
2. Example 2:\[ \color{red} { \sqrt[3]{x} \cdot \sqrt[3]{x} \cdot \sqrt[3]{x} = x } \]
3. Example 3: \[ \color{red} { (a - \sqrt{b})(a + \sqrt{b}) = (a)^2 - (\sqrt{b})^2 = a^2 - b } \]

More examples on how to Rationalize Denominators of Radical Expressions.

Questions

1. \( \sqrt{128} \cdot \sqrt{32} \)
2. \( \sqrt{2} \cdot \sqrt{6} + 3\sqrt{12} \)
3. \( \dfrac{3\sqrt{14} + 4\sqrt{63}}{3\sqrt{7}} \)
4. \( \sqrt[3]{32} \sqrt[3]{16} \)
5. \( \sqrt[3]{\dfrac{64}{7}} \)
6. \( \dfrac{\sqrt[3]{2x} - \sqrt[3]{54x}}{\sqrt[3]{2}} \)
7. \( \dfrac{\sqrt{x} + \sqrt{x + 1}}{\sqrt{x} - \sqrt{x + 1}} \)

Answers to the above questions

1. 
   Write 128 and 32 as products of prime factors: \( 128 = 2^7, 32 = 2^5 \) hence \[ \sqrt{128} \cdot \sqrt{32} = \sqrt{2^7} \cdot \sqrt{2^5} = \sqrt{2^7 \cdot 2^5} = \sqrt{2^{12}} = \sqrt{(2^6)^2} = |2^6| = 64 \]
2. 
   Use product rule to write that } \sqrt{2} \cdot \sqrt{6} = \sqrt{12} \[ \sqrt{2} \cdot \sqrt{6} + 3\sqrt{12} = \sqrt{2 \cdot 6} + 3\sqrt{12} = \sqrt{12} + 3\sqrt{12} = \sqrt{12}(1 + 3) = 4\sqrt{12} \] \[ = 4\sqrt{4 \cdot 3} = 4\sqrt{4}\sqrt{3} = 4 \cdot 2 \cdot \sqrt{3} = 8\sqrt{3} \]
3. 
   Write 14 and 63 as products of prime numbers \( 14 = 2 \times 7, 63 = 3^2 \times 7 \) and substitute \[ \dfrac{3\sqrt{14} + 4\sqrt{63}}{3\sqrt{7}} = \dfrac{3\sqrt{2 \cdot 7} + 4\sqrt{3^2 \cdot 7}}{3\sqrt{7}} \] \[ = \dfrac{3\sqrt{2}\sqrt{7} + 4 \cdot 3\sqrt{7}}{3\sqrt{7}} \] \[ = \dfrac{\sqrt{7}(3\sqrt{2} + 12)}{3\sqrt{7}} \] \[ = \dfrac{3\sqrt{2} + 12}{3} = \sqrt{2} + 4 \]
4. 
   Write 32 and 16 as products of prime numbers \( 32 = 2^5, 16 = 2^4 \) and substitute} \[ \sqrt[3]{32} \sqrt[3]{16} = \sqrt[3]{2^5} \sqrt[3]{2^4} = \sqrt[3]{2^5 \cdot 2^4} = \sqrt[3]{2^9} = \sqrt[3]{(2^3)^3} = 2^3 = 8 \]
5. 
   Write 64 as products of prime numbers \( 64 = 2^6 \) and substitute} \[ \sqrt[3]{\dfrac{64}{7}} = \sqrt[3]{\dfrac{2^6}{7}} = \sqrt[3]{\dfrac{(2^2)^3}{7}} = \dfrac{4}{\sqrt[3]{7}} \] Rationalize the denominator by multiplying numerator and denominator by \(\left(\sqrt[3]{7}\right)^2\) \[ = \dfrac{4}{\sqrt[3]{7}} \cdot \dfrac{\left(\sqrt[3]{7}\right)^2}{\left(\sqrt[3]{7}\right)^2} = \dfrac{4\left(\sqrt[3]{7}\right)^2}{7} = \dfrac{4\sqrt[3]{49}}{7} \]
6. 
   Write 54 as products of prime numbers \( 54 = 2 \times 3^3 \) and substitute \[ \dfrac{\sqrt[3]{2x} - \sqrt[3]{54x}}{\sqrt[3]{2}} = \dfrac{\sqrt[3]{2x} - \sqrt[3]{2 \cdot 3^3 x}}{\sqrt[3]{2}} = \dfrac{\sqrt[3]{2x} - \sqrt[3]{2x} \cdot \sqrt[3]{3^3}}{\sqrt[3]{2}} = \dfrac{\sqrt[3]{2x} - 3\sqrt[3]{2x}}{\sqrt[3]{2}} \] \[ = \dfrac{-2\sqrt[3]{2x}}{\sqrt[3]{2}} = -2\sqrt[3]{\dfrac{2x}{2}} = -2\sqrt[3]{x} \]
7. 
   Multiply the denominator and numerator by the conjugate of the denominator \[ \dfrac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} - \sqrt{x+1}} = \dfrac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} - \sqrt{x+1}} \cdot \dfrac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} + \sqrt{x+1}} \] Expand and simplify \[ = \dfrac{(\sqrt{x})^2 + (\sqrt{x+1})^2 + 2\sqrt{x}\sqrt{x+1}}{(\sqrt{x})^2 - (\sqrt{x+1})^2} \] \[ = \dfrac{x + (x + 1) + 2\sqrt{x}\sqrt{x+1}}{x - (x + 1)} \] \[ = \dfrac{2x + 1 + 2\sqrt{x(x+1)}}{-1} \] \[ = -2x - 1 - 2\sqrt{x(x+1)} \]

More Questions With Answers

1. \(\sqrt[3]{25} \cdot \sqrt[3]{125}\)
2. \((5\sqrt[3]{64})(-3\sqrt[3]{16})\)
3. \((7\sqrt{\dfrac{2}{5}})(2\sqrt{10})\)
4. \(\sqrt[3]{2\dfrac{10}{27}}\)
5. \((\sqrt{17x})(\sqrt{34x})\)
6. \(\sqrt{4y^2}\)
7. \(\sqrt{8y^4}\)
8. \(\sqrt{25+144}\)
9. \(\sqrt[2n]{x^{2n}}\), \(n\) a positive integer
10. \((\sqrt{x-2})(4\sqrt{x-2})\)
11. \(\dfrac{\sqrt[3]{27a^3b^5}}{\sqrt[3]{8a^6b^2}}\)
12. \(\dfrac{\sqrt{x}-\sqrt{x-2}}{\sqrt{x}+\sqrt{x-2}}\)

Solutions to the Above Questions

1. \(\sqrt[3]{25} \cdot \sqrt[3]{125} = \sqrt[3]{3125} = 5\sqrt[3]{25}\)
2. \((5\sqrt[3]{64})(-3\sqrt[3]{16}) = -15\sqrt[3]{1024} = -15\sqrt[3]{64 \cdot 16} = -15 \cdot 4 \cdot \sqrt[3]{16} = -60\sqrt[3]{16}\)
3. \((7\sqrt{\dfrac{2}{5}})(2\sqrt{10}) = 14\sqrt{\dfrac{20}{5}} = 14\sqrt{4} = 14 \cdot 2 = 28\)
4. \(\sqrt[3]{2\dfrac{10}{27}} = \sqrt[3]{\dfrac{64}{27}} = \dfrac{4}{3}\)
5. \((\sqrt{17x})(\sqrt{34x}) = \sqrt{578x^2} = x\sqrt{578} = x\sqrt{289 \cdot 2} = 17x\sqrt{2}\)
6. \(\sqrt{4y^2} = 2|y|\)
7. \(\sqrt{8y^4} = 2y^2\sqrt{2}\)
8. \(\sqrt{25+144} = \sqrt{169} = 13\)
9. \(\sqrt[2n]{x^{2n}} = |x|\)
10. \((\sqrt{x-2})(4\sqrt{x-2}) = 4(x-2)\)
11. \(\dfrac{\sqrt[3]{27a^3b^5}}{\sqrt[3]{8a^6b^2}} = \dfrac{3ab\sqrt[3]{b^2}}{2a^2b\sqrt[3]{1}} = \dfrac{3b}{2a} \)
12. \(\dfrac{\sqrt{x}-\sqrt{x-2}}{\sqrt{x}+\sqrt{x-2}} = \dfrac{(\sqrt{x}-\sqrt{x-2})^2}{x - (x-2)} = \dfrac{x + (x-2) - 2\sqrt{x(x-2)}}{2} = \dfrac{2x - 2 - 2\sqrt{x(x-2)}}{2} = x - 1 - \sqrt{x(x-2)}\)

Links and References

• High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers
• Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
• Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
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