Multiplying and Dividing Radical Expressions
In order to multiply or divide radical expressions, the indices of the roots must be equal.
Where defined,
$$ \Large \sqrt[n]{a } \cdot \sqrt[n]{b } = \sqrt[n]{a \cdot b } $$
$$ \Large \frac{ \sqrt[n]{ a} }{ \sqrt[n]{ b}} = \sqrt[n]{ \frac{a}{b} } $$
Warning:
If the indices of the roots are different, multiplication or division can be performed by finding a common index.
$$ \sqrt[n]{a } \cdot \sqrt[m]{b } = \sqrt[n.m]{a^m } \cdot \sqrt[m.n]{b^n } = \sqrt[m.n]{a^m \cdot b^n } $$
$$ \frac{\sqrt[n]{a } }{\sqrt[m]{b } } = \frac{\sqrt[n.m]{a^m } }{\sqrt[m.n]{b^n } } = \sqrt[m \cdot n]{ \frac{a^m}{b^n} } \quad (b\neq 0)$$
Example:
\( \bullet \quad \displaystyle\frac{\sqrt[3]{2 } \cdot \sqrt[3]{3 } }{\sqrt[3]{5 } } = \sqrt[3]{ \frac{2 \cdot 3}{5} } = \sqrt[3]{\frac{6}{5} } \)
Question 11
What is the result of the operation? \[ \sqrt[3]{ 2} \cdot \sqrt[5]{16 } \]
\[ \text{A)} 2\sqrt[15]{4 } \quad \text{B) }\sqrt[15]{4 } \quad \text{C)} \sqrt[15]{2 } \quad \text{D) } 1 \quad \text{E)} 2 \]
Solution:
If the indices of the roots are made equal to 3. 5 = 15,
$$ \sqrt[3]{ 2} \cdot \sqrt[5]{16 } = \sqrt[3]{ 2} \cdot \sqrt[5]{2^4 } $$
$$ \sqrt[3.5]{ 2^5} \cdot \sqrt[5.3]{2^{4.3} }$$
$$ \sqrt[3.5]{ 2^5 \cdot 2^{12}} = \sqrt[15]{ 2^{17}} $$
$$ \sqrt[15]{ 2^{15} \cdot 2^2} = 2 \sqrt[15]{ 4} $$
\(\textbf{Answer: A} \)
Question 12
\[ \large \frac{\sqrt[4]{27 } \cdot \sqrt[3]{ 9} }{\sqrt{ 3} } \] What is the result of the operation?
\[ \text{A)} \sqrt[12]{3^5 } \quad \text{B) }\sqrt[12]{3^7 }\quad \text{C)} \sqrt[12]{3^{11} } \quad \text{D) } 3 \quad \text{E)} 1 \]
Solution:
If the indices of the roots 2, 3, and 4 \(\to\) are made equal to 12,
\[ \displaystyle \frac{\sqrt[4]{27} \cdot \sqrt[3]{9}}{\sqrt[3]{3}} = \frac{\sqrt[4]{3^3} \cdot \sqrt[3]{3^2}}{\sqrt[3]{3}} \]
\[ =\displaystyle \frac{\sqrt[4 \cdot 3]{3^{3 \cdot 3}} \cdot \sqrt[3 \cdot 4]{3^{2 \cdot 4}}}{\sqrt[2 \cdot 6]{3^6}} \]
\[ = \displaystyle\frac{\sqrt[12]{3^9} \cdot \sqrt[12]{3^8}}{\sqrt[12]{3^6}} \newline = \sqrt[12]{\frac{3^9 \cdot 3^8}{3^6}} \]
\[ = \displaystyle\sqrt[12]{3^{9+8 \ – \ 6}} \newline = \sqrt[12]{3^{11}} \]
\(\textbf{Answer: C} \)
Rationalizing the Denominator:
- For \( n> m \quad b \neq0 \), in expressions of the form \(\Large \frac{a}{\sqrt[n]{b^m } } \), the numerator and denominator are multiplied by \(\large \sqrt[n]{b^{n \ – \ m} } \) to eliminate the radical from the denominator.
$$ \displaystyle \frac{a}{\sqrt[n]{b^m}} = \frac{a}{\sqrt[n]{b^m}} \cdot \frac{\sqrt[n]{b^{n \ – \ m}}}{\sqrt[n]{b^{n \ – \ m}}} = \frac{a \cdot \sqrt[n]{b^{n \ – \ m}}}{b} $$
Examples:
\( \bullet \quad \displaystyle\displaystyle \frac{a}{\sqrt{b}} =\displaystyle \frac{a}{\sqrt{b}} \cdot \displaystyle\frac{\sqrt{b}}{\sqrt{b}} = \displaystyle\frac{a\sqrt{b}}{b} \)
\(\bullet \quad \displaystyle\frac{1}{\sqrt[7]{32}} = \displaystyle\frac{1}{\sqrt[7]{2^5}} \cdot \displaystyle\frac{\sqrt[7]{2^2}}{\sqrt[7]{2^2}} =\displaystyle \frac{\sqrt[7]{4}}{2} \\ \)
\( \bullet \quad \displaystyle\frac{1}{\sqrt[3]{2} \cdot \sqrt[4]{3^3}} = \displaystyle\frac{1}{\sqrt[3]{2} \cdot \sqrt[4]{3^3}} \cdot \frac{\sqrt[3]{2^2} \cdot \sqrt[4]{3}}{\sqrt[3]{2^2} \cdot \sqrt[4]{3}} =\displaystyle \frac{\sqrt[3]{4} \cdot \sqrt[4]{3}}{2 \cdot 3} = \displaystyle\frac{\sqrt[3]{4} \cdot \sqrt[4]{3}}{6} \)
2. In expressions of the form \( \Large \frac{a}{\sqrt{b }\ – \ \sqrt{c } } \), the numerator and denominator are multiplied by \( \sqrt{b } + \sqrt{c } \), and in expressions of the form \( \Large \frac{a}{\sqrt{b }+ \sqrt{c } } \), they are multiplied by \( \sqrt{b } \ – \ \sqrt{c } \).
Since $$ (x \ – \ y) . (x+y) = x^2 \ – \ y^2 $$,
$$(\sqrt{ b} \ – \ \sqrt{ c}) \cdot (\sqrt{ b} + \sqrt{ c}) = (\sqrt{ b} )^2 – (\sqrt{ c} )^2 = b-c $$. In this way, a difference of squares is obtained in the denominator, and the radical is eliminated from the denominator.
\(\bullet \quad \displaystyle \frac{a}{\sqrt{b} \ – \ \sqrt{c}} = \frac{a}{\sqrt{b} \ – \ \sqrt{c}} \cdot \frac{\sqrt{b} + \sqrt{c}}{\sqrt{b} + \sqrt{c}} = \frac{a (\sqrt{b} + \sqrt{c})}{b \ – \ c} \)
\( \bullet \quad \displaystyle\frac{a}{\sqrt{b} + \sqrt{c}} = \frac{a}{\sqrt{b} + \sqrt{c}} \cdot \frac{\sqrt{b} \ – \ \sqrt{c}}{\sqrt{b} \ – \ \sqrt{c}} = \frac{a (\sqrt{b} \ – \ \sqrt{c})}{b \ – \ c} \)
Examples:
\( \bullet \quad \displaystyle\frac{1}{\sqrt{5} \ – \ 2} = \frac{1}{\sqrt{5} \ -\ 2} \cdot \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{\sqrt{5} + 2}{(\sqrt{5})^2 \ – \ 2^2} = \frac{\sqrt{5} + 2}{5 \ – \ 4} = \\ \sqrt{5} + 2 \)
\( \bullet \quad \displaystyle \frac{2}{\sqrt{5} + \sqrt{3}} = \frac{2}{\sqrt{5} + \sqrt{3}} \cdot \frac{\sqrt{5} \ – \ \sqrt{3}}{\sqrt{5} \ – \ \sqrt{3}} = \frac{2 (\sqrt{5} \ – \ \sqrt{3})}{(\sqrt{5})^2 \ – \ (\sqrt{3})^2} = \frac{2 (\sqrt{5} \ – \ \sqrt{3})}{5 \ – \ 3} = \\ \sqrt{5} \ – \ \sqrt{3} \)
Question 13
\[ \frac{2}{3 + 2\sqrt{2}} \ – \ \frac{1}{3 \ – \ \sqrt{8}} + \frac{12}{\sqrt{2}} \] What is the result of the operation?
\[ \text{A)} 1 \quad \text{B) } 2 \quad \text{C)} 3 \quad \text{D) } 4 \quad \text{E)} 5 \]
Çözüm:
$$ =\frac{2}{3 + 2\sqrt{2}} \ – \ \frac{1}{3 \ – \ \sqrt{8}} + \frac{12}{\sqrt{2}} $$
$$= \frac{2}{3 + 2\sqrt{2}} \cdot \frac{3 \ – \ 2\sqrt{2}}{3 \ – \ 2\sqrt{2}} \ – \ \frac{1}{3 \ – \ \sqrt{8}} \cdot \frac{3 + \sqrt{8}}{3 + \sqrt{8}} + \frac{12}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \\ $$
$$= \frac{2 (3 \ – \ 2\sqrt{2})}{3^2 \ – \ (2\sqrt{2})^2} \ – \ \frac{1 (3 + \sqrt{8})}{3^2 \ – \ (\sqrt{8})^2} + \frac{12\sqrt{2}}{2} $$
$$ = \frac{2 (3 \ – \ 2\sqrt{2})}{9 \ – \ 8} \ – \ \frac{3 + \sqrt{8}}{9 \ – \ 8} + 6\sqrt{2} $$
$$= 6 \; – \; 4\sqrt{2} \; – \; 3 \; – \; 2\sqrt{2} \; + \; 6\sqrt{2} \\ = 3 $$
\(\textbf{Answer: C} \)
Question 14
What is the equivalent of the expression? \[\frac{3}{\sqrt{4 \ – \ \sqrt{7 } } } \]
\[ \text{A)} \sqrt{4 \ – \ \sqrt{7 } } \quad \text{B) } \sqrt{4 +\sqrt{7 } } \quad \text{C)} \sqrt{ 3} \quad \text{D) } 2 \quad \text{E)} 1 \]
Solution:
\[ \frac{3}{\sqrt{4 \ – \ \sqrt{7}}} = \frac{3}{\sqrt{4 \ – \ \sqrt{7}}} \cdot \frac{\sqrt{4 + \sqrt{7}}}{\sqrt{4 + \sqrt{7}}} = \frac{3 \cdot \sqrt{4 + \sqrt{7}}} {\sqrt{ 4^2 \ – \ (\sqrt{ 7} )^2} } = \frac {3 \cdot \sqrt{4 + \sqrt{7}}}{\sqrt{ 9} } \]
$$ = \sqrt{4 + \sqrt{ 7} } $$
\(\textbf{Answer: B} \)
Question 15
If \[ \frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{5}} = \frac{a \sqrt{3} + b \sqrt{2} + c \sqrt{30}}{12} \], find \( a + b + c \).
\[ \text{A) } 1 \quad \text{B) } 2 \quad \text{C) } 3 \quad \text{D) } 4 \quad \text{E) } 5 \]
Solution:
\[ \frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{5}} = \frac{1}{(\sqrt{2} + \sqrt{3}) + \sqrt{5}} \cdot \frac{(\sqrt{2} + \sqrt{3}) \ – \ \sqrt{5}}{(\sqrt{2} + \sqrt{3}) \ – \ \sqrt{5}} \]
$$= \frac{ \sqrt{2 } + \sqrt{ 3} \ – \ \sqrt{ 5} }{ (\sqrt{2 } + \sqrt{3 })^2 \ – \ (\sqrt{ 5} )^2 } = \frac{\sqrt{2 } + \sqrt{ 3} \ – \ \sqrt{ 5} }{2 + 3+ 2 \sqrt{6 } \ – \ 5} $$
$$ =\frac{\sqrt{2 } + \sqrt{ 3} \ – \ \sqrt{ 5} }{2 \sqrt{6 }} \cdot \frac{\sqrt{ 6} }{\sqrt{6 } } = \frac{2 \sqrt{ 3} + 3 \sqrt{2 } \; – \; \sqrt{ 30} }{12} $$
From here, \( a = 2, \; b= 3, \; c= -1 \) are found. Thus \(a + b+ c = 4 \).
\(\textbf{Answer: D} \)
Warning:
For \( n \in Z^+ \), if there is an expression of the form \(\sqrt[2n]{a } \ – \ \sqrt[2n]{b } \) in the denominator, the numerator and denominator are multiplied by \(\sqrt[2n]{a } + \sqrt[2n]{b } \). If there is an expression of the form \(\sqrt[2n]{a } + \sqrt[2n]{b } \) in the denominator, the numerator and denominator are multiplied by \(\sqrt[2n]{a } \ – \ \sqrt[2n]{b } \).
Question 16
What is the equivalent of the expression? \[ \frac{1}{\sqrt[4]{ 2} \ – \ 1 } \]
\[ \text{A) } \sqrt[4]{2} + 1 \quad
\text{B) } \sqrt[4]{2} \ – \ 1 \quad
\text{C) } \sqrt{2} \ – \ 1 \quad
\text{D) } (\sqrt[4]{2} + 1)(\sqrt{2} + 1) \quad
\text{E) } (\sqrt[4]{2} \ – \ 1)(\sqrt{2} \ – \ 1)
\]
Solution:
\[ \frac{1}{\sqrt[4]{2} \ – \ 1} = \frac{1}{\sqrt[4]{2} – 1} \cdot \frac{\sqrt[4]{2} + 1}{\sqrt[4]{2} + 1} \]
\[ = \frac{\sqrt[4]{2} + 1}{(\sqrt[4]{2})^2 \ – \ 1^2} = \frac{\sqrt[4]{2} + 1}{\sqrt{2} \ – \ 1} \]
\[ = \frac{\sqrt[4]{2} + 1}{\sqrt{2} \ – \ 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} \]
\[ = (\sqrt[4]{2} + 1)(\sqrt{2} + 1) \]
\(\textbf{Answer: D} \)
3. In expressions of the form \( \displaystyle \frac{a}{\sqrt[3]{b } \ – \ \sqrt[3]{c } } \), the numerator and denominator are multiplied by \( \sqrt[3]{b^2 } + \sqrt[3]{bc } +\sqrt[3]{c^2 } \).
Since \( (x \ – \ y) \cdot (x^2+ xy + y^2) = (x^3 \ – \ y^3)\),
\( (\sqrt[3]{b } \ – \ \sqrt[3]{ c} ) \cdot( \sqrt[3]{b ^2 } + \sqrt[3]{bc } + \sqrt[3]{c^2 } )= (\sqrt[3]{ b} )^3 \ – \ (\sqrt[3]{c })^3 = b \ – \ c \).
In this way, a difference of cubes is obtained in the denominator, and the radical is eliminated from the denominator.
$$ \frac{a}{( \sqrt[3]{b } \ – \ \sqrt[3]{ c} ) } = \frac{a}{( \sqrt[3]{b } \ – \ \sqrt[3]{ c} ) } \cdot \frac{\sqrt[3]{b ^2 } + \sqrt[3]{bc } + \sqrt[3]{c^2 } }{\sqrt[3]{b ^2 } + \sqrt[3]{bc } + \sqrt[3]{c^2 } } $$
$$ = \frac{a \cdot \sqrt[3]{b ^2 } + \sqrt[3]{bc } + \sqrt[3]{c^2 } }{b \ – \ c} $$
In expressions of the form \(\displaystyle\frac{a}{\sqrt[3]{ b} + \sqrt[3]{ c} } \), the numerator and denominator are multiplied by \( \sqrt[3]{b^2 } \; – \;\sqrt[3]{ bc} + \;\; \sqrt[3]{ c^2} \).
Since \( (x+y) \cdot (x^2 \ – \ xy + y^2) = (x^3 +y^3)\),
\( (\sqrt[3]{b } + \sqrt[3]{ c} ) \cdot (\sqrt[3]{b ^2 } \ – \ \sqrt[3]{bc } + \sqrt[3]{c^2 }) = (\sqrt[3]{ b} )^3 – (\sqrt[3]{c })^3 = b+c \).
In this way, a sum of cubes is obtained in the denominator, and the radical is eliminated from the denominator.
$$\displaystyle \frac{a}{\sqrt[3]{ b} + \sqrt[3]{ c} } = \frac{a}{\sqrt[3]{ b} + \sqrt[3]{ c} } \cdot \frac{\sqrt[3]{b ^2 }\; -\; \sqrt[3]{bc }\; + \;\sqrt[3]{c^2 } }{\sqrt[3]{b ^2 } \;- \;\sqrt[3]{bc } \;+\; \sqrt[3]{c^2 } }$$
$$ = \displaystyle\frac{a \cdot (\sqrt[3]{b ^2 }\; – \;\sqrt[3]{ bc} \;+\; \sqrt[3]{ c^2} ) } {b+ c } $$
Examples:
\( \bullet \quad \displaystyle\frac{1}{\sqrt[3]{5} – \sqrt[3]{3}} = \displaystyle\frac{1}{\sqrt[3]{5} \;- \; \sqrt[3]{3}}\; \cdot \; \displaystyle\frac{\sqrt[3]{5^2}\; \; +\; \sqrt[3]{5 \;\cdot\; 3}\; + \; \sqrt[3]{3^2}}{\sqrt[3]{5^2} \;+ \; \sqrt[3]{5\; \cdot \; 3}\; +\; \sqrt[3]{3^2}} \)
\(= \displaystyle\frac{\sqrt[3]{25 } \;+ \;\sqrt[3]{ 15 }\; + \; \sqrt[3]{ 9}} {(\sqrt[3]{ 5})^3 \; -\; (\sqrt[3]{3 } )^3 } = \displaystyle\frac{\sqrt[3]{25} \; + \; \sqrt[3]{ 15} \;+ \; \sqrt[3]{ 9} }{2} \)
\( \bullet \quad \displaystyle \frac{1}{\sqrt[3]{3} + \sqrt{2}} = \frac{1}{\sqrt[3]{3} + \sqrt{2}} \cdot\displaystyle \frac{\sqrt[3]{3} \ – \ \sqrt{2}}{\sqrt[3]{3} \ – \ \sqrt{2}} =\displaystyle \frac{\sqrt[3]{3} – \sqrt{2}}{(\sqrt[3]{3})^2 – (\sqrt{2})^2} = \frac{\sqrt[3]{3} – \sqrt{2}}{\sqrt[3]{9} – 2}\)
\(= \displaystyle\frac{ \sqrt[3]{3} \; – \; \sqrt{ 2} }{\sqrt[3]{9 }\; – \;2 } \cdot \frac{ \sqrt[3]{9^2 } \;+\; 2 \sqrt[3]{9 }\; + \;4 }{\sqrt[3]{9^2 } \;+ 2\; \sqrt[3]{9 }\; + \;4 } = \displaystyle\frac{ ( \sqrt[3]{ 3} \ – \ \sqrt{ 2} ) (\sqrt[3]{ 81}\; + \; 2 \sqrt[3]{9 } \;+\;4 ) }{(\sqrt[3]{ 9} )^3 \;- \; 2^3 } \)
\(= (\sqrt[3]{ 3} \ – \ \sqrt{ 2} ) \cdot (3 \sqrt[3]{ 3} + 2 \sqrt[3]{ 9} + 4 ) \)
Question 17
What is the equivalent of the expression? \[ \frac{1}{\sqrt[3]{9 } \;+ \; \sqrt[3]{ 6}\; +\; \sqrt[3]{ 4} } \]
\[ \text{A) } \sqrt[3]{3} \;+ \; \sqrt[3]{ 2} \quad
\text{B) }\sqrt[3]{3}\; -\; \sqrt[3]{ 2} \quad
\text{C) } \sqrt{3 } \; + \; \sqrt{2 } \quad
\text{D) } \sqrt{3 } \; – \; \sqrt{2 } \quad
\text{E) } \frac{1}{5}
\]
Solution:
\[ \frac{1}{\sqrt[3]{9 } \;+ \; \sqrt[3]{ 6}\; +\; \sqrt[3]{ 4} } = \frac{1}{\sqrt[3]{3^2 } \;+ \; \sqrt[3]{ 3.2}\; +\; \sqrt[3]{ 2^2} } \cdot \frac{\sqrt[3]{ 3} \ – \ \sqrt[3]{ 2} }{\sqrt[3]{ 3} \ – \ \sqrt[3]{ 2} } \]
\[= \frac { \sqrt[3]{ 3} \; -\; \sqrt[3]{2 } }{(\sqrt[3]{ 3})^3 \; -\; (\sqrt[3]{2 })^3 } = \sqrt[3]{ 3} \; -\; \sqrt[3]{2 } \]
\(\textbf{Answer: B} \)
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