Nested Radicals

 

\( 1. \quad \sqrt{x \;+ \;2\sqrt{y } } \) or \( \sqrt{x\; -\; 2\sqrt{y } } \) In expressions of this form, we check whether the expression inside the radical is a perfect square. For this,

\[ \begin{cases} x = a + b \\ y = a \cdot b \end{cases} \quad \text{where,} \]

\[ \sqrt{\underbrace{x}_ {a+b}\;+ \;2\sqrt{\underbrace{y}_{a . b}}} = \sqrt{(\sqrt{a}\; +\; \sqrt{b})^2} = |\sqrt{a} \;+ \;\sqrt{b}| \]

 

Warning:

 

The coefficient of the inner radical expression must be 2.

 

Examples:

 

\( \bullet \quad \sqrt{\underbrace{4}_{3+1}\; – \;2{\sqrt{\underbrace{3}_{3.1}}}} = \sqrt{3 } \;- \;1 \)

 

\( \bullet \quad \sqrt{\underbrace{7}_{4+3} \;- \; 2{\sqrt{\underbrace{12}_{4.3}}}} = |\sqrt{4} – \sqrt{3}| ={2 }\; -\; \sqrt{3 } \)

 

Question 15

 

What is the result of the operation ? \[ \sqrt{3 \ – \ \sqrt{ 8} } \ – \ \sqrt{2 } \]

\[
\text{A) } -1 \quad
\text{B) } 0 \quad
\text{C) } 1 \quad
\text{D) } \sqrt{2 } \quad
\text{E) } 2
\]

 

Solution:

 

Let’s make the coefficient of the inner radical equal to 2.

\[\begin{array}{l l }
\sqrt{3 \ – \ \sqrt{ 8} } \; -\; \sqrt{2 } = \sqrt{\underbrace{3}_{2\;+\;1} – \underbrace{2}_{2.1}\sqrt{ 2} } \; -\; \sqrt{2 }
\end{array}\]

\[ |\sqrt{2}\; – \;\sqrt{1}| \;- \; \sqrt{2 } =\sqrt{2 }\; -\; 1 -\;\sqrt{2 } \]

\[ = -1 \]

\(\textbf{Answer: A} \)

 

Question 16

 

What is the result of the operation ? \[ \sqrt{15 \ – \ 6\sqrt{ 6} } + \sqrt{6 } \]

\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E) } 5
\]

 

Solution:

 

Let’s make the coefficient of the inner radical equal to 2.

\[ \sqrt{15 \ – \ 6\sqrt{ 6} } + \sqrt{6 } = \sqrt{15 \ – \ 2 \cdot 3 \sqrt{ 6} } + \sqrt{6 } \]

\[ = \sqrt{15 \ – \ 2 \cdot 3 \sqrt{ 6} } + \sqrt{6 } = \sqrt{15 \ – \ 2 \sqrt{ 3^2 \cdot 6} } + \sqrt{6 } = \sqrt{\underbrace{15}_{9+ 6} – 2 \sqrt{ \underbrace{54}_{9. 6}} } + \sqrt{6 } \]

\[ = |\sqrt{9 }\; -\; \sqrt{ 6} | + \sqrt{ 6} = 3\;- \;\sqrt{ 6} \;+ \; \sqrt{ 6} = 3 \]

\(\textbf{Answer: C} \)

 

Question 17

 

What is the result of the operation ? \[ \sqrt{4 + \sqrt{7 } } \]

\[
\text{A) } \frac{\sqrt{3 } +1 }{ \sqrt{ 2} } \quad
\text{B) } \frac{\sqrt{2 } +1 }{ 2 } \quad
\text{C) } \frac{\sqrt{7 } +1 }{ \sqrt{ 2} } \quad
\text{D) } \frac{\sqrt{7 } }{ 2 } \quad
\text{E) } \frac{1}{2}
\]

 

Solution:

 

 

Warning:

 

For \( a> 0 , b> 0 \), and \( a^2> b \),

\[ \sqrt{ a\; + \; \sqrt{b } } = \sqrt \displaystyle \frac{{ a\; +\; \sqrt{ a^2 \;-\; b} } }{2}\; +\; \sqrt \displaystyle\frac{{ a \;-\; \sqrt{ a^2 \;-\; b} } }{2} \]

\[ \sqrt{ a \;- \; \sqrt{b } } = \sqrt\displaystyle \frac{{ a\; +\; \sqrt{ a^2 \;-\; b} } }{2}\; -\; \sqrt\displaystyle \frac{{ a \;-\; \sqrt{ a^2 \;-\; b} } }{2} \]

 

Example:

 

\[\sqrt{2 + \sqrt{ 3} } = \sqrt \displaystyle\frac{{ 2 \;+\; \sqrt{ 4 \;-\; 3} } }{2} + \sqrt \displaystyle\frac{{ 2 \;-\; \sqrt{ 4 \;-\; 3} } }{2} \]

\[ = \displaystyle\frac{\sqrt{ 3} }{\sqrt{2 } } + \displaystyle\frac{1}{\sqrt{2 } } = \displaystyle\frac{\sqrt{ 3}+ 1 }{\sqrt{ 2} } \]

 

\( 2. \quad \sqrt[m]{ \sqrt[n]{ \sqrt[t]{a } } } \; = \; \sqrt[ {m.n.t } ]{ a} \)

 

Example:

 

\( \displaystyle\sqrt{\sqrt[3]{\sqrt[5]{2}}} =\displaystyle \sqrt[3 \cdot 4 \cdot 2]{2} = \displaystyle\sqrt[24]{2} \)

 

Question 19

 

What is the result of the operation ? \[ \sqrt{ 2\sqrt[3]{2\sqrt[5]{2}}} \]

\[
\text{A) } \sqrt[5]{ 8} \quad
\text{B) } \sqrt[5]{ 4} \quad
\text{C) } \sqrt[5]{ 2} \quad
\text{D) } \sqrt[10]{ 128} \quad
\text{E) } \sqrt[10]{ 8}
\]

 

Solution:

 

Let’s move the factors between the radicals into the innermost radical.

\[ \sqrt{ 2\sqrt[3]{2\sqrt[5]{2}}} = \sqrt[2]{ \sqrt[3]{ 2^3 \cdot 2 \sqrt[5]{2 } } } = \sqrt[2]{ \sqrt[3]{ \sqrt[5]{ 2^{20} \cdot 2} } } \]

\[ \sqrt[2 \cdot 3 \cdot 5]{2^{21}} = \sqrt[10]{2^7} = \sqrt[10]{128} \]

\(\textbf{Answer: D} \)

 

Question 20

 

If \[ a = 2^{\frac{12}{11} } \],

what is the result of the operation ?

\[ \sqrt{a \sqrt[3]{a^2 \sqrt{a } } } \]

\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } 4 \quad
\text{D) } 8 \quad
\text{E) } 16
\]

 

Solution:

 

$$ \sqrt{a \sqrt[3]{a^2 \sqrt{a } } } = \sqrt{\underbrace{}_{a \to a^3} \sqrt[3]{a^3 \cdot a^2 \sqrt{a } } }= \sqrt{\sqrt[3]{a^5 \cdot \sqrt{a}}} $$

$$ \sqrt{\sqrt[3]{a^5 \cdot \sqrt{a}}} = \sqrt{\sqrt[3]{\underbrace{}_{a^3 \to a^{5 \cdot 2}} \cdot \sqrt{ a^{10} \cdot a}}} = \sqrt[2. 3. 2 ]{ a^{11}} $$

$$ a^{\frac{11}{12} } = \left( 2^{\frac{12}{11} } \right)^{\frac{11}{12} } = 2$$

\(\textbf{Answer: B} \)

 

 

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