Question

Q13. If N = $\it{\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}$, then N equals ___.​

Answer 1

(Minor edit in the calculation process.)

Question

If $N=\dfrac{\sqrt{\sqrt{5} +2} +\sqrt{\sqrt{5} -2}}{\sqrt{\sqrt{5} +1} }-\sqrt{3-2\sqrt{2} }$, then $N$ equals _.

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Required Knowledge

• Radical Notation

A radical consists of radicand, index, and radical symbol. It is the opposite process to the exponent. Let's consider an example, $\sqrt[3]{8}$.

→ Radicand 8

→ Index 3

This means the cube of the number is 8.

One more important thing about the index is that an even index considers only the positive root.

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Solution

Consider the two terms as

• $A=\dfrac{\sqrt{\sqrt{5} +2} +\sqrt{\sqrt{5} -2}}{\sqrt{\sqrt{5} +1} }$
• $B=\sqrt{3-2\sqrt{2} }$

Simplifying $A$

First, the value of $A^2$ is $2$. Then, $A=\sqrt{2}$ since $A>0$.

$\implies A^2=\dfrac{\sqrt{5} +2+2\sqrt{(\sqrt{5} +2)(\sqrt{5} -2)} +\sqrt{5} -2}{\sqrt{5} +1}$

$\implies A^2=\dfrac{2\sqrt{5} +2\sqrt{5-4} }{\sqrt{5} +1}$

$\implies A^2=\dfrac{2\sqrt{5} +2}{\sqrt{5} +1}=\boxed{2}$

$\therefore A=\sqrt{2}$

Simplifying $B$

Then, the value of $B^2$ is $3-2\sqrt{2}$. Then, $B=\sqrt{2} -1$ since $B>0$.

Considering $B$ as a sum of irrational and rational parts,

$B^2= 2-2\sqrt{2} +1=(\sqrt{2} -1)^2$

$\therefore B=\sqrt{2} -1$

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If $N=\dfrac{\sqrt{\sqrt{5} +2} +\sqrt{\sqrt{5} -2}}{\sqrt{\sqrt{5} +1} }-\sqrt{3-2\sqrt{2} }$, then $N$ equals $1$.

This is the required answer.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:N = \sf{\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}$

Let assume that

$\red{\bf :\longmapsto\:N = A - B} - - - - (1)$

where,

$\rm :\longmapsto\:A = \sf{\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}}$

and

$\rm :\longmapsto\:B \: = \: \sqrt{3 - 2 \sqrt{2} }$

Consider,

$\blue{\bf :\longmapsto\:B \: = \: \sqrt{3 - 2 \sqrt{2} } }$

$\rm \: = \: \: \sqrt{2 + 1 - 2 \sqrt{2} }$

$\rm \: = \: \: \sqrt{ {( \sqrt{2})}^{2} + {(1)}^{2} - 2 \times \sqrt{2} \times 1}$

$\rm \: = \: \: \sqrt{ {( \sqrt{2} - 1) }^{2} }$

$\rm \: = \: \: \sqrt{2} - 1$

$\blue{\bf\implies \: \sqrt{3 - 2 \sqrt{2}} = \sqrt{2} - 1 - - - - (2)}$

Now,

Consider,

$\green{\bf :\longmapsto\:A = \sf{\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}}}$

$\rm \: = \: \: \dfrac{ \sqrt{ \sqrt{5} + 2 } }{ \sqrt{ \sqrt{5} + 1} } + \dfrac{ \sqrt{ \sqrt{5} - 2 } }{ \sqrt{ \sqrt{5} + 1} }$

$\rm \: = \: \: C + D$

where,

$\rm :\longmapsto\:C = \dfrac{ \sqrt{ \sqrt{5} + 2 } }{ \sqrt{ \sqrt{5} + 1} }$

and

$\rm :\longmapsto\:D = \dfrac{ \sqrt{ \sqrt{5} - 2 } }{ \sqrt{ \sqrt{5} + 1} }$

$\rm :\implies\:A = C + D - - - (3)$

Consider,

$\rm :\longmapsto\:C = \dfrac{ \sqrt{ \sqrt{5} + 2 } }{ \sqrt{ \sqrt{5} + 1} }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:C = \dfrac{ \sqrt{ \sqrt{5} + 2 } }{ \sqrt{ \sqrt{5} + 1} } \times \dfrac{ \sqrt{ \sqrt{5} - 1 } }{ \sqrt{ \sqrt{5} - 1} }$

$\rm :\longmapsto\:C = \dfrac{ \sqrt{ (\sqrt{5} + 2)( \sqrt{5} - 1) } }{ \sqrt{ (\sqrt{5} + 1)( \sqrt{5} - 1)} }$

$\rm :\longmapsto\:C = \dfrac{ \sqrt{5 + 2 \sqrt{5} - \sqrt{5} - 2 } }{ \sqrt{5 - 1} }$

$\: \: \: \: \: \red{\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2}\bigg \}}$

$\rm :\longmapsto\:C = \dfrac{ \sqrt{3 + \sqrt{5} } }{ \sqrt{4} }$

$\rm :\longmapsto\:C = \sqrt{\dfrac{3 + \sqrt{5} }{4} }$

$\rm :\longmapsto\:C = \sqrt{\dfrac{6 + 2\sqrt{5} }{8} }$

$\rm :\longmapsto\:C = \sqrt{\dfrac{5 +1 + 2\sqrt{5} }{8} }$

$\rm :\longmapsto\:C = \sqrt{\dfrac{ {( \sqrt{5})}^{2} + {(1)}^{2} + 2\sqrt{5} }{8} }$

$\rm :\longmapsto\:C = \sqrt{\dfrac{ {( \sqrt{5} + 1)}^{2} }{8} }$

$\bf\implies \:C = \dfrac{ \sqrt{5} + 1}{2 \sqrt{2} } - - - (4)$

Consider,

$\rm :\longmapsto\:D = \dfrac{ \sqrt{ \sqrt{5} - 2 } }{ \sqrt{ \sqrt{5} + 1} }$

On rationalizing the denominator, we get

$\rm :\longmapsto\:D = \dfrac{ \sqrt{ \sqrt{5} - 2 } }{ \sqrt{ \sqrt{5} + 1} } \times \dfrac{ \sqrt{ \sqrt{5} - 1 } }{ \sqrt{ \sqrt{5} - 1} }$

$\rm :\longmapsto\:D = \dfrac{ \sqrt{ (\sqrt{5} - 2)( \sqrt{5} - 1) } }{ \sqrt{ (\sqrt{5} + 1)( \sqrt{5} - 1)} }$

$\rm :\longmapsto\:D = \dfrac{ \sqrt{5 - 2 \sqrt{5} - \sqrt{5} + 2 } }{ \sqrt{5 - 1} }$

$\rm :\longmapsto\:D = \dfrac{ \sqrt{7 - 3\sqrt{5} } }{ \sqrt{4} }$

$\rm :\longmapsto\:D = \sqrt{\dfrac{7 - 3\sqrt{5} }{4} }$

$\rm :\longmapsto\:D = \sqrt{\dfrac{14 - 6\sqrt{5} }{8} }$

$\rm :\longmapsto\:D = \sqrt{\dfrac{9 + 5 - 6\sqrt{5} }{8} }$

$\rm :\longmapsto\:D = \sqrt{\dfrac{ {(3)}^{2} + {( \sqrt{5}) }^{2} - 2 \times 3 \times \sqrt{5} }{8} }$

$\rm :\longmapsto\:D = \sqrt{\dfrac{ {(3 - \sqrt{5} )}^{2} }{8} }$

$\bf\implies \:D = \dfrac{ 3 - \sqrt{5}}{2 \sqrt{2} } - - - (5)$

On substituting the values of C and D in equation (3), we get

$\rm :\longmapsto\:A = \dfrac{ \sqrt{5} + 1 }{2 \sqrt{2} } + \dfrac{3 - \sqrt{5} }{2 \sqrt{2} }$

$\rm :\longmapsto\:A = \dfrac{ \sqrt{5} + 1 + 3 - \sqrt{5} }{2 \sqrt{2} }$

$\rm :\longmapsto\:A = \dfrac{ 4 }{2 \sqrt{2} }$

$\rm :\longmapsto\:A = \dfrac{ 2 \times \sqrt{2} \times \sqrt{2} }{2 \sqrt{2} }$

$\green{\bf\implies \:A = \sqrt{2} - - - (6)}$

On substituting the values of A and B from equation (2) and equation (6), we get

$\red{\rm :\longmapsto\:N = \sqrt{2} - ( \sqrt{2} - 1)}$

$\red{\rm :\longmapsto\:N = \sqrt{2} - \sqrt{2} + 1}$

$\red{\bf\implies \:N = 1}$