Question

$\sf\red{QuEsTiOn}$

$\displaystyle \rm \frac{\sqrt{\sqrt{5} +2}+\sqrt{\sqrt{5} -2} }{\sqrt{\sqrt{5} +1} } -\sqrt{3-2\sqrt{2} }$
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Answer 1

Question:-

$\sf \large \frac{\sqrt{\sqrt{5} +2}+\sqrt{\sqrt{5} -2} }{\sqrt{\sqrt{5} +1} } -\sqrt{3-2\sqrt{2} }$

Given:-

$\sf \large \frac{\sqrt{\sqrt{5} +2}+\sqrt{\sqrt{5} -2} }{\sqrt{\sqrt{5} +1} } -\sqrt{3-2\sqrt{2} }$

To Find:-

• The value of the Given function.

Solution:-

$\sf \large Let, \frac{\sqrt{\sqrt{5} +2}+\sqrt{\sqrt{5} -2} }{\sqrt{\sqrt{5} +1} } -\sqrt{3-2\sqrt{2} } \: \: = x$

On squaring both sides, we get

$\bigg( \frac{ \sqrt{ \sqrt{5} + 2 } + \sqrt{ \sqrt{5} - 2} }{ \sqrt{ \sqrt{5} + 1} } \bigg) \sf \large ^{2} = x {}^{2}$

$\sf \large⇒x {}^{2} = \frac{ \sqrt{5} + 2 + \sqrt{5} -2 + 2 \sqrt{( \sqrt{5} + 2)( \sqrt{5} - 2)} }{ \sqrt{5} + 1}$

$\sf \large⇒x {}^{2} = \frac{2 \sqrt{5} + 2 \sqrt{5 - 4} }{ \sqrt{5} + 1 }$

$\sf \large⇒x {}^{2} = \frac{2( \sqrt{5} + 1 )}{ \sqrt{5} + 1}$

$\sf \large⇒x {}^{2} = 2$

$\sf \large⇒x {}^{2} = \sqrt{2}$

$\sf \large Also, \sqrt{3 - 2 \sqrt{2} } \: can \: be \: written \\ \sf \large as \: \sqrt{1 {}^{2} + ( \sqrt{2}) {}^{2} - 2 \times 1 \times \sqrt{2} } \: i.e., \: \sqrt{2} - 1$

Answer:-

$\sf \large \color{blue} \therefore \frac{\sqrt{\sqrt{5} +2}+\sqrt{\sqrt{5} -2} }{\sqrt{\sqrt{5} +1} } -\sqrt{3-2\sqrt{2} } = 1.$

Hope you have satisfied. ⚘