Question

Can someone help me in this question?

=$\dfrac{\sqrt{\sqrt5 - 2} + \sqrt{\sqrt5 + 2}}{\sqrt{\sqrt5 + 1}}$

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Answer 1

$\small{\underline{\sf{Question:-}}}$

$\small{ \dfrac{\sqrt{\sqrt5 - 2} + \sqrt{\sqrt5 + 2}}{\sqrt{\sqrt5 + 1}}}$

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

Consider,

• The required answer be x

= $\dfrac{\sqrt{\sqrt5 - 2} + \sqrt{\sqrt5 + 2}}{\sqrt{\sqrt5 + 1}}$ = x

Now,

• Squaring on both sides

= ${\dfrac{\sqrt{\sqrt5 - 2} + \sqrt{\sqrt5 + 2}}{\sqrt{\sqrt5 + 1}}}^{2}$ = x²

Using identity,

• (a + b)² = a² + 2ab + b²

$= \small{\dfrac{\sqrt{5} - 2 + 2 \sqrt{(\sqrt{5} - 2) × (\sqrt{5} + 2)} + \sqrt{5} + 2}{\sqrt{5} + 1}} = \sf{ {x}^{2}}$

$= {\dfrac{\sqrt{5} \cancel{- 2} + 2 \sqrt{(\sqrt{5} - 2) × (\sqrt{5} + 2)} + \sqrt{5} \cancel{+ 2} } {\sqrt{5} + 1}} = \sf{ {x}^{2}}$

= $\dfrac{\sqrt{5} + 2 \sqrt{ 5 - 4} + \sqrt{5}}{\sqrt{5} + 1}$ = x²

= $\dfrac{2 \sqrt{5} + 2 \sqrt{ 1 }} {\sqrt{5} + 1}$ = x²

Now,

• Take 2 as common and factorize

= $\dfrac{2( \sqrt{5} + 1 ) } {\sqrt{5} + 1}$ = x²

= $\dfrac{2 \cancel{( \sqrt{5} + 1 ) }} {\cancel{\sqrt{5} + 1}}$ = x²

= 2 = x²

= √2 = x

Thus, the required answer is √2 respectively!!

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