Question

If x= 9+4√5 find √x+1/√x??

Answer 1

$x = 9 + 4 \sqrt{5} \\ \\ \frac{1}{x} = \frac{1}{9 + 4 \sqrt{5} } \\ \\ \frac{1}{x} = \frac{1}{9 + 4 \sqrt{5} } \times \frac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} } \\ \\ \frac{1}{x} = \frac{(9 - 4 \sqrt{5} )}{81 - 80} \\ \\ = 9 - 4 \sqrt{5}$

Now,

x + 1/x = 9 + 4root5 + 9 - 4 root5

=> x + 1/x = 18

On adding 2 in both sides, we get

x + 1/x + 2 = 18 + 2

${( \sqrt{x} )}^{2} + { (\frac{1}{ \sqrt{x} }) }^{2} + 2( \sqrt{x} )( \frac{1}{ \sqrt{x} } ) = 20 \\ \\ {( \sqrt{x} + \frac{1}{ \sqrt{x} }) }^{2} = 20 \\ \\ \sqrt{x} + \frac{1}{ \sqrt{x} } = 2 \sqrt{5}$

Answer 2

$Given : x = 9 + 4\sqrt{5}$

$= > \sqrt{x} = \sqrt{9 + 4\sqrt{5}}$

$= > \sqrt{x} = \sqrt{(\sqrt{5})^2 + 2^2 + 2 * 2 * \sqrt{5}}$

$= > \sqrt{x} = \sqrt{(\sqrt{5} + 2)^2}$

$= > \sqrt{x} = \sqrt{5} + 2$

Now,

$= > \frac{1}{\sqrt{5} + 2} * \frac{\sqrt{5} - 2}{\sqrt{5} - 2}$

$= > \frac{\sqrt{5} - 2}{5 - 4}$

$= > \sqrt{5} - 2$

Therefore:

$= > \sqrt{x} + \frac{1}{\sqrt{x}} = (\sqrt{5} + 2) + (\sqrt{5} - 2)$

$= > 2\sqrt{5}$

Hope this helps!