Question

If x=9+4root5 ,find the value of rootx-1 upon root

Answer 1

Hi...☺

Here is your answer...✌

GIVEN THAT,

$x = 9 + 4 \sqrt{5} \\ \\ x = 4 + 5 + 4 \sqrt{5} \\ \\ x = {2}^{2} + {( \sqrt{5})}^{2} + 2 \times 2 \times \sqrt{5} \\ \\ x ={ (2 + \sqrt{5} )}^{2}$

Square rooting both sides
We get,

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

And,

$\frac{1}{ \sqrt{x} } = \frac{1}{2 + \sqrt{5} } \times \frac{2 - \sqrt{5} }{2 - \sqrt{5} } \\ \\ \frac{1}{ \sqrt{x} } = \frac{2 - \sqrt{5} }{ {2}^{2} - {(\sqrt{5})}^{2} } \\ \\ \frac{1}{ \sqrt{x} } = \frac{2 - \sqrt{5} }{4 - 5} = \frac{2 - \sqrt{5} }{ - 1} \\ \\ \frac{1}{ \sqrt{x} } = \sqrt{5} - 2$

Now,

$\sqrt{x} - \frac{1}{ \sqrt{x} } = 2 + \sqrt{5} - (\sqrt{5} - 2) \\ \\ = 2 + \sqrt{5} - \sqrt{5} + 2 \\ \\ = 4$

Answer 2

Answer:

answer in image pls refer this