Question

if a equal to root 5 + root 10 by root 10 minus root 5 and b root 10 minus root 5 by root 10 + root 5 show that root a minus root b minus 2 root a b equal to zero​

Answer 1

Step-by-step explanation:

$a = \frac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} - \sqrt{5} } \\ \\ b = \frac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} } \\ \\ a \times b = ab = 1 \\ \\ now \\ \\ a = \frac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} - \sqrt{5} } \\ \\ a = ( \frac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} - \sqrt{5} } )( \frac{ \sqrt{10} + \sqrt{5} }{ \sqrt{10} + \sqrt{5} } ) \\ \\ a = \frac{ {( \sqrt{10} + \sqrt{5} )}^{2} }{10 - 5} \\ \\ a = \frac{ {( \sqrt{10} + \sqrt{5} )}^{2} }{ {( \sqrt{5} )}^{2} } \\ \\ a = {( \frac{ \sqrt{10} + \sqrt{5} }{ \sqrt{5} }) }^{2} \\ \\ a = {( \sqrt{2} + 1)}^{2} \\ \\ \sqrt{a} = \sqrt{2} + 1 \\ \\ also \\ \\ b = \frac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} } \\ \\ b = ( \frac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} + \sqrt{5} } )( \frac{ \sqrt{10} - \sqrt{5} }{ \sqrt{10} - \sqrt{5} } ) \\ \\ b = \frac{ {( \sqrt{10} - \sqrt{5} ) }^{2} }{10 - 5} \\ \\ b = \frac{ {( \sqrt{10} - \sqrt{5}) }^{2} }{5} \\ \\ b = \frac{ {( \sqrt{10} - \sqrt{5} ) }^{2} }{ {( \sqrt{5} )}^{2} } \\ \\ b = {( \frac{ \sqrt{10} - \sqrt{5} }{ \sqrt{5} }) }^{2} \\ \\ b = {( \sqrt{2} - 1)}^{2} \\ \\ \sqrt{b} = \sqrt{2} - 1 \\ \\ also \\ \\ ab = 1 \\ \\ \sqrt{ab} = 1 \\ \\ now \\ \\ \sqrt{a} - \sqrt{b} - 2 \sqrt{ab} \\ \\ = (\sqrt{2} + 1) - ( \sqrt{2} - 1) - 2(1) \\ \\ = \sqrt{2} + 1 - \sqrt{2} + 1 - 2 \\ \\ = \sqrt{2} - \sqrt{2} + 1 + 1 - 2 \\ \\ = 0 \\ \\ = > \sqrt{a} - \sqrt{b} - 2 \sqrt{ab} = 0$

hope it helps