if x =2 -under root 5 …
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Payal Singh 6 years, 10 months ago
Given : x = {tex}2-\sqrt 5{/tex}
Now
{tex}{1\over x} = {1\over 2-\sqrt 5}{/tex}
Rationalize the denominator,
We get
{tex}{1\over x} = {1\over 2-\sqrt 5}\times {2+\sqrt 5\over 2+\sqrt 5}{/tex}
{tex}=> {1\over x} = {2+\sqrt 5\over (2)^2-(\sqrt 5)^2} {/tex}
{tex}=> {1\over x} = {2+\sqrt 5\over 4-5}= -2-\sqrt 5{/tex}
Now,
{tex}x^2+{1\over x^2}= (2-\sqrt 5)^2+(-2-\sqrt5)^2{/tex}
= {tex}4+5-4\sqrt5+4+5+4\sqrt5{/tex}= 18
1Thank You