KS15 wrote:

Can you show this using the formula a^2-b^2 i.e (a-b)(a+b) and then solving? I know it is easy to solve using numbers but not able to get using this formula

Thanks

KS15(x – 2)^2 > 2

=> (x – 2)^2 - 2 >0

=>\((x – 2)^2 - \sqrt{2} >0\)

=> \((x-2+\sqrt{2}) (x-2-\sqrt{2}) >0\)

This will term either both term are positive or both terms are negative

both positive :

\(x-2+\sqrt{2} > 0\) and \(x-2-\sqrt{2} > 0\)

\(x>2-\sqrt{2}\) and \(x >2+\sqrt{2}\)

OR

both negative :

\(x-2+\sqrt{2} < 0\) and \(x-2-\sqrt{2} < 0\)

\(x<2-\sqrt{2}\)and \(x <2+\sqrt{2}\)

Combining all 4 equations we can write :

\(x >2+\sqrt{2}\) OR \(x<2-\sqrt{2}\)