Answer :
Answer:
This is a Difference of Squares so [tex](x+1)(x-1) = x^{2} -1[/tex]
Therefore, [tex]x^{2} -1 > 2[/tex] or x>±[tex]\sqrt{3}[/tex].
Answer:
This is a Difference of Squares so [tex](x+1)(x-1) = x^{2} -1[/tex]
Therefore, [tex]x^{2} -1 > 2[/tex] or x>±[tex]\sqrt{3}[/tex].