Question

x^2>64

Answer 1

x has to be greater than 8

Answer 2

x>8

Answer 3

But @mirrorimagemirage what about \(-9\) ? Surely \(-9\)^2 = 81> 64.

We use this property for \(a>0\) ( it is trivially true for \(a<0\) ).

Far all real \(x\) we have \(x^2>a\) implies \(x>\sqrt{a}\) \(\underline{\text{AND}}\) \(x<-\sqrt{a}\)
So for this problem, we have

\(x^2>64\implies x>\sqrt{64} = 8 \ \text{and } x<-\sqrt{64}=-8\), so we have \[\{x \ | \ x> 8 \text{ and } x<-8\}\] or we could say \(x\in (-\infty,-8)\cup(8,\infty)\).

I hope this makes sense.