I have the following equation, the problem is that there comes a time when I only have the x ^ 4 left and I don't know what to do next.
x^4-1 = 0
x^4 = 1
...
Does anyone know how the equation is still solved?
I had never done anything similar with exponents.
On
Square root both sides: $\sqrt{x^4}=x^2=\pm 1\implies x^2=-1~\mbox{or}~x^2=1$.
$x^2=-1\implies \sqrt{x^2}=x=\pm\sqrt{-1}=\pm i$
and
$x^2=1\implies \sqrt{x^2}=x=\pm 1$
There are two real solutions ($1$ and $-1$) and two complex solutions ($i$ and $-i$).
Also could have factored using difference of squares: $x^4-1=0\implies(x^2+1)(x^2-1)=0$ then solve each factor separately.
$x^2+1=0\implies x^2=-1$ and $x^2-1=0\implies x^2=1$, which are solved above.
So you have to solve $x^4 = 1$. Substitute $u = x^2$ to get $u^2 = 1$. So $u = \pm 1$.
Thus, you now have to solve $x^2 = 1$ and $x^2 = -1$. Can you take it from here?