# Why does solving this equation a certain way yield only complex roots instead of real ones?

For the system of equations $$4x^2 + y = 4,\quad x^4 - y = 1$$ if I attempt to solve by solving each equation for $y$ and setting them equal to each other, I obtain $$4 - 4x^2 = x^4 - 1$$ $$-4(x^2 - 1) = (x^2 - 1)(x^2 + 1)$$ $$-4 = x^2 + 1$$ $$x = ±\sqrt5 i$$ However, it can be shown by graphing the equations, and by following the method of elimination, that the system has the real solutions $(1,0)$ and $(-1,0)$.

Why is it that one method yields only complex roots while another yields the real roots?

You cannot divide the sides by $x^2-1$ since there are roots you are cancelling there. The correct way to solve the equation is $$x^4+4x^2-4-1=0\Rightarrow (x^2)^2+4(x^2)-5=0 \Rightarrow x^2=-5,1\Rightarrow x=\pm i\sqrt{5},x=\pm 1$$
Because you are losing roots when doing cancelation of $(x^2-1)$.
You divided by $x^2-1$ in the third step. That expression is not necessarily non-zero..