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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?

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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 $$x^4+4x^2-4-1=0\Rightarrow (x^2)^2+4(x^2)-5=0 \Rightarrow x^2=-5,1\Rightarrow x=\pm i\sqrt{5},x=\pm 1$$

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Because you are losing roots when doing cancelation of $(x^2-1)$.

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You divided by $x^2-1$ in the third step. That expression is not necessarily non-zero..

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