<s>[INST] $(X^2-1)^2+(X^2-1)-12=0$ has 2 solutions only or 3? I am asked to solve for X using substitution:
$(X^2-1)^2+(X^2-1)-12=0$
Let $U = X^2-1$
$U^2+U-12=0$
$U^2+4U-3U-12=0$
$U(U+4)-3(U+4)=0$
$(U+4)(U-3)$
Then:
$U-3=0$
$U=3$
$X^2-1=3$
$X^2=4$
$X=\pm2$
This is the provided solution by my textbook.
However:
$U+4=0$
$U=-4$
$X^2-1=-4$
$X^2=-3$
$X=i\sqrt{3}$
My textbook does not provide $X=i\sqrt{3}$ as a solution. Why is $X=i\sqrt{3}$ not a solution in addition to $\pm2$?
 [/INST] Note that if $X^2=-3$, this does not mean that $X=i\sqrt 3$. There is another solution $X=-i\sqrt 3$ that you have neglected. In either case, the reason your textbook might not have included $X=\pm 2,\pm i\sqrt 3$ as all four solutions is because it was only seeking for real solutions, of which there are only $\pm 2$.
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