Find all values of $z$ such that $z^4 - 4z^2 + 3 = 0$.  Enter all the solutions, separated by commas.
If we let $y=z^2$, then our equation becomes a simple quadratic equation:
$$y^2-4y+3=0.$$Indeed, this equation factors easily as $(y-3)(y-1)=0$, so either $y-3=0$ or $y-1=0$.

We now explore both possibilities.

If $y-3=0$, then $y=3$, so $z^2=3$, so $z=\pm\sqrt 3$.

If $y-1=0$, then $y=1$, so $z^2=1$, so $z=\pm 1$.

Thus we have four solutions to the original equation: $z=\boxed{-\sqrt{3},-1,1,\sqrt{3}}$.
-\sqrt{3},-1,1,\sqrt{3}
Level 2
Intermediate Algebra
Problem: Find all values of $z$ such that $z^4 - 4z^2 + 3 = 0$.  Enter all the solutions, separated by commas.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation z^4 - 4z^2 + 3 = 0, we can make a substitution to simplify the expression. Let w = z^2. Then the equation transforms into w^2 - 4w + 3 = 0. This is a quadratic equation in w.

Next, we apply the quadratic formula, w = (4 ± √(16 - 12)) / 2. This simplifies to w = (4 ± 2) / 2, yielding two solutions: w = 3 and w = 1.

Now, we revert back to z by substituting w back with z^2. For w = 3, we have z^2 = 3, which gives us z = ±√3. For w = 1, we have z^2 = 1, which gives us z = ±1.

Thus, the complete set of solutions for z includes both the positive and negative roots derived from the values of w. Therefore, the final answer is -√3, -1, 1, √3.<|END_OF_LATENT|> The final answer is $-\sqrt{3},-1,1,\sqrt{3}$.