### Instruction: Find all values of $z$ such that $z^4 - 4z^2 + 3 = 0$.  Enter all the solutions, separated by commas.### Response: I notice that this equation is a quartic, but it only has even powers of $z$.
That makes me think of substituting $w = z^2$, so I can reduce it to a quadratic equation in $w$.
If I do that, I get $w^2 - 4w + 3 = 0$.
This looks familiar, and I can factor it as $(w - 3)(w - 1) = 0$.
Now, to find the values of $w$, I just set each factor equal to zero and solve.
I get $w = 3$ or $w = 1$.
But I want to find the values of $z$, not $w$.
So I need to undo the substitution I made.
I can do that by taking the square root of both sides of $w = 3$ and $w = 1$, remembering to include both the positive and negative roots.
For $w = 3$, I get $z = \pm \sqrt{3}$.
For $w = 1$, I get $z = \pm 1$.
So the four values of $z$ that satisfy the equation are $\sqrt{3}$, $-\sqrt{3}$, $1$, and $-1$.