[Math] Solving for all values of k if a quadratic equation containing it has two distinct real roots

Question

The equation $$x^2-2kx+1=0$$ has two distinct real roots. Find the set of all possible values of k.

What I did:

1. $$b-4ac=0$$
2. $$(-2k)^2-4=0$$
3. $$(-2k)^2=4$$
4. $$-2k=2$$
5. $$k=(-1)$$

From this, I'm unsure where to go– it says find the set. I understand that $$k=1$$ as well as -1 but how is that a set? Like I think the question wants me to give my answer in set notation but I'm unsure how to do that.

Answer 1

For distinct real roots, $$b^2-4ac\gt0$$ $$(-2k)^2-4\gt0$$ $$k^2-1\gt0$$ $$(k+1)(k-1)\gt0$$ $$k\in(-\infty,-1)\cup(1,\infty)$$